From 9b7a881014f9b47ee96298b21f55b64fea8e49d5 Mon Sep 17 00:00:00 2001 From: lcerrato Date: Thu, 12 Dec 2024 14:28:53 -0500 Subject: [PATCH] (grc_conversion) EpiDoc and CTS conversion tlg1799 translation work #1399 --- data/tlg1799/tlg001/__cts__.xml | 10 +- .../tlg001/tlg1799.tlg001.perseus-eng1.xml | 3108 ++--- .../tlg001/tlg1799.tlg001.perseus-eng2.xml | 10456 ++++++++++++++++ .../tlg001/tlg1799.tlg001.perseus-grc2.xml | 192 +- manifest.txt | 1 + 5 files changed, 11306 insertions(+), 2461 deletions(-) create mode 100644 data/tlg1799/tlg001/tlg1799.tlg001.perseus-eng2.xml diff --git a/data/tlg1799/tlg001/__cts__.xml b/data/tlg1799/tlg001/__cts__.xml index c8343259b..f17684519 100644 --- a/data/tlg1799/tlg001/__cts__.xml +++ b/data/tlg1799/tlg001/__cts__.xml @@ -1,8 +1,14 @@ Elementa + - Elementa - Euclid. Euclidis Opera omnia, Volume 1-5. Heiberg, Johan Ludvig, editor. Leipzig: Teubner, 1883-88. + Στοιχεῖα + Euclid. Euclidis Opera omnia, Vols. 1-5. Heiberg, Johan Ludvig, editor. Leipzig: Teubner, 1883-88. + + + The Thirteen Books of Euclid's Elements + Euclid. The Thirteen Books of Euclid's Elements, Vols. 1-3. Heath, Thomas Little, editor. Cambridge, UK: Cambridge University Press, 1908. + \ No newline at end of file diff --git a/data/tlg1799/tlg001/tlg1799.tlg001.perseus-eng1.xml b/data/tlg1799/tlg001/tlg1799.tlg001.perseus-eng1.xml index 7ded2265a..973071a7f 100644 --- a/data/tlg1799/tlg001/tlg1799.tlg001.perseus-eng1.xml +++ b/data/tlg1799/tlg001/tlg1799.tlg001.perseus-eng1.xml @@ -1,1681 +1,107 @@ - - - -%PersProse; -]> - - + + + + - Elements + The Thirteen Books of Euclid's Elements Euclid - Thomas L. Heath - + Thomas L. Heath Perseus Project, Tufts University Gregory Crane Prepared under the supervision of Lisa Cerrato - William Merrill - Elli Mylonas David Smith - The National Science Foundation + The National Science Foundation Trustees of Tufts University Medford, MA - Perseus Project + Perseus Digital Library Project + Perseus 3.0 + tlg1799.tlg001.perseus-eng2.xml + + Available under a Creative Commons Attribution-ShareAlike 4.0 International License + + + - - 40542855 - 0486600882 - 0486600890 - 0486600904 - Euclid - Euclid's Elements - Sir Thomas Little Heath + The Thirteen Books of Euclid's Elements + Thomas Little Heath - New York - Dover - 1956 + Cambridge, UK + Cambridge University Press + 1908 + 1-3 - +HathiTrust +HathiTrust + HathiTrust + - + + - + + +

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+ + + + + - + --> + - English - Greek - Latin + English + Greek + Latin + + Epidoc and CTS conversion and header review for best practices + + - - - - - CHAPTER I. - EUCLID AND THE TRADITIONS ABOUT HIM. -

As in the case of the other great mathematicians of Greece, so in Euclid's case, we have only the most meagre particulars of the life and personality of the man.

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Most of what we have is contained in the passage of Proclus' summary relating to him, which is as followsProclus, ed. Friedlein, p. 68, 6-20.:

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Not much younger than these (sc. Hermotimus of Colophon and Philippus of Medma) is Euclid, who put together the Elements, collecting many of Eudoxus' theorems, perfecting many of Theaetetus', and also bringing to irrefragable demonstration the things which were only somewhat loosely proved by his predecessors. This man livedThe word ge/gone must apparently mean flourished, as Heiberg understands it (Litterargeschichtliche Studien über Euklid, 1882, p. 26), not was born, as Hankel took it : otherwise part of Proclus' argument would lose its cogency. in the time of the first Ptolemy. For Archimedes, who came immediately after the first (Ptolemy)So Heiberg understands e)pibalw\n tw=| prw/tw| (sc. *ptolemai/w|). Friedlein's text has kai\ between e)pibalw\n and tw=| prw/tw|; and it is right to remark that another reading is kai\ e)n tw=| prw/tw| (without e)pibalw/n) which has been translated in his first book, by which is understood On the Sphere and Cylinder I., where (1) in Prop. 2 are the words let BC be made equal to D by the second (proposition) of the first of Euclid's (books), - and (2) in Prop. 6 the words For these things are handed down in the Elements - (without the name of Euclid). Heiberg thinks the former passage is referred to, and that Proclus must therefore have had before him the words by the second of the first of Euclid: - a fair proof that they are genuine, though in themselves they would be somewhat suspicious., makes mention of Euclid: and, further, they say that Ptolemy once asked him if there was in geometry any shorter way than that of the elements, and he answered that there was no royal road to geometryThe same story is told in Stobaeus, Ecl. (II. p. 228, 30, ed. Wachsmuth) about Alexander and Menaechmus. Alexander is represented as having asked Menaechmus to teach him geometry concisely, but he replied : O king, through the country there are royal roads and roads for common citizens, but in geometry there is one road for all.. He is then younger than the pupils of Plato but older than Eratosthenes and Archimedes; for the latter were contemporary with one another, as Eratosthenes somewhere says.

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This passage shows that even Proclus had no direct knowledge of Euclid's birthplace or of the date of his birth or death. He proceeds by inference. Since Archimedes lived just after the first Ptolemy, and Archimedes mentions Euclid, while there is an anecdote about some Ptolemy and Euclid, therefore Euclid lived in the time of the first Ptolemy.

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We may infer then from Proclus that Euclid was intermediate between the first pupils of Plato and Archimedes. Now Plato died in 347/6, Archimedes lived 287-212, Eratosthenes c. 284-204 B.C. Thus Euclid must have flourished c. 300 B.C., which date agrees well with the fact that Ptolemy reigned from 306 to 283 B.C.

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It is most probable that Euclid received his mathematical training in Athens from the pupils of Plato; for most of the geometers who could have taught him were of that school, and it was in Athens that the older writers of elements, and the other mathematicians on whose works Euclid's Elements depend, had lived and taught. He may himself have been a Platonist, but this does not follow from the statements of Proclus on the subject. Proclus says namely that he was of the school of Plato and in close touch with that philosophyProclus, p. 68, 20, kai\ th=| proaire/sei de\ *platwniko/s e)sti kai\ th=| filosofi/a| tau/th| oi\kei=os.. But this was only an attempt of a New Platonist to connect Euclid with his philosophy, as is clear from the next words in the same sentence, for which reason also he set before himself, as the end of the whole Elements, the construction of the so-called Platonic figures. - It is evident that it was only an idea of Proclus' own to infer that Euclid was a Platonist because his Elements end with the investigation of the five regular solids, since a later passage shows him hard put to it to reconcile the view that the construction of the five regular solids was the end and aim of the Elements with the obvious fact that they were intended to supply a foundation for the study of geometry in general, to make perfect the understanding of the learner in regard to the whole of geometryibid. p. 71, 8.. - To get out of the difficulty he saysibid. p. 70, 19 sqq. that, if one should ask him what was the aim (skopo/s) of the treatise, he would reply by making a distinction between Euclid's intentions (1) as regards the subjects with which his investigations are concerned, (2) as regards the learner, and would say as regards (1) that the whole of the geometer's argument is concerned with the cosmic figures. This latter statement is obviously incorrect. It is true that Euclid's Elements end with the construction of the five regular solids; but the planimetrical portion has no direct relation to them, and the arithmetical no relation at all; the propositions about them are merely the conclusion of the stereometrical division of the work.

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One thing is however certain, namely that Euclid taught, and founded a school, at Alexandria. This is clear from the remark of Pappus about ApolloniusPappus, VII. p. 678, 10-12, susxola/sas toi=s u(po\ *eu)klei/dou maqhtai=s e)n *)alecandrei/a| plei=ston xro/non, o(/qen e)/sxe kai\ th\n toiau/thn e)/cin ou)k a)maqh=. : he spent a very long time with the pupils of Euclid at Alexandria, and it was thus that he acquired such a scientific habit of thought.

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It is in the same passage that Pappus makes a remark which might, to an unwary reader, seem to throw some light on the personality of Euclid. He is speaking about Apollonius' preface to the first book of his Conics, where he says that Euclid had not completely worked out the synthesis of the three- and four-line locus, - which in fact was not possible without some theorems first discovered by himself. Pappus says on thisPappus, VII. pp. 676, 25-678, 6. Hultsch, it is true, brackets the whole passage pp. 676, 25-678, 15, but apparently on the ground of the diction only.: Now Euclid— regarding Aristaeus as deserving credit for the discoveries he had already made in conics, and without anticipating him or wishing to construct anew the same system (such was his scrupulous fairness and his exemplary kindliness towards all who could advance mathematical science to however small an extent), being moreover in no wise contentious and, though exact, yet no braggart like the other [Apollonius] —wrote so much about the locus as was possible by means of the conics of Aristaeus, without claiming completeness for his demonstrations. It is however evident, when the passage is examined in its context, that Pappus is not following any tradition in giving this account of Euclid: he was offended by the terms of Apollonius' reference to Euclid, which seemed to him unjust, and he drew a fancy picture of Euclid in order to show Apollonius in a relatively unfavourable light.

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Another story is told of Euclid which one would like to believe true. According to StobaeusStobaeus, l.c., some one who had begun to read geometry with Euclid, when he had learnt the first theorem, asked Euclid, ’But what shall I get by-learning these things?’ Euclid called his slave and said ’Give him threepence, since he must make gain out of what he learns.’

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In the middle ages most translators and editors spoke of Euclid as Euclid of Megara. This description arose out of a confusion between our Euclid and the philosopher Euclid of Megara who lived about 400 B.C. The first trace of this confusion appears in Valerius Maximus (in the time of Tiberius) who saysVIII. 12, ext. 1. that Plato, on being appealed to for a solution of the problem of doubling the cubical altar, sent the inquirers to Euclid the geometer. - There is no doubt about the reading, although an early commentator on Valerius Maximus wanted to correct Eucliden - into Eudoxum, - and this correction is clearly right. But, if Valerius Maximus took Euclid the geometer for a contemporary of Plato, it could only be through confusing him with Euclid of Megara. The first specific reference to Euclid as Euclid of Megara belongs to the 14th century, occurring in the u(pomnhmatismoi/ of Theodorus Metochita (d. 1332) who speaks of Euclid of Megara, the Socratic philosopher, contemporary of Plato, - as the author of treatises on plane and solid geometry, data, optics etc. : and a Paris MS. of the 14th century has Euclidis philosophi Socratici liber elementorum. - The misunderstanding was general in the period from Campanus' translation (Venice 1482) to those of Tartaglia (Venice 1565) and Candalla (Paris 1566). But one Constantinus Lascaris (d. about 1493) had already made the proper distinction by saying of our Euclid that he was different from him of Megara of whom Laertius wrote, and who wrote dialogues -Letter to Fernandus Acuna, printed in Maurolycus, Historia Siciliae, fol. 21 r. (see Heiberg, Euklid-Studien, pp. 22-3, 25).; and to Commandinus belongs the credit of being the first translatorPreface to translation (Pisauri, 1572). to put the matter beyond doubt: Let us then free a number of people from the error by which they have been induced to believe that our Euclid is the same as the philosopher of Megara - etc.

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Another idea, that Euclid was born at Gela in Sicily, is due to the same confusion, being based on Diogenes Laertius' descriptionDiog. L. II. 106, p. 58 ed. Cobet. of the philosopher Euclid as being of Megara, or, according to some, of Gela, as Alexander says in the *diadoxai/. -

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In view of the poverty of Greek tradition on the subject even as early as the time of Proclus (410-485 A.D.), we must necessarily take cum grano the apparently circumstantial accounts of Euclid given by Arabian authors; and indeed the origin of their stories can be explained as the result (1) of the Arabian tendency to romance, and (2) of misunderstandings.

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We readCasiri, Bibliotheca Arabico-Hispana Escurialensis, I. p. 339. Casiri's source is alQifti (d. 1248), the author of the Ta'rīkh al-Ḥukamā, a collection of biographies of philosophers, mathematicians, astronomers etc. that Euclid, son of Naucrates, grandson of ZenarchusThe Fihrist says son of Naucrates, the son of Berenice (?) (see Suter's translation in Abhandlungen zur Gesch. d. Math. VI. Heft, 1892, p. 16)., called the author of geometry, a philosopher of somewhat ancient date, a Greek by nationality domiciled at Damascus, born at Tyre, most learned in the science of geometry, published a most excellent and most useful work entitled the foundation or elements of geometry, a subject in which no more general treatise existed before among the Greeks: nay, there was no one even of later date who did not walk in his footsteps and frankly profess his doctrine. Hence also Greek, Roman and Arabian geometers not a few, who undertook the task of illustrating this work, published commentaries, scholia, and notes upon it, and made an abridgment of the work itself. For this reason the Greek philosophers used to post up on the doors of their schools the well-known notice: ’Let no one come to our school, who has not first learned the elements of Euclid.’ - The details at the beginning of this extract cannot be derived from Greek sources, for even Proclus did not know anything about Euclid's father, while it was not the Greek habit to record the names of grandfathers, as the Arabians commonly did. Damascus and Tyre were no doubt brought in to gratify a desire which the Arabians always showed to connect famous Greeks in some way or other with the East. Thus Naṣīraddīn, the translator of the Elements, who was of Ṭūs in Khurāsān, actually makes Euclid out to have been Thusinus - alsoThe same predilection made the Arabs describe Pythagoras as a pupil of the wise Salomo, Hipparchus as the exponent of Chaldaean philosophy or as the Chaldaean, Archimedes as an Egyptian etc. (Ḥăjī Khalfa, Lexicon Bibliographicum, and Casiri).. The readiness of the Arabians to run away with an idea is illustrated by the last words of the extract. Everyone knows the story of Plato's inscription over the porch of the Academy: let no one unversed in geometry enter my doors -; the Arab turned geometry into Euclid's geometry, and told the story of Greek philosophers in general and their Academies. -

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Equally remarkable are the Arabian accounts of the relation of Euclid and ApolloniusThe authorities for these statements quoted by Casiri and Ḥājī Khalfa are al-Kindi's tract de instituto libri Euclidis (al-Kindī died about 873) and a commentary by Qāḍīzāde ar-Rūmī (d. about 1440) on a book called Ashkāl at-ta' sīs (fundamental propositions) by Ashraf Shamsaddīn as-Samarqandī (c. 1276) consisting of elucidations of 35 propositions selected from the first books of Euclid. Naṣīraddīn likewise says that Euclid cut out two of 15 books of elements then existing and published the rest under his own name. According to Qāḍīzāde the king heard that there was a celebrated geometer named Euclid at Tyre: Naṣīraddīn says that he sent for Euclid of Ṭūs.. According to them the Elements were originally written, not by Euclid, but by a man whose name was Apollonius, a carpenter, who wrote the work in 15 books or sectionsSo says the Fihrist. Suter (op. cit. p. 49) thinks that the author of the Fihrist did not suppose Apollonius of Perga to be the writer of the Elements, as later Arabian authorities did, but that he distinguished another Apollonius whom he calls a carpenter. - Suter's argument is based on the fact that the Fihrist's article on Apollonius (of Perga) says nothing of the Elements; and that it gives the three great mathematicians, Euclid, Archimedes and Apollonius, in the correct chronological order.. In the course of time some of the work was lost and the rest became disarranged, so that one of the kings at Alexandria who desired to study geometry and to master this treatise in particular first questioned about it certain learned men who visited him and then sent for Euclid who was at that time famous as a geometer, and asked him to revise and complete the work and reduce it to order. Euclid then re-wrote it in 13 books which were thereafter known by his name. (According to another version Euclid composed the 13 books out of commentaries which he had published on two books of Apollonius on conics and out of introductory matter added to the doctrine of the five regular solids.) To the thirteen books were added two more books, the work of others (though some attribute these also to Euclid) which contain several things not mentioned by Apollonius. According to another version Hypsicles, a pupil of Euclid at Alexandria, offered to the king and published Books XIV. and XV., it being also stated that Hypsicles had discovered - the books, by which it appears to be suggested that Hypsicles had edited them from materials left by Euclid.

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We observe here the correct statement that Books XIV. and XV. were not written by Euclid, but along with it the incorrect information that Hypsicles, the author of Book XIV., wrote Book XV. also.

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The whole of the fable about Apollonius having preceded Euclid and having written the Elements appears to have been evolved out of the preface to Book XIV. by Hypsicles, and in this way; the Book must in early times have been attributed to Euclid, and the inference based upon this assumption was left uncorrected afterwards when it was recognised that Hypsicles was the author. The preface is worth quoting:

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Basilides of Tyre, O Protarchus, when he came to Alexandria and met my father, spent the greater part of his sojourn with him on account of their common interest in mathematics. And once, when examining the treatise written by Apollonius about the comparison between the dodecahedron and the icosahedron inscribed in the same sphere, (showing) what ratio they have to one another, they thought that Apollonius had not expounded this matter properly, and accordingly they emended the exposition, as I was able to learn from my father. And I myself, later, fell in with another book published by Apollonius, containing a demonstration relating to the subject, and I was greatly interested in the investigation of the problem. The book published by Apollonius is accessible to all— for it has a large circulation, having apparently been carefully written out later—but I decided to send you the comments which seem to me to be necessary, for you will through your proficiency in mathematics in general and in geometry in particular form an expert judgment on what I am about to say, and you will lend a kindly ear to my disquisition for the sake of your friendship to my father and your goodwill to me. -

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The idea that Apollonius preceded Euclid must evidently have been derived from the passage just quoted. It explains other things besides. Basilides must have been confused with basileu/s, and we have a probable explanation of the Alexandrian king, - and of the learned men who visited - Alexandria. It is possible also that in the Tyrian - of Hypsicles' preface we have the origin of the notion that Euclid was born in Tyre. These inferences argue, no doubt, very defective knowledge of Greek: but we could expect no better from those who took the Organon of Aristotle to be instrumentum musicum pneumaticum, - and who explained the name of Euclid, which they variously pronounced as Uclides or Icludes, to be compounded of Ucli a key, and Dis a measure, or, as some say, geometry, so that Uclides is equivalent to the key of geometry!

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Lastly the alternative version, given in brackets above, which says that Euclid made the Elements out of commentaries which he wrote on two books of Apollonius on conics and prolegomena added to the doctrine of the five solids, seems to have arisen, through a like confusion, out of a later passageHeiberg's Euclid, vol. V. p. 6. in Hypsicles' Book XIV.: And this is expounded by Aristaeus in the book entitled ’Comparison of the five figures,’ and by Apollonius in the second edition of his comparison of the dodecahedron with the icosahedron. - The doctrine of the five solids - in the Arabic must be the Comparison of the five figures - in the passage of Hypsicles, for nowhere else have we any information about a work bearing this title, nor can the Arabians have had. The reference to the two books of Apollonius on conics will then be the result of mixing up the fact that Apollonius wrote a book on conics with the second edition of the other work mentioned by Hypsicles. We do not find elsewhere in Arabian authors any mention of a commentary by Euclid on Apollonius and Aristaeus: so that the story in the passage quoted is really no more than a variation of the fable that the Elements were the work of Apollonius.

- - - - CHAPTER II. - EUCLID'S OTHER WORKS. -

In giving a list of the Euclidean treatises other than the Elements, I shall be brief: for fuller accounts of them, or speculations with regard to them, reference should be made to the standard histories of mathematicsSee, for example, Loria, Le scienze esatte nell' antica Grecia, 1914, pp. 245-268; 1. L. Heath, History of Greek Mathematics, 1921, I. pp. 421-446. Cf. Heiberg, Litterargeschichtliche Studien über Euklid, pp. 36-153; Euclidis opera omnia, ed. Heiberg and Menge, Vols. VI.—VIII..

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I will take first the works which are mentioned by Greek authors.

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I. The Pseudaria.

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I mention this first because Proclus refers to it in the general remarks in praise of the Elements which he gives immediately after the mention of Euclid in his summary. He saysProclus, p. 70, 1-18.: But, inasmuch as many things, while appearing to rest on truth and to follow from scientific principles, really tend to lead one astray from the principles and deceive the more superficial minds, he has handed down methods for the discriminative understanding of these things as well, by the use of which methods we shall be able to give beginners in this study practice in the discovery of paralogisms, and to avoid being misled. This treatise, by which he puts this machinery in our hands, he entitled (the book) of Pseudaria, enumerating in order their various kinds, exercising our intelligence in each case by theorems of all sorts, setting the true side by side with the false, and combining the refutation of error with practical illustration. This book then is by way of cathartic and exercise, while the Elements contain the irrefragable and complete guide to the actual scientific investigation of the subjects of geometry. -

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The book is considered to be irreparably lost. We may conclude however from the connexion of it with the Elements and the reference to its usefulness for beginners that it did not go outside the domain of elementary geometryHeiberg points out that Alexander Aphrodisiensis appears to allude to the work in his commentary on Aristotle's Sophistici Elenchi (fol. 25 b): Not only those (e)/legxoi) which do not start from the principles of the science under which the problem is classed...but also those which do start from the proper principles of the science but in some respect admit a paralogism, e.g. the Pseudographemata of Euclid. - Tannery (Bull. des sciences math. et astr. 2e Série, VI., 1882, Ière Partie, p. 147) conjectures that it may be from this treatise that the same commentator got his information about the quadratures of the circle by Antiphon and Bryson, to say nothing of the lunules of Hippocrates. I think however that there is an objection to this theory so far as regards Bryson; for Alexander distinctly says that Bryson's quadrature did not start from the proper principles of geometry, but from some principles more general..

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2. The Data.

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The Data (dedome/na) are included by Pappus in the Treasury of Analysis (to/pos a)naluo/menos), and he describes their contentsPappus, VII. p. 638. They are still concerned with elementary geometry, though forming part of the introduction to higher analysis. Their form is that of propositions proving that, if certain things in a figure are given (in magnitude, in species, etc.), something else is given. The subjectmatter is much the same as that of the planimetrical books of the Elements, to which the Data are often supplementary. We shall see this later when we come to compare the propositions in the Elements which give us the means of solving the general quadratic equation with the corresponding propositions of the Data which give the solution. The Data may in fact be regarded as elementary exercises in analysis.

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It is not necessary to go more closely into the contents, as we have the full Greek text and the commentary by Marinus newly edited by Menge and therefore easily accessibleVol. VI. in the Teubner edition of Euclidis opera omnia by Heiberg and Menge. A translation of the Data is also included in Simson's Euclid (though naturally his text left much to be desired)..

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3. The book On divisions (of figures).

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This work (peri\ diaire/sewn bibli/on) is mentioned by ProclusProclus, p. 69, 4.. In one place he is speaking of the conception or definition (lo/gos) of figure, and of the divisibility of a figure into others differing from it in kind; and he adds: For the circle is divisible into parts unlike in definition or notion (a)no/moia tw=| lo/gw|), and so is each of the rectilineal figures; this is in fact the business of the writer of the Elements in his Divisions, where he divides given figures, in one case into like figures, and in another into unlikeibid. 144, 22-26.. - Like - and unlike - here mean, not similar - and dissimilar - in the technical sense, but like - or unlike in definition or notion - (lo/gw|): thus to divide a triangle into triangles would be to divide it into like - figures, to divide a triangle into a triangle and a quadrilateral would be to divide it into unlike - figures.

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The treatise is lost in Greek but has been discovered in the Arabic. First John Dee discovered a treatise De divisionibus by one Muhammad BagdadinusSteinschneider places him in the 10th c. H. Suter (Bibliotheca Mathematica, IV_{3}, 1903, pp. 24, 27) identifies him with Abū (Bekr) Muh. b. 'Abdalbāqī al-Baġdādī, Qād|ī (Judge) of Māristān (circa 1070-1141), to whom he also attributes the Liber judei (? judicis) super decimum Euclidis translated by Gherard of Cremona. and handed over a copy of it (in Latin) in 1563 to Commandinus, who published it, in Dee did not himself translate the tract from the Arabic; he in 1570De superficierum divisionibus liber Machometo Bagdadino adscriptus, nunc primum Ioannis Dee Londinensis et Federici Commandini Urbinatis opera in lucem editus, Pisauri, 1570, afterwards included in Gregory's Euclid (Oxford, 1703).. Dee did not himself translate the tract from the Arabic; he found it in Latin in a MS. which was then in his own possession but was about 20 years afterwards stolen or destroyed in an attack by a mob on his house at MortlakeR. C. Archibald, Euclid's Book on the Division of Figures with a restoration based on Woepck's text and on the Practica geometriae of Leonardo Pisano, Cambridge, 1915, pp. 4-9.. Dee, in his preface addressed to Commandinus, says nothing of his having translated the book, but only remarks that the very illegible MS. had caused him much trouble and (in a later passage) speaks of the actual, very ancient, copy from which I wrote out... - (in ipso unde descripsi vetustissimo exemplari). The Latin translation of this tract from the Arabic was probably made by Gherard of Cremona (1114-1187), among the list of whose numerous translations a liber divisionum - occurs. The Arabic original cannot have been a direct translation from Euclid, and probably was not even a direct adaptation of it; it contains mistakes and unmathematical expressions, and moreover does not contain the propositions about the division of a circle alluded to by Proclus. Hence it can scarcely have contained more than a fragment of Euclid's work.

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But Woepcke found in a MS. at Paris a treatise in Arabic on the division of figures, which he translated and published in 1851Fournal Asiatique, 1851, p. 233 sqq.. It is expressly attributed to Euclid in the MS. and corresponds to the description of it by Proclus. Generally speaking, the divisions are divisions into figures of the same kind as the original figures, e.g. of triangles into triangles; but there are also divisions into unlike - figures, e.g. that of a triangle by a straight line parallel to the base. The missing propositions about the division of a circle are also here: to divide into two equal parts a given figure bounded by an arc of a circle and two straight lines including a given angle - and to draw in a given circle two parallel straight lines cutting off a certain part of the circle. - Unfortunately the proofs are given of only four propositions (including the two last mentioned) out of 36, because the Arabic translator found them too easy and omitted them. To illustrate the character of the problems dealt with I need only take one more example: To cut off a certain fraction from a (parallel-) trapezium by a straight line which passes through a given point lying inside or outside the trapezium but so that a straight line can be drawn through it cutting both the parallel sides of the trapezium. - The genuineness of the treatise edited by Woepcke is attested by the facts that the four proofs which remain are elegant and depend on propositions in the Elements, and that there is a lemma with a true Greek ring: to apply to a straight line a rectangle equal to the rectangle contained by AB, AC and deficient by a square. - Moreover the treatise is no fragment, but finishes with the words end of the treatise, - and is a well-ordered and compact whole. Hence we may safely conclude that Woepcke's is not only Euclid's own work but the whole of it. A restoration of the work, with proofs, was attempted by OfterdingerL. F. Ofterdinger, Beiträge zur Wiederherstellung der Schrift des Euklides über die Theilung der Figuren, Ulm, 1853., Who however does not give Woepcke's props. 30, 31, 34, 35, 36. We have now a satisfactory restoration, with ample notes and an introduction, by R. C. Archibald, who used for the purpose Woepcke's text and a section of Leonardo of Pisa's Practica geometriae (1220). There is a remarkable similarity between the propositions of Woepcke's text and those of Leonardo, suggesting that Leonardo may have had before him a translation (perhaps by Gherard of Cremona) of the Arabic tract..

-

4.The Porisms.

-

It is not possible to give in this place any account of the controversies about the contents and significance of the three lost books of Porisms, or of the important attempts by Robert Simson and Chasles to restore the work. These may be said to form a whole literature, references to which will be found most abundantly given by Heiberg and Loria, the former of whom has treated the subject from the philological point of view, most exhaustively, while the latter, founding himself generally on Heiberg, has added useful details, from the mathematical side, relating to the attempted restorations, etc.Heiberg, Euklid-Studien, pp. 56-79, and Loria, op. cit., pp. 253-265. It must suffice here to give an extract from the only original source of information about the nature and contents of the Porisms, namely PappusPappus, ed. Hultsch, VII. pp. 648-660. I put in square brackets the words bracketed by Hultsch.. In his general preface about the books composing the Treasury of Analysis (to/pos a)naluo/menos) he says:

-

“After the Tangencies (of Apollonius) come, in three books, the Porisms of Euclid, [in the view of many] a collection most ingeniously devised for the analysis of the more weighty problems, [and] although nature presents and unlimited number of such porismsI adopt Heiberg's reading of a comma here instead of a full stop., [they have added nothing to what was written originally by Euclid, except that some before my time have shown their want of taste by adding to a few (of the propositions) second proofs, each (proposition) admitting of a definite number of demonstrations, as we have shown, and Euclid having given one for each, namely that which is the most lucid. These porisms embody a theory subtle, natural, necessary, and of considerable generality, which is fascinating to those who can see and produce results].

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“Now all the varieties of porisms belong, neither to theorems nor problems, but to a species occupying a sort of intermediate position [so that their enunciations can be formed like those of either theorems or problems], the result being that, of the great number of geometers, some regarded them as of the class of theorems, and others of problems, looking only to the form of the proposition. But that the ancients knew better the difference between these three things is clear from the definitions. For they said that a theorem is that which is proposed with a view to the demonstration of the very thing proposed, a problem that which is thrown out with a view to the construction of the very thing proposed, and a porism that which is proposed with a view to the producing of the very thing proposed. [But this definition of the porism was changed by the more recent writers who could not produce everything, but used these elements and proved only the fact that that which is sought really exists, but did not produce itHeiberg points out that Props. 5-9 of Archimedes' treatise On Spirals are porisms in this sense. To take Prop. 5 as an example, DBF is a tangent to a circle with centre K. It is then possible, says Archimedes, to draw a straight line KHF, meeting the circumference in H and the tangent in F, such that -

where c is the circumference of any circle. To prove this he assumes the following construction. E being any straight line greater than c, he says: let KG be parallel to DF, and let the line GH equal to E be placed verging to the point B. - Archimedes must of course have known how to effect this construction, which requires conics. But that it is possible requires very little argument, for if we draw any straight line BHG meeting the circle in H and KG in G, it is obvious that as G moves away from C, HG becomes greater and greater and may be made as great as we please. The later writers - would no doubt have contented themselves with this consideration without actually constructing HG. and were accordingly confuted by the definition and the whole doctrine. They based their definition on an incidental characteristic, thus: A porism is that which falls short of a locustheorem in respect of its hypothesisAs Heiberg says, this translation is made certain by a preceding passage of Pappus (p. 648, 1-3) where he compares two enunciations, the latter of which falls short of the former in hypothesis but goes beyond it in requirement. - E.g. the first enunciation requiring us, given three circles, to draw a circle touching all three, the second may require us, given only two circles (one less datum), to draw a circle touching them and of a given size (an extra requirement).. Of this kind of porisms loci are a species, and they abound in the Treasury of Analysis; but this species has been collected, named and handed down separately from the porisms, because it is more widely diffused than the other species]. But it has further become characteristic of porisms that, owing to their complication, the enunciations are put in a contracted form, much being by usage left to be understood; so that many geometers understand them only in a partial way and are ignorant of the more essential features of their contents.

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“[Now to comprehend a number of propositions in one enunciation is by no means easy in these porisms, because Euclid himself has not in fact given many of each species, but chosen, for examples, one or a few out of a great multitudeI translate Heiberg's reading with a full stop here followed by pro\s a)rxh=| de\ o(/mws [pro\s a)rxh\n (dedome/non) Hultsch] tou= prw/tou bibli/ou..... But at the beginning of the first book he has given some propositions, to the number of ten, of one species, namely that more fruitful species consisting of loci.] Consequently, finding that these admitted of being comprehended in one enunciation, we have set it out thus: -

If, in a system of four straight linesThe four straight lines are described in the text as (the sides) u(pti/ou h)\ parupti/ou, i.e. sides of two sorts of quadrilaterals which Simson tries to explain (see p. 120 of the Index Graecitatis of Hultsch's edition of Pappus). which cut each other two and two, three points on one straight line be given while the rest except one lie on different straight lines given in position, the remaining point also will lie on a straight line given in position.In other words (Chasles, p. 23; Loria, p. 256), if a triangle be so deformed that each of its sides turns about one of three points in a straight line, and two of its vertices lie on two straight lines given in position, the third vertex will also lie on a straight line..

- -

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“This has only been enunciated of four straight lines, of which not more than two pass through the same point, but it is not known (to most people) that it is true of any assigned number of straight lines if enunciated thus: -

If any number of straight lines cut one another, not more than two (passing) through the same point, and all the points (of intersection situated) on one of them be given, and if each of those which are on another (of them) lie on a straight line given in position—

- or still more generally thus: if any number of straight lines cut one another, not more than two (passing) through the same point, and all the points (of intersection situated) on one of them be given, while of the other points of intersection in multitude equal to a triangular number a number corresponding to the side of this triangular number lie respectively on straight lines given in position, provided that of these latter points no three are at the angular points of a triangle (sc. having for sides three of the given straight lines)—each of the remaining points will lie on a straight line given in positionLoria (p. 256, n. 3) gives the meaning of this as follows, pointing out that Simson was the discoverer of it: If a complete n-lateral be deformed so that its sides respectively turn about n points on a straight line, and (n-1) of its n(n-1)/2 vertices move on as many straight lines, the other (n-1)(n-2)/2 of its vertices likewise move on as many straight lines: but it is necessary that it should be impossible to form with the (n-1) vertices any triangle having for sides the sides of the polygon. -.

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“It is probable that the writer of the Elements was not unaware of this but that he only set out the principle; and he seems, in the case of all the porisms, to have laid down the principles and the seed only [of many important things], the kinds of which should be distinguished according to the differences, not of their hypotheses, but of the results and the things sought. [All the hypotheses are different from one another because they are entirely special, but each of the results and things sought, being one and the same, follow from many different hypotheses.]

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“We must then in the first book distinguish the following kinds of things sought:

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“At the beginning of the bookReading, with Heiberg, tou= bibli/ou [tou= z Hultsch]. is this proposition: -

I. ’If from two given points straight lines be drawn meeting on a straight line given in position, and one cut off from a straight line given in position (a segment measured) to a given point on it, the other will also cut off from another (straight line a segment) having to the first a given ratio.’

-

-

“Following on this (we have to prove) -

- II. that such and such a point lies on a straight line given in position;

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- III. that the ratio of such and such a pair of straight lines is given;” - etc. etc. (up to XXIX.).

-

-

The three books of the porisms contain 38 lemmas; of the theorems themselves there are 171. -

-

Pappus further gives lemmas to the Porisms (pp. 866-918, ed. Hultsch).

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With Pappus' account of Porisms must be compared the passages of Proclus on the same subject. Proclus distinguishes two senses in which the word po/risma is used. The first is that of corollary where something appears as an incidental result of a proposition, obtained without trouble or special seeking, a sort of bonus which the investigation has presented us withProclus, pp. 212, 14; 301, 22.. The other sense is that of Euclid's Porismsibid. p. 212, 12. The term porism is used of certam problems, like the Porisms written by Euclid. -. In this senseibid. pp. 301, 25 sqq. porism is the name given to things which are sought, but need some finding and are neither pure bringing into existence nor simple theoretic argument. For (to prove) that the angles at the base of isosceles triangles are equal is a matter of theoretic argument, and it is with reference to things existing that such knowledge is (obtained). But to bisect an angle, to construct a triangle, to cut off, or to place—all these things demand the making of something; and to find the centre of a given circle, or to find the greatest common measure of two given commensurable magnitudes, or the like, is in some sort between theorems and problems. For in these cases there is no bringing into existence of the things sought, but finding of them, nor is the procedure purely theoretic. For it is necessary to bring that which is sought into view and exhibit it to the eye. Such are the porisms which Euclid wrote, and arranged in three books of Porisms. -

-

Proclus' definition thus agrees well enough with the first, older, - definition of Pappus. A porism occupies a place between a theorem and a problem: it deals with something already existing, as a theorem does, but has to find it (e.g. the centre of a circle), and, as a certain operation is therefore necessary, it partakes to that extent of the nature of a problem, which requires us to construct or produce something not previously existing. Thus, besides III. I of the Elements and X. 3, 4 mentioned by Proclus, the following propositions are real porisms: III. 25, VI. 11-13, VII. 33, 34, 36, 39, VIII. 2, 4, X. 10, XIII. 18. Similarly in Archimedes On the Sphere and Cylinder I. 2-6 might be called porisms.

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The enunciation given by Pappus as comprehending ten of Euclid's propositions may not reproduce the form of Euclid's enunciations; but, comparing the result to be proved, that certain points lie on straight lines given in position, with the class indicated by II. above, where the question is of such and such a point lying on a straight line given in position, and with other classes, e.g. (V.) that such and such a line is given in position, (VI.) that such and such a line verges to a given point, (XXVII.) that there exists a given point such that straight lines drawn from it to such and such (circles) will contain a triangle given in species, we may conclude that a usual form of a porism was to prove that it is possible to find a point with such and such a property - or a straight line on which lie all the points satisfying given conditions - etc.

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Simson defined a porism thus: Porisma est propositio in qua proponitur demonstrare rem aliquam, vel plures datas esse, cui, vel quibus, ut et cuilibet ex rebus innumeris, non quidem datis, sed quae ad ea quae data sunt eandem habent relationem, convenire ostendendum est affectionem quandam communem in propositione descriptamThis was thus expressed by Chasles: Le porisme est une proposition dans laquelle on demande de démontrer qu'une chose ou plusieurs choses sont données, qui, ainsi que l'une quelconque d'une infinité d'autres choses non données, mais dont chacune est avec des choses données dans une même relation, ont une certaine propriété commune, décrite dans la proposition. -. -

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From the above it is easy to understand Pappus' statement that loci constitute a large class of porisms. A locus is well defined by Simson thus: Locus est proposition in qua propositum est datam esse demonstrare, vel invenire lineam aut superficiem cuius quodlibet punctum, vel superficiem in qua quaelibet linea data lege descripta, communem quandam habet proprietatem in propositione descriptam. - Heiberg cites an excellent instance of a locus which is a porism, namely the following proposition quoted by EutociusCommentary on Apollonius' Conics (vol. II. p. 180, ed. Heiberg). from the Plane Loci of Apollonius:

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Given two points in a plane, and a ratio between unequal straight lines, it is possible to draw, in the plane, a circle such that the straight lines drawn from the given points to meet on the circumference of the circle have (to one another) a ratio the same as the given ratio. -

-

A difficult point, however, arises on the passage of Pappus, which says that a porism is that which, in respect of its hypothesis, falls short of a locus-theorem - (topikou= qewrh/matos). Heiberg explains it by comparing the porism from Apollonius' Plane Loci just given with Pappus' enunciation of the same thing, to the effect that, if from two given points two straight lines be drawn meeting in a point, and these straight lines have to one another a given ratio, the point will lie on either a straight line or a circumference of a circle given in position. Heiberg observes that in this latter enunciation something is taken into the hypothesis which was not in the hypothesis of the enunciation of the porism, viz. that the ratio of the straight lines is the same. - I confess this does not seem to me satisfactory: for there is no real difference between the enunciations, and the supposed difference in hypothesis is very like playing with words. Chasles says: Ce qui constitue le porisme est ce qui manque à l' hypothèse d'un théorème local (en d'autres termes, le porisme est inférieur, par l'hypothèse, au théorème local; c'est-à-dire que quand quelques parties d'une proposition locale n'ont pas dans l'énoncé la détermination qui leur est propre, cette proposition cesse d'être regardée comme un theéorème et devient un porisme). - But the subject still seems to require further elucidation.

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While there is so much that is obscure, it seems certain (1) that the Porisms were distinctly part of higher geometry and not of elementary geometry, (2) that they contained propositions belonging to the modern theory of transversals and to projective geometry. It should be remembered too that it was in the course of his researches on this subject that Chasles was led to the idea of anharmonic ratios.

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Lastly, allusion should be made to the theory of ZeuthenDie Lehre von den Kegelschnitten im Altertum, chapter VIII. on the subject of the porisms. He observes that the only porism of which Pappus gives the complete enunciation, If from two given points straight lines be drawn meeting on a straight line given in position, and one cut off from a straight line given in position (a segment measured) towards a given point on it, the other will also cut off from another (straight line a segment) bearing to the first a given ratio, - is also true if there be substituted for the first given straight line a conic regarded as the locus with respect to four lines, - and that this extended porism can be used for completing Apollonius' exposition of that locus. Zeuthen concludes that the Porisms were in part byproducts of the theory of conics and in part auxiliary means for the study of conics, and that Euclid called them by the same name as that applied to corollaries because they were corollaries with respect to conics. But there appears to be no evidence to confirm this conjecture.

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5. The Surface-loci (to/poi pro\s e)pifanei/a|).

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The two books on this subject are mentioned by Pappus as part of the Treasury of AnalysisPappus, VII. p. 636.. As the other works in the list which were on plane subjects dealt only with straight lines, circles, and conic sections, it is a priori likely that among the loci in this treatise (loci which are surfaces) were included such loci as were cones, cylinders and spheres. Beyond this all is conjecture based on two lemmas given by Pappus in connexion with the treatise.

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(1) The first of these lemmasibid. VII. p. 1004. and the figure attached to it are not satisfactory as they stand, but a possible restoration is indicated by TanneryBulletin des sciences math. et astron., 2o Série, VI. 149.. If the latter is right, it suggests that one of the loci contained all the points on the elliptical parallel sections of a cylinder and was therefore an oblique circular cylinder. Other assumptions with regard to the conditions to which the lines in the figure may be subject would suggest that other loci dealt with were cones regarded as containing all points on particular elliptical parallel sections of the conesFurther particulars will be found in The Works of Archimedes, pp. lxii—lxiv, and in Zeuthen, Die Lehre von den Kegelschnitten, p. 425 sqq..

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(2) In the second lemma Pappus states and gives a complete proof of the focus-and-directrix property of a conic, viz. that the locus of a point whose distance from a given point is in a given ratio to its distance from a fixed line is a conic section, which is an ellipse, a parabola or a hyperbola according as the given ratio is less than, equal to, or greater than unityPappus, VII. pp. 1006-1014, and Hultsch's Appendix, pp. 1270-3.. Two conjectures are possible as to the application of this theorem in Euclid's Surface-loci. (a) It may have been used to prove that the locus of a point whose distance from a given straight line is in a given ratio to its distance from a given plane is a certain cone. (b) It may have been used to prove that the locus of a point whose distance from a given point is in a given ratio to its distance from a given plane is the surface formed by the revolution of a conic about its major or conjugate axisFor further details see The Works of Archimedes, pp. lxiv, lxv, and Zeuthen, l. c.. Thus Chasles may have been correct in his conjecture that the Surface-loci dealt with surfaces of revolution of the second degree and sections of the sameAperçu historique, pp. 273-4..

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6. The Conics.

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Pappus says of this lost work: The four books of Euclid's Conics were completed by Apollonius, who added four more and gave us eight books of Conics -Pappus, VII. p. 672.. - It is probable that Euclid's work was lost even by Pappus'time, for he goes on to speak of Aristaeus, who wrote the still extant five books of Solid Loci connected with the conics. - Speaking of the relation of Euclid's work to that of Aristaeus on conics regarded as loci, Pappus says in a later passage (bracketed however by Hultsch) that Euclid, regarding Aristaeus as deserving credit for the discoveries he had already made in conics, did not (try to) anticipate him or construct anew the same system. We may no doubt conclude that the book by Aristaeus on solid loci preceded Euclid's on conics and was, at least in point of originality, more important. Though both treatises dealt with the same subject-matter, the object and the point of view were different; had they been the same, Euclid could scarcely have refrained, as Pappus says he did, from attempting to improve upon the earlier treatise. No doubt Euclid wrote on the general theory of conics as Apollonius did, but confined himself to those properties which were necessary for the analysis of the Solid Loci of Aristaeus. The Conics of Euclid were evidently superseded by the treatise of Apollonius.

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As regards the contents of Euclid's Conics, the most important source of our information is Archimedes, who frequently refers to propositions in conics as well known and not needing proof, adding in three cases that they are proved in the elements of conics - or in the conics, - which expressions must clearly refer to the works of Aristaeus and EuclidFor details of these propositions see my Apollonius of Perga, pp. xxxv, xxxvi.

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Euclid still used the old names for the conics (sections of a rightangled, acute-angled, or obtuse-angled cone), but he was aware that an ellipse could be obtained by cutting a cone in any manner by a plane not parallel to the base (assuming the section to lie wholly between the apex of the cone and its base) and also by cutting a cylinder. This is expressly stated in a passage from the Phaenomena of Euclid about to be mentionedPhaenomena, ed. Menge, p. 6: If a cone or a cylinder be cut by a plane not parallel to the base, the section is a section of an acute-angled cone, which is like a shield (qureo/s). -.

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7. The Phaenomena.

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This is an astronomical work and is still extant. A much interpolated version appears in Gregory's Euclid. An earlier and better recension is however contained in the MS. Vindobonensis philos. Gr. 103, though the end of the treatise, from the middle of prop. 16 to the last (18), is missing. The book, now edited by MengeEuclidis opera omnia, vol. VIII., 1916, pp. 2-156.,consists of propositions in spheric geometry. Euclid based it on Autolycus' work peri\ kinoume/nhs sfai/ras, but also, evidently, on an earlier textbook of Sphaerica of exclusively mathematical content. It has been conjectured that the latter textbook may have been due to EudoxusHeiberg, Euklid-Studien, p. 46; Hultsch, Autolycus, p. XII; A. A. Björnbo, Studien über Menelaos' Sphärik (Abhandlungen zur Geschichte der mathematischen Wissenschaften, XIV. 1902), p. 56sqq..

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8. The Optics.

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This book needs no description, as it has been edited by Heiberg recentlyEuclidis opera omnia, vol. VII. (1895).,both in its genuine form and in the recension by Theon. The Catoptrica published by Heiberg in the same volume is not genuine, and Heiberg suspects that in its present form it may be Theon's. It is not even certain that Euclid wrote Catoptrica at all, as Proclus may easily have had Theon's work before him and inadvertently assigned it to EuclidHeiberg, Euclid's Optics, etc. p. l..

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9. Besides the above-mentioned works, Euclid is said to have written the Elements of MusicProclus, p. 69, 3. (ai( kata\ mousikh\n stoixeiw/seis). Two treatises are attributed to Euclid in our MSS. of the Musici, the katatomh\ kano/nos, Sectio canonis (the theory of the intervals), and the ei)sagwgh\ a(rmonikh/ (introduction to harmony)Both treatises edited by Jan in Musici Scriptores Graeci, 1895. pp. 113-166, 167-207, and by Menge in Euclidis opera omnia, vol. VIII., 1916, pp. 157-183, 185-223..The first, resting on the Pythagorean theory of music, is mathematical, and the style and diction as well as the form of the propositions mostly agree with what we find in the Elements. Jan thought it genuine, especially as almost the whole of the treatise (except the preface) is quoted in extenso, and Euclid is twice mentioned by name, in the commentary on Ptolemy's Harmonica published by Wallis and attributed by him to Porphyry. Tannery was of the opposite opinionComptes rendus de l' Acad. des inscriptions et belles-lettres, Paris, 1904, pp. 439-445. Cf. Bibliotheca Mathematica, vI3, 1905-6, p. 225, note I..The latest editor, Menge, suggests that it may be a redaction by a less competent hand from the genuine Euclidean Elements of Music. The second treatise is not Euclid's, but was written by Cleonides, a pupil of AristoxenusHeiberg, Euklid-Studien, pp. 52-55; Jan, Musici Scriptores Graeci, pp. 169-174..

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Lastly, it is worth while to give the Arabians' list of Euclid's works. I take this from Suter's translation of the list of philosophers and mathematicians in the Fihrist, the oldest authority of the kind that we possessH. Suter, Das Mathematiker-Verzeichniss im Fihrist in Abhandlungen zur Geschichte der Mathematik, VI., 1892, pp. 1-87 (see especially p. 17). Cf. Casiri, 1. 339, 340, and Gartz, De interpretibus et explanatoribus Euclidis Arabicis, 1823, pp. 4, 5..To the writings of Euclid belong further [in addition to the Elements]: the book of Phaenomena; the book of Given Magnitudes [Data]; the book of Tones, known under the name of Music, not genuine; the book of Division, emended by Thābit; the book of Utilisations or Applications [Porisms], not genuine; the book of the Canon; the book of the Heavy and Light; the book of Synthesis, not genuine; and the book of Analysis, not genuine. -

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It is to be observed that the Arabs already regarded the book of Tones (by which must be meant the ei)sagwgh\ a(rmonikh/) as spurious. The book of Division is evidently the book on Divisions (of figures). The next book is described by Casiri as liber de utilitate suppositus. - Suter gives reason for believing the Porisms to be meantSuter, op. cit. pp. 49, 50. Wenrich translated the word as utilia. - Suter says that the nearest meaning of the Arabic word as of porism - is use, gain (Nutzen, Gewinn), while a further meaning is explanation, observation, addition: a gain arising out of what has preceded (cf. Proclus' definition of the porism in the sense of a corollary).,but does not apparently offer any explanation of why the work is supposed to be spurious. The book of the Canon is clearly the katatomh\ kano/nos. The book on the Heavy and Light - is apparently the tract De levi et ponderoso, included in the Basel Latin translation of 1537, and in Gregory's edition. The fragment, however, cannot safely be attributed to Euclid, for (1) we have nowhere any mention of his having written on mechanics, (2) it contains the notion of specific gravity in a form so clear that it could hardly be attributed to anyone earlier than ArchimedesHeiberg, Euklid-Studien, pp. 9, 10..Suter thinksSuter, op. cit. p. 50.that the works on Analysis and Synthesis (said to be spurious in the extract) may be further developments of the Data or Porisms, or may be the interpolated proofs of Eucl. XIII. 1-5, divided into analysis and synthesis, as to which see the notes on those propositions.

- - - - CHAPTER III. - GREEK COMMENTATORS ON THE ELEMENTS OTHER THAN PROCLUS. -

That there was no lack of commentaries on the Elements before the time of Proclus is evident from the terms in which Proclus refers to them; and he leaves us in equally little doubt as to the value which, in his opinion, the generality of them possessed. Thus he says in one place (at the end of his second prologue)Proclus, p. 84, 8.:

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Before making a beginning with the investigation of details, I warn those who may read me not to expect from me the things which have been dinned into our ears ad nauseam (diateqru/lhtai) by those who have preceded me, viz. lemmas, cases, and so forth. For I am surfeited with these things and shall give little attention to them. But I shall direct my remarks principally to the points which require deeper study and contribute to the sum of philosophy, therein emulating the Pythagoreans who even had this common phrase for what I mean ’a figure and a platform, but not a figure and sixpencei.e. we reach a certain height, use the platform so attained as a base on which to build another stage, then use that as a base and so on..’ -

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In another placeProclus, p. 200, 10. he says: Let us now turn to the elucidation of the things proved by the writer of the Elements, selecting the more subtle of the comments made on them by the ancient writers, while cutting down their interminable diffuseness, giving the things which are more systematic and follow scientific methods, attaching more importance to the working-out of the real subject-matter than to the variety of cases and lemmas to which we see recent writers devoting themselves for the most part. -

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At the end of his commentary on Eucl. I. Prochis remarksibid. p. 432, 15. that the commentaries then in vogue were full of all sorts of confusion, and contained no account of causes, no dialectical discrimination, and no philosophic thought.

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These passages and two others in which Proclus refers to the commentatorsibid. p. 289, II; p. 328, 16. - suggest that these commentators were numerous. He does not however give many names; and no doubt the only important commentaries were those of Heron, Porphyry, and Pappus.

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- 1. Heron. -

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Proclus alludes to Heron twice as Heron mechanicusProclus, p. 305, 24; p. 346, 13., in another placeibid. p. 41, 10. he associates him with Ctesibius, and in the three other passagesibid. p. 196, 16; p. 323, 7; p. 429, 13. where Heron is mentioned there is no reason to doubt that the same person is meant, namely Heron of Alexandria. The date of Heron is still a vexed question. In the early stages of the controversy much was made of the supposed relation of Heron to Ctesibius. The best MS. of Heron's Belopoeica has the heading *hrwnos *kthsi*bi/ou belopoii+ka/, and an anonymous Byzantine writer of the tenth century, evidently basing himself on this title, speaks of Ctesibius as Heron's kaqhghth/s, master - or teacher. - We know of two men of the name of Ctesibius. One was a barber who lived in the time of Ptolemy Euergetes II, i.e. Ptolemy VII, called Physcon (died 117 B.C.), and who is said to have made an improved water-organAthenaeus, Deipno-Soph. iv., c. 75, p. 174 b—c.. The other was a mechanician mentioned by Athenaeus as having made an elegant drinking-horn in the time of Ptolemy Philadelphus (285-247 B.C.)ibid. XI., c. 97, p. 497 b—c.. MartinMartin, Recherches sur la vie et les ouvrages d'Héron d'Alexandrie, Paris, 1854, p. 27. took the Ctesibius in question to be the former and accordingly placed Heron at the beginning of the first century B.C., say 126-50 B.C. But Philo of ByzantiumPhilo, Mechan. Synt., p. 50, 38, ed. Schöne., who repeatedly mentions Ctesibius by name, says that the first mechanicians had the advantage of being under kings who loved fame and supported the arts. Hence our Ctesibius is more likely to have been the earlier Ctesibius who was contemporary with Ptolemy II Philadelphus.

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But, whatever be the date of Ctesibius, we cannot safely conclude that Heron was his immediate pupil. The title Heron's (edition of) Ctesibius's Belopoeica - does not, in fact, justify any inferenee as to the interval of time between the two works.

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We now have better evidence for a terminus post quem. The Metrica of Heron, besides quoting Archimedes and Apollonius, twice refers to the books about straight lines (chords) in a circle - (e)n toi=s peri\ tw=n e)n ku/klw| eu)qeiw=n). Now we know of no work giving a Table of Chords earlier than that of Hipparchus. We get, therefore, at once, 150 B.C. or thereabouts as the terminus post quem. But, again, Heron's Mechanica quotes a definition of centre of gravity - as given by Posidonius, a Stoic -: and, even if this Posidonius lived before Archimedes, as the context seems to imply, it is certain that another work of Heron's, the Definitions, owes something to Posidonius of Apamea or Rhodes, Cicero's teacher (135-51 B.C.). This brings Heron's date down to the end of the first century B.C., at least.

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We have next to consider the relation, if any, between Heron and Vitruvius. In his De Architectura, brought out apparently in 14 B.C., Vitruvius quotes twelve authorities on machinationes including Archytas (second), Archimedes (third), Ctesibius (fourth) and Philo of Byzantium (sixth), but does not mention Heron. Nor is it possible to establish inter-dependence between Vitruvius and Heron; the differences between them seem on the whole more numerous and important than the resemblances (e.g. Vitruvius uses 3 as the value of p, while Heron always uses the Archimedean value 3 1/7). The inference is that Heron can hardly have written earlier than the first century A.D.

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The most recent theory of Heron's date makes him later than Claudius Ptolemy the astronomer (100-178 A.D.). The arguments are mainly these. (1) Ptolemy claims as a discovery of his own a method of measuring the distance between two places (as an arc of a great circle on the earth's surface) in the case where the places are neither on the same meridian nor on the same parallel circle. Heron, in his Dioptra, speaks of this method as of a thing generally known to experts. (2) The dioptra described in Heron's work is a fine and accurate instrument, much better than anything Ptolemy had at his disposal. (3) Ptolemy, in his work *peri\ r(opw=n, asserted that water with water round it has no weight and that the diver, however deep he dives, does not feel the weight of the water above him. Heron, strangely enough, accepts as true what Ptolemy says of the diver, but is dissatisfied with the explanation given by some, - namely that it is because water is uniformly heavy—this seems to be equivalent to Ptolemy's dictum that water in water has no weight—and he essays a different explanation based on Archimedes. (4) It is suggested that the Dionysius to whom Heron dedicated his Definitions is a certain Dionysius who was praefectus urbi in 301 A.D.

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On the other hand Heron was earlier than Pappus, who was writing under Diocletian (284-305 A.D.), for Pappus alludes to and draws upon the works of Heron. The net result, then, of the most recent research is to place Heron in the third century A.D. and perhaps little earlier than Pappus. HeibergHeronis Alexandrini opera, vol. v. (Teubner, 1914), p. IX. accepts this conclusion, which may therefore, perhaps, be said to hold the field for the present.Fuller details of the various arguments will be found in my History of Greek Mathematics, 1921, vol. II., pp. 298-306..

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That Heron wrote a systematic commentary on the Elements might be inferred from Proclus, but it is rendered quite certain by references to the commentary in Arabian writers, and particularly in an-Nairīzī's commentary on the first ten Books of the Elements. The Fihrist says, under Euclid, that Heron wrote a commentary on this book [the Elements], endeavouring to solve its difficultiesDas Mathematiker- Verzeichniss im Fihrist (tr. Suter), p. 16. -; and under Heron, He wrote: the book of explanation of the obscurities in Euclidibid. p. 22..... - An-Nairīzī's commentary quotes Heron by name very frequently, and often in such a way as to leave no doubt that the author had Heron's work actually before him. Thus the extracts are given in the first person, introduced by Heron says - (Dixit Yrinus - or Heron -); and in other places we are told that Heron says nothing, - or is not found to have said anything, - on such and such a proposition. The commentary of an-Nairīzī is in part edited by Besthorn and Heiberg from a Leiden MS. of the translation of the Elements by al-Hajjāj with the commentary attachedCodex Leidensis 399, 1. Euclidis Elementa ex interprctatione al-Hadschdschadschii cum commentariis al-Narizii. Five parts carrying the work to the end of Book iv. were issued in 1893, 1897, 1900, 1905 and 1910 respectively.. But this MS. only contains six Books, and several pages in the first Book, which contain the comments of Simplicius on the first twenty-two definitions of the first Book, are missing. Fortunately the commentary of an-Nairīzī has been discovered in a more complete form, in a Latin translation by Gherardus Cremonensis of the twelfth century, which contains the missing comments by Simplicius and an-Nairīzī's comments on the first ten Books. This valuable work has recently been edited by CurtzeAnaritii in decem libros priores elementorum Euclidis commentarii ex interpretatione Gherardi Cremonensis...edidit Maximilianus Curtze (Teubner, Leipzig, 1899)..

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Thus from the three sources, Proclus, and the two versions of an-Nairīzī, which supplement one another, we are able to form a very good idea of the character of Heron's commentary. In some cases observations given by Proclus without the name of their author are seen from an-Nairīzī to be Heron's; in a few cases notes attributed by Proclus to Heron are found in an-Nairīzī without Heron's name; and, curiously enough, one alternative proof (of I. 25) given as Heron's by Proclus is introduced by the Arab with the remark that he has not been able to discover who is the author.

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Speaking generally, the comments of Heron do not seem to have contained much that can be called important. We find

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(1) A few general notes, e.g. that Heron would not admit more than three axioms.

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(2) Distinctions of a number of particular cases of Euclid's propositions according as the figure is drawn in one way or in another.

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Of this class are the different cases of I. 35, 36, III. 7, 8 (where the chords to be compared are drawn on different sides of the diameter instead of on the same side), III. 12 (which is not Euclid's, but Heron's own, adding the case of external contact to that of internal contact in III. 11), VI. 19 (where the triangle in which an additional line is drawn is taken to be the smaller of the two), VII. 19 (where he gives the particular case of three numbers in continued proportion, instead of four proportionals).

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(3) Alternative proofs. Of these there should be mentioned (a) the proofs of II. 1-10 without a figure, - being simply the algebraic forms of proof, easy but uninstructive, which are so popular nowadays, the proof of III. 25 (placed after III. 30 and starting from the arc instead of the chord), III. 10 (proved by III. 9), III. 13 (a proof preceded by a lemma to the effect that a straight line cannot meet a circle in more than two points). Another class of alternative proof is (b) that which is intended to meet a particular objection (e)/nstasis) which had been or might be raised to Euclid's construction. Thus in certain cases he avoids producing a particular straight line, where Euclid produces it, in order to meet the objection of any one who should deny our right to assume that there is any space availableCf. Proclus, 275, 7 ei\ de\ le/goi tis to/pon mh\ ei\de/nai..., 289, 18 le/gei ou=)n tis o)/ti ou)k e)/sti to/pos..... Of this class are Heron's proofs of I. II, 1.20, and his note on 1. 16. Similarly on 1. 48 he supposes the right-angled triangle which is constructed to be constructed on the same side of the common side as the given triangle is. A third class (c) is that which avoids reductio ad absurdum. Thus, instead of indirect proofs, Heron gives direct proofs of I. 19 (for which he requires, and gives, a preliminary lemma), and of I. 25.

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(4) Heron supplies certain converses of Euclid's propositions, e.g. converses of II. 12, 13, VIII. 27.

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(5) A few additions to, and extensions of, Euclid's propositions are also found. Some are unimportant, e.g. the construction of isosceles and scalene triangles in a note on I. 1, the construction of two tangents in III. 17, the remark that VII. 3 about finding the greatest common measure of three numbers can be applied to as many numbers as we please (as Euclid tacitly assumes in VII. 31). The most important extension is that of III. 20 to the case where the angle at the circumference is greater than a right angle, and the direct deduction from this extension of the result of III. 22. Interesting also are the notes on I. 37 (on I. 24 in Proclus), where Heron proves that two triangles with two sides of one equal to two sides of the other and with the included angles supplementary are equal, and compares the areas where the sum of the two included angles (one being supposed greater than the other) is less or greater than two right angles, and on I. 47, where there is a proof (depending on preliminary lemmas) of the fact that, in the figure of the proposition, the straight lines AL, BK, CF meet in a point. After iv. 16 there is a proof that, in a regular polygon with an even number of sides, the bisector of one angle also bisects its opposite, and an enunciation of the corresponding proposition for a regular polygon with an odd number of sides.

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Van PeschDe Procli fontibus, Lugduni-Batavorum, 1900, gives reason for attributing to Heron certain other notes found in Proclus, viz. that they are designed to meet the same sort of points as Heron had in view in other notes undoubtedly written by him. These are (a) alternative proofs of I. 5, I. 17, and I. 32, which avoid the producing of certain straight lines, (b) an alternative proof of 1.9 avoiding the construction of the equilateral triangle on the side of BC opposite to A; (c) partial converses of I. 35-38, starting from the equality of the areas and the fact of the parallelograms or triangles being in the same parallels, and proving that the bases are the same or equal, may also be Heron's. Van Pesch further supposes that it was in Heron's commentary that the proof by Menelaus of I. 25 and the proof by Philo of I. 8 were given.

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The last reference to Heron made by an-Nairīzī occurs in the note on VIII. 27, so that the commentary of the former must at least have reached that point.

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II. Porphyry.

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The Porphyry here mentioned is of course the Neo-Platonist who lived about 232-304 A.D. Whether he really wrote a systematic commentary on the Elements is uncertain. The passages in Proclus which seem to make this probable are two in which he mentions him (1) as having demonstrated the necessity of the words not on the same side - in the enunciation of I. 14Proclus, pp. 297, 1-298, 10., and (2) as having pointed out the necessity of understanding correctly the enunciation of I. 26, since, if the particular injunctions as to the sides of the triangles to be taken as equal are not regarded, the student may easily fall into erroribid. p. 352, 13, 14 and the pages preceding,. These passages, showing that Porphyry carefully anaiysed Euclid's enunciations in these cases, certainly suggest that his remarks were part of a systematic commentary. Further, the list of mathematicians in the Fihrist gives Porphyry as having written a book on the Elements. - It is true that Wenrich takes this book to have been a work by Porphyry mentioned by Suidas and Proclus (Theolog. Platon.), peri\ a)rxw=n libri II.Fihrist (tr. Suter), p. 9, 10 and p. 45 (note 5)..

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There is nothing of importance in the notes attributed to Porphyry by Proclus.

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(1) Three alternative proofs of I. 20, which avoid producing a side of the triangle, are assigned to Heron and Porphyry without saying which belonged to which. If the first of the three was Heron's, I agree with van Pesch that it is more probable that the two others were both Porphyry's than that the second was Heron's and only the third Porphyry's. For they are similar in character, and the third uses a result obtained in the secondVan Pesch, De Procli fontibus, pp. 129, 130. Heiberg assigned them as above in his Euklid-Studien (p. 160), but seems to have changed his view later. (See Besthorn-Heiberg. Codex Leidensis, p. 93, note 2.).

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(2) Porphyry gave an alternative proof of I. 18 to meet a childish objection which is supposed to require the part of AC equal to AB to be cut off from CA and not from AC.

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Proclus gives a precisely similar alternative proof of I. 6 to meet a similar supposed objection; and it may well be that, though Proclus mentions no name, this proof was also Porphyry's, as van Pesch suggestsVan Pesch, op. cit. pp. 130-1..

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Two other references to Porphyry found in Proclus cannot have anything to do with commentaries on the Elements. In the first a work called the *summikta/ is quoted, while in the second a philosophical question is raised.

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III. Pappus.

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The references to Pappus in Proclus are not numerous; but we have other evidence that he wrote a commentary on the Elements. Thus a scholiast on the definitions of the Data uses the phrase as Pappus says at the beginning of his (commentary) on the 10th (book) of EuclidEuclid's Data, ed. Menge, p. 262.. - Again in the Fihrist we are told that Pappus wrote a commentary to the tenth book of Euclid in two partsFihrist (tr. Suter), p. 22.. Fragments of this still survive in a MS. described by WoepckeMémoirés présentés à l'académie des sciences, 1856, XIV. pp. 658-719., Paris. No. 952. 2 (supplément arabe de la Bibliothèque impériale), which contains a translation by Abū `Uthmān (beginning of 10th century) of a Greek commentary on Book X. It is in two books, and there can now be no doubt that the author of the Greek commentary was PappusWoepcke read the name of the author, in the title of the first book, as B. los (the dot representing a missing vowel). He quotes also from other MSS. (e.g. of the Ta)rīkh alHukamā and of the Fihrist) where he reads the name of the commentator as B. lis, B.n.s or B.l.s. Woepcke takes this author to be Valens, and thinks it possible that he may be the same as the astrologer Vettius Valens. This Heiberg (Euklid-Studien, pp. 169, 170) proves to be impossible, because, while one of the mss. quoted by Woepcke says that B.n.s, le RoÛmi - (late-Greek) was later than Claudius Ptolemy and the Fihrist says B.l.s, le RoÛmi - wrote a commentary on Ptolemy's Planisphaerium, Vettius Valens seems to have lived under Hadrian, and must therefore have been an eldercontemporary of Ptolemy. But Suter shows (Fihrist, p. 22 and p. 54, note 92) that Banos is only distinguished from Babos by the position of a certain dot, and Balos may also easily have arisen from an original Babos (there is no P in Arabic), so that Pappus must be the person meant. This is further confirmed by the fact that the Fihrist gives this author and Valens as the subjects of two separate paragraphs, attributing to the latter astrological works only.. Again Eutocius, in his note on Archimedes, On the Sphere and Cylinder I. 13, says that Pappus explained in his commentary on the Elements how to inscribe in a circle a polygon similar to a polygon inscribed in another circle; and this would presumably come in his commentary on Book XII., just as the problem is solved in the second scholium on Eucl. XII. I. Thus Pappus' commentary on the Elements must have been pretty complete, an additional confirmation of this supposition being forthcoming in the reference of Marinus (a pupil and follower of Proclus) in his preface to the Data to the commentaries of Pappus on the bookHeiberg, Euklid-Studien, p. 173; Euclid's Data, edd. Menge, pp. 256, lii.. -

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The actual references to Pappus in Proclus are as follows:

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(1) On the Postulate (4) that all right angles are equal, Pappus is quoted as saying that the converse, viz. that all angles equal to a right angle are right, is not trueProclus, pp. 189, 190., since the angle included between the arcs of two semicircles which are equal, and have their diameters at right angles and terminating at one point, is equal to a right angle, but is not a right angle.

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(2) On the axioms Pappus is quoted as saying that, in addition to Euclid's axioms, others are on record as well (sunanagra/fesqai) about unequals added to equals and equals added to unequalsibid. p. 197, 6-10.; these, says Proclus, follow from the Euclidean axioms, while others given by Pappus are involved by the definitions, namely those which assert that all parts of the plane and of the straight line coincide with one another, - that a point divides a straight line, a line a surface, and a surface a solid, - and that the infinite is (obtained) in magnitudes both by addition and diminutionibid. p. 198, 3-15.. -

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(3) Pappus gave a pretty proof of 1. 5. This proof has, I think, been wrongly understood; on this point see my note on the proposition.

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(4) On 1. 47 Proclus saysProclus, p. 429, 9-15.: As the proof of the writer of the Elements is manifest, I think that it is not necessary to add anything further, but that what has been said is sufficient, since indeed those who have added more, like Heron and Pappus, were obliged to make use of what is proved in the sixth book, without attaining any important result. - We shall see what Heron's addition consisted of; what Pappus may have added we do not know, unless it was something on the lines of his extension of 1. 47 found in the Synagoge (IV. p. 176, ed. Hultsch).

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We may fairly conclude, with van PeschVan Pesch, De Procli fontibus, p. 134 sqq., that Pappus is drawn upon in various other passages of Proclus where he quotes no authority, but where the subject-matter reminds us of other notes expressly assigned to Pappus or of what we otherwise know to have been favourite questions with him. Thus:

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1. We are reminded of the curvilineal angle which is equal to but not a right angle by the note on 1. 32 to the effect that the converse (that a figure with its interior angles together equal to two right angles is a triangle) is not true unless we confine ourselves to rectilineal figures. This statement is supported by reference to a figure formed by four semicircles whose diameters form a square, and one of which is turned inwards while the others are turned outwards. The figure forms two angles equal to - right angles in the sense described by Pappus on Post. 4, while the other curvilineal angles are not considered to be angles at all, and are left out in summing the internal angles. Similarly the allusions in the notes on 1. 4, 23 to curvilineal angles of which certain moon-shaped angles (mhnoeidei=s) are shown to be equal to - rectilineal angles savour of Pappus.

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2. On 1. 9 Proclus saysProclus, p. 272, 10. that Others, starting from the Archimedean spirals, divided any given rectilineal angle in any given ratio. - We cannot but compare this with Pappus iv. p. 286, where the spiral is so used; hence this note, including remarks immediately preceding about the conchoid and the quadratrix, which were used for the same purpose, may very well be due to Pappus.

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3. The subject of isoperimetric figures was a favourite one with Pappus, who wrote a recension of Zenodorus' treatise on the subjectPappus, v. pp. 304-350; for Zenodorus' own treatise see Hultsch's Appendix, pp. 1189 —1211.. Now on I. 35 Proclus speaksProclus, pp. 396-8. about the paradox of parallelograms having equal area (between the same parallels) though the two sides between the parallels may be of any length, adding that of parallelograms with equal perimeter the rectangle is greatest if the base be given, and the square greatest if the base be not given etc. He returns to the subject on 1. 37 about trianglesibid. pp. 403-4.. Compareibid. pp. 236-7. also his note on 1. 4. These notes may have been taken from Pappus.

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4. Again, on 1. 21, Proclus remarks on the paradox that straight lines may be drawn from the base to a point within a triangle which are (1) together greater than the two sides, and (2) include a less angle, provided that the straight lines may be drawn from points in the base other than its extremities. The subject of straight lines satisfying condition (1) was treated at length, with reference to a variety of cases, by PappusPappus, 111. pp. 104-130., after a collection of paradoxes - by Erycinus, of whom nothing more is known. Proclus gives Pappus' first case, and adds a rather useless proof of the possibility of drawing straight lines satisfying condition (2) alone, adding that the proposition stated has been proved by me without using the parallels of the commentatorsProclus, p. 328, 15.. - By the commentators - Pappus is doubtless meant.

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5. Lastly, the four-sided triangle, - called by Zenodorus the hollow-angled, -Proclus, p. 165, 24; cf. pp. 328, 329. is mentioned in the notes on 1. Def. 24-29 and I. 21. As Pappus wrote on Zenodorus' work in which the term occurredSee Pappus, ed. Hultsch, pp. 1154, 1206., Pappus may be responsible for these notes.

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IV. Simplicius.

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According to the FihristFihrist (tr. Suter), p. 21., Simplicius the Greek wrote a commentary to the beginning of Euclid's book, which forms an introduction to geometry. - And in fact this commentary on the definitions, postulates and axioms (including the postulate known as the ParallelAxiom) is preserved in the Arabic commentary of an-NairĩzĩAn-Nairĩzĩ, ed. Besthorn-Heiberg, pp. 9-41, 119-133, ed. Curtze, pp. 1-37, 65-73. The Codex Leidensis, from which Besthorn and Heiberg's edition is taken, has unfortunately lost some leaves, so that there is a gap from Def. 1 to Def. 23 (parallels). The loss is, however, made good by Curtze's edition of the translation by Gherard of Cremona.. On two subjects this commentary of Simplicius quotes a certain .Aganis, - the first subject being the definition of an angle, and the second the definition of parallels and the parallel-postulate. Simplicius gives word for word, in a long passage placed by an-Nairīzī after 1. 29, an attempt by Aganis - to prove the parallel-postulate. It starts from a definition of parallels which agrees with Geminus' view of them as given by ProclusProclus, p. 177, 21., and is closely connected with the definition given by Posidoniusibid. p. 176, 7.. Hence it has been assumed that Aganis - is none other than Geminus, and the historical importance of the commentary of Simplicius has been judged accordingly. But it has been recently shown by Tannery that the identification of Aganis - with Geminus is practically impossibleBibliotheca Mathematica, 113, 1900, pp. 9-11. In the translation of Besthorn-Heiberg Aganis is called by Simplicius in one place philosophus Aganis, - in another magister noster Aganis, - in Gherard's version he is socius Aganis - and socius noster Aganis. - These expressions seem to leave no doubt that Aganis was a contemporary and friend, if not master, of Simplicius; and it is impossible to suppose that Simplicius (fl. about 500 A.D.) could have used them of a man who lived four and a half centuries before his time. A phrase in Simplicius' word-forword quotation from Aganis leads to the same conclusion. He speaks of people who objected even in ancient times - (iam antiquitus) to the use by geometers of this postulate. This would not have been an appropriate phrase had Geminus been the writer. I do not think that this difficulty can he got over by Suter's suggestionZeitschrift für Math. u. Physik, XLIV., hist.-litt. Abth. p. 61. that the passages in question may have been taken out of Heron's commentary, and that an-Nairĩzĩ may have forgotten to name the author; it seems clear that Simplicius is the person who described Aganis. - Hence we are driven to suppose that Aganis was not Geminus, but some unknown contemporary of SimpliciusThe above argument seems to me quite insuperable. The other arguments of Tannery do not, however, carry conviction to my mind. I do not follow the reasoning based on Aganis' definition of an angle. It appears to me a pure assumption that Geminus would have seen that Posidonius' definition of parallels was not admissible. Nor does it seem to me to count for much that Proclus, while telling us that Geminus held that the postulate ought to be proved and warned the unwary against hastily concluding that two straight lines approaching one another must necessarily meet (cf. a curve and its asymptote), gives no hint that Geminus did try to prove the postulate. It may well be that Proclus omitted Geminus' proof - (if he wrote one) because he preferred Ptolemy's attempt which he gives (pp. 365-7). Considerable interest will however continue to attach to the comments of Simplicius so fortunately preserved.

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Proclus tells us that one Aegaeas (? Aenaeas) of Hierapolis wrote an epitome of the ElementsProclus, p. 361, 21.; but we know nothing more of him or of it.

- - - - CHAPTER IV. - PROCLUS AND HIS SOURCESMy task in this chapter is made easy by the appearance, in the nick of time, of the dissertation De Procli fontibus by J. G. van Pesch (Lugduni-Batavorum, Apud L. van Nifterik, mdcccc). The chapters dealing directly with the subject show a thorough acquaintance on the part of the author with all the literature bearing on it; he covers the whole field and he exercises a sound and sober judgment in forming his conclusions. The same cannot always be said of his only predecessor in the same inquiry, Tannery (in La Géométrie grecque, 1887), who often robs his speculations of much of their value through his proneness to run away with an idea; he does so in this case, basing most of his conclusions on an arbitrary and unwarranted assumption as to the significance of the words oi( peri/ tina (e.g. *h(/rwna, *poseidw/nion etc.) as used in Proclus.. -

It is well known that the commentary of Proclus on Eucl. Book I. is one of the two main sources of information as to the history of Greek geometry which we possess, the other being the Collection of Pappus. They are the more precious because the original works of the forerunners of Euclid, Archimedes and Apollonius are lost, having probably been discarded and forgotten almost immediately after the appearance of the masterpieces of that great trio.

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Proclus himself lived 410-485 A.D., so that there had already passed a sufficient amount of time for the tradition relating to the pre-Euclidean geometers to become obscure and defective. In this connexion a passage is quoted from SimpliciusSimplicius on Aristotle's Physics, ed. Diels, pp. 54-69. who, in his account of the quadrature of certain lunes by Hippocrates of Chios, while mentioning two authorities for his statements, Alexander Aphrodisiensis (about 220 A.D.) and Eudemus, says in one placeibid. p. 68, 32., As regards Hippocrates of Chios we must pay more attention to Eudemus, since he was nearer the times, being a pupil of Aristotle. -

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The importance therefore of a critical examination of Proclus' commentary with a view to determining from what original sources he drew need not be further emphasised.

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Proclus received his early training in Alexandria, where Olympiodorus was his instructor in the works of Aristotle, and mathematics was taught him by one HeronCf. Martin, Recherches sur la vie et les ouvrages d'Héron d'Alexandrie, pp. 240-2. (of course a different Heron from the mechanicus Hero - of whom we have already spoken). He afterwards went to Athens where he was imbued by Plutarch, and by Syrianus, with the Neo-Platonic philosophy, to which he then devoted heart and soul, becoming one of its most prominent exponents. He speaks everywhere with the highest respect of his masters, and was in turn regarded with extravagant veneration by his contemporaries, as we learn from Marinus his pupil and biographer. On the death of Syrianus he was put at the head of the Neo-Platonic school. He was a man of untiring industry, as is shown by the number of books which he wrote, including a large number of commentaries, mostly on the dialogues of Plato. He was an acute dialectician, and pre-eminent among his contemporaries in the range of his learningZeller calls him Der Gelehrte, dem kein Feld damaligen Wissens verschlossen ist. -; he was a competent mathematician; he was even a poet. At the same time he was a believer in all sorts of myths and mysteries and a devout worshipper of divinities both Greek and Oriental.

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Though he was a competent mathematician, he was evidently much more a philosopher than a mathematicianVan Pesch observes that in his commentaries on the Tïmaeus (pp. 671-2) he speaks as no real mathematician could have spoken. In the passage referred to the question is whether the sun occupies a middle place among the planets. Proclus rejects the view of Hipparchus and Ptolemy because o( qeourgo/s - (sc. the Chaldean, says Zeller) thinks otherwise, whom it is not lawful to disbelieve. - Martin says rather neatly, Pour Proclus, les Éléments d'Euclide ont l'heureuse chance de n'e=tre contredits ni par les Oracles chaldaïques, ni par les spéculations des pythagoriciens anciens et nouveaux...... -. This is shown even in his commentary on Eucl. I., where, not only in the Prologues (especially the first), but also in the notes themselves, he seizes any opportunity for a philosophical digression. He says himself that he attaches most importance to the things which require deeper study and contribute to the sum of philosophyProclus, p. 84, 13. -; alternative proofs, cases, and the like (though he gives many) have no attraction for him; and, in particular, he attaches no value to the addition of Heron to 1. 47ibid. p. 429, 12., which is of considerable mathematical interest. Though he esteemed mathematics highly, it was only as a handmaid to philosophy. He quotes Plato's opinion to the effect that mathematics, as making use of hypotheses, falls short of the non-hypothetical and perfect scienceibid. p. 31, 20. -... Let us then not say that Plato excludes mathematics from the sciences, but that he declares it to be secondary to the one supreme scienceibid. p. 32, 2.. - And again, while mathematical science must be considered desirable in itself, though not with reference to the needs of daily life, - if it is necessary to refer the benefit arising from it to something else, we must connect that benefit with intellectual knowledge (noera\n gnw=sin), to which it leads the way and is a propaedeutic, clearing the eye of the soul and taking away the impediments which the senses place in the way of the knowledge of universals (tw=n o(/lwn)ibid. p. 27, 27 to 28. 7; cf. also p. 21, 25, pp. 46, 47.. -

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We know that in the Neo-Platonic school the younger pupils learnt mathematics; and it is clear that Proclus taught this subject, and that this was the origin of the commentary. Many passages show him as a master speaking to scholars. Thus we have illustrated and made plain all these things in the case of the first problem, but it is necessary that my hearers should make the same inquiry as regards the others as wellProclus, p. 210, 18., - and I do not indicate these things as a merely incidental matter but as preparing us beforehand for the doctrine of the Timaeusibid. p. 384, 2.. - Further, the pupils whom he was addressing were beginners in mathematics; for in one place he says that he omits for the present - to speak of the discoveries of those who employed the curves of Nicomedes and Hippias for trisecting an angle, and of those who used the Archimedean spiral for dividing an angle in any given ratio, because these things would be too difficult for beginners (dusqewrh/tous toi=s ei)sagome/nois)ibid. p. 272, 12.. Again, if his pupils had not been beginners, it would not have been necessary for Proclus to explain what is meant by saying that sides subtend certain anglesibid. p. 238, 12., the difference between adjacent and uertical anglesibid. p. 298, 14. etc., or to exhort them, as he often does, to work out other particular cases for themselves, for practice (gumnasi/as e(/neka)Cf. p. 224, 15 (on 1. 2)..

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The commentary seems then to have been founded on Proclus' lectures to beginners in mathematics. But there are signs that it was revised and re-edited for a larger public; thus he gives notice in one placeibid. p. 84, 9. to those who shall come upon - his work (toi=s e)nteucome/nois). There are also passages which could not have been understood by the beginners to whom he lectured, e.g. passages about the cylindrical helixibid. p. 105., conchoids and cissoidsibid. p. 113.. These passages may have been added in the revised edition, or, as van Pesch conjectures, the explanations given in the lectures may have been much fuller and more comprehensible to beginners, and they may have been shortened on revision.

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In his comments on the propositions of Euclid, Proclus generally proceeds in this way: first he gives explanations regarding Euclid's proofs, secondly he gives a few different cases, mainly for the sake of practice, and thirdly he addresses himself to refuting objections raised by cavillers to particular propositions. The latter class of note he deems necessary because of sophistical cavils - and the attitude of the people who rejoiced in finding paralogisms and in causing annoyance to scientific menibid. p. 375, 8.. His commentary does not seem to have been written for the purpose of correcting or improving Euclid. For there are very few passages of mathematical content in which Proclus can be supposed to be propounding anything of his own; nearly all are taken from the works of others, mostly earlier commentators, so that, for the purpose of improving on or correcting Euclid, there was no need for his commentary at all. Indeed only in one place does he definitely bring forward anything of his own to get over a difficulty which he finds in Euclidibid. pp. 368-373.; this is where he tries to prove the parallel-postulate, after first giving Ptolemy's attempt and then pointing out objections to it. On the other hand, there are a number of passages in which he extols Euclid; thriceProclus, p. 280, 9; p. 282, 20; pp. 335, 336. also he supports Euclid against Apollonius where the latter had given proofs which he considered better than Euclid's (1. 10, 11, and 23).

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Allusion must be made to the debated question whether Proclus continued his commentaries beyond Book 1. His intention to do so is clear from the following passages. Just after the words above quoted about the trisection etc. of an angle by means of certain curves he says, For we may perhaps more appropriately examine these things on the third book, where the writer of the Elements bisects a given circumferenceibid. p. 272, 14.. - Again, after saying that of all parallelograms which have the same perimeter the square is the greatest and the rhomboid least of all, - he adds: But this we will prove in another place; for it is more appropriate to the (discussion of the) hypotheses of the second bookibid. p. 398, 18.. - Lastly, when alluding (on I. 45) to the squaring of the circle, and to Archimedes' proposition that any circle is equal to the right-angled triangle in which the perpendicular is equal to the radius of the circle and the base to its perimeter, he adds, But of this elsewhereibid. p. 423, 6. -; this may imply an intention to treat of the subject on Eucl. XII., though Heiberg doubts itHeiberg, Euklid-Studien, p. 165, note.. But it is clear that, at the time when the commentary on Book 1. was written, Proclus had not yet begun to write on the other Books and was uncertain whether he would be able to do so: for at the end he saysProclus, p. 432, 9., For my part, if I should be able to discuss the other booksThe words in the Greek are: ei) me\n dunhqei/hmen kai\ toi=s loipoi=s to\n au)to\n tro/pon e)celqei=n. For e)celqei=n Heiberg would read e)peclqei=n in the same manner, I should give thanks to the gods; but, if other cares should draw me away, I beg those who are attracted by this subject to complete the exposition of the other books as well, following the same method, and addressing themselves throughout to the deeper and better defined questions involved - (to\ pragmateiw=des pantaxou= kai\ eu)diai/reton metadiw/kontas).

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There is in fact no satisfactory evidence that Proclus did actually write any more commentaries than that on Book 1.True, a Vatican Ms. has a collection of scholia on Books I. (extracts from the extant commentary of Proclus), II., V., VI., X. headed *ei)s ta\ *eu)klei/dou stoixei=a prolambano/mena e)k tw=n *pro/klou .spora/dhn kai\ kat) e)pitomh/n. Heiberg holds that this title itself suggests that the authorship of Proclus was limited to the scholia on Book I.; for prolambano/mena e)k tw=n *pro/klou suits extracts from Proclus' prologues, but hardly scholia to later Books. Again, a certain scholium (Heiberg in Hermes XXXVIII., 1903, p. 341, No. 17) purports to quote words from the end of a scholium of Proclus - on X. 9. The words quoted are from the scholium X. No. 62, one of the Scholia Vaticana. But none of the other, older, sources connect Proclus' name with X. No. 62; it is probable therefore that a Byzantine, who had in his Ms. of Euclid the collection of Schol. Vat. and knew that those on Book I. came from Proclus, himself attached Proclus' name to the others. The contrary view receives support from two facts pointed out by Heiberg, viz. (1) that the scholiast's copy of Proclus was not so much better than our MSS. as to suggest that the scholiast had further commentaries of Proclus which have vanished for usWhile one class of scnona (Schol. Vat.) have some better readings than our MSS. of Proclus have, and partly fill up the gaps at 1. 36, 37 and 1. 41-43, the other class (Schol. Vind.) derive from an inferior Proclus MS. which also had the same lacunae.; (2) that there is no trace in the scholia of the notes which Proclus promised in the passages quoted above.

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Coming now to the question of the sources of Proclus, we may say that everything goes to show that his commentary is a compilation, though a compilation in the better sense - of the termKnoche, Untersuchungen über des Proklus Diadochus Commentar zu Euklid's Elementen (1862), p. 11.. He does not even give us to understand that we shall find in it much of his own; let us, - he says, now turn to the exposition of the theorems proved by Euclid, selecting the more subtle of the comments made on them by the ancient writers, and cutting down their interminable diffuseness...Proclus, p. 200, 10-13. -: not a word about anything of his own. At the same time, he seems to imply that he will not necessarily on each occasion quote the source of each extract from an earlier commentary; and, in fact, while he quotes the name of his authority in many places, especially where the subject is important, in many others, where it is equally certain that he is not giving anything of his own, he mentions no authority. Thus he quotes Heron by name six times; but we now know, from the commentary of an-Nairīzī, that a number of other passages, where he mentions no name, are taken from Heron, and among them the not unimportant addition of an alternative proof to 1. 19. Hence we can by no means conclude that, where no authority is mentioned, Proclus is giving notes of his own. The presumption is génerally the other way; and it is often possible to arrive at a conclusion, either that a particular note is not Proclus' own, or that it is definitely attributable to someone else, by applying the ordinary principles of criticism. Thus, where the note shows an unmistakable affinity to another which Proclus definitely attributes to some commentator by name, especially when both contain some peculiar and distinctive idea, we cannot have much doubt in assigning both to the same commentatorInstances of the application of this criterion will be found in the discussion of Proclus' indebtedness to the commentaries of Heron, Porphyry and Pappus.. Again, van Pesch finds a criterion in the form of a note, where the explanation is so condensed as to be only just intelligible; the note is that in which a converse of 1. 32 is provedVan Pesch attributes this converse and proof to Pappus, arguing from the fact that the proof is followed by a passage which, on comparison with Pappus' note on the postulate that all right angles are equal, he feels justified in assigning to Pappus. I doubt if the evidence is sufficient. the proposition namely that a rectilineal figure which has all its interior angles together equal to two right angles is a triangle.

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It is not safe to attribute a passage to Proclus himself because he uses the first person in such expressions as I say - or I will prove - —for he was in the habit of putting into his own words the substance of notes borrowed from others—nor because, in speaking of an objection raised to a particular proposition, he uses such expressions as perhaps someone may object - (i)/sws d) a)/n tines e)nstai=en...): for sometimes other words in the same passage indicate that the objection had actually been taken by someoneVan Pesch illustrates this by an objection refuted in the note on 1. 9, p. 273, 11 sqq. After using the above expression to introduce the objection, Proclus uses further on (p. 273, 25) the term they say -(fasi/n).. Speaking generally, we shall not be justified in concluding that Proclus is stating something new of his own unless he indicates this himself in express terms.

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As regards the form of Proclus' references to others by name, van Pesch notes that he very seldom mentions the particular work from which he is borrowing. If we leave out of account the references to Plato's dialogues, there are only the following references to books: the Bacchae of PhilolausProclus, p. 22, 15., the Symmikta of Porphyryibid. p. 56, 25., Archimedes On the Sphere and Cylinderibid. p. 71, 18., Apollonius On the cochliasibid. p. 105, 5., a book by Eudemus on The Angleibid. p. 125, 8., a whole book of Posidonius directed against Zeno of the Epicurean sectibid. p. 200, 2., Carpus' Astronomyibid. p. 241, 19., Eudemus' History of Geometryibid. p. 352, 15., and a tract by Ptolemy on the parallel-postulateibid. p. 362, 15..

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Again, Proclus does not always indicate that he is quoting something at second-hand. He often does so, e.g. he quotes Heron as the authority for a statement about Philippus, Eudemus as attributing a certain theorem to Oenopides etc.; but he says on 1. 12 that Oenopides first investigated this problem, thinking it useful for astronomy - when he cannot have had Oenopides' work before him.

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It has been said above that Proclus was in the habit of stating in his own words the substance of the things which he borrowed. We are prepared for this when we find him stating that he will select the best things from ancient commentaries and cut short their interminable diffuseness, - that he will briefly describe - (suntomws i(storh=sai) the other proofs of 1. 20 given by Heron and Porphyry and also the proofs of 1. 25 by Menelaus and Heron. But the best evidence is of course to be found in the passages where he quotes works still extant, e.g. those of Plato, Aristotle and Plotinus. Examination of these passages shows great divergences from the original; even where he purports to quote textually, using the expressions Plato says, - or Plotinus says, - he by no means quotes word for wordSee the passages referred to by van Pesch (p. 70). The most glaring case is a passage (p. 21, 19) where he quotes Plotinus, using the expression Plotinus says...... - Comparison with Plotinus, Ennead. 1. 3, 3, shows that very few words are those of Plotinus himself; the rest represent Plotinus' views in Proclus' own language.. In fact, he seems to have had a positive distaste for quoting textually from other works. He cannot conquer this even when quoting from Euclid; he says in his note on 1. 22, we will follow the words of the geometer - but fails, nevertheless, to reproduce the text of Euclid unchangedProclus, P. 330, 19 sqq.

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We now come to the sources themselves from which Proclus drew in writing his commentary. Three have already been disposed of, viz. Heron, Porphyry and Pappus, who had all written commentaries on the ElementsSee pp. 20 to 27 above.. We go on to

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Eudemus, the pupil of Aristotle, who, among other works, wrote a history of arithmetic, a history of astronomy, and a history of geometry. The importance of the last mentioned work is attested by the frequent use made of it by ancient writers. That there was no other history of geometry written after the time of Eudemus seems to be proved by the remark of Proclus in the course of his famous summary: Those who compiled histories bring the development of this science up to this point. Not much younger than these is EuclidProclus, p. 68, 4-7..... - The loss of Eudemus' history is one of the gravest which fate has inflicted upon us, for it cannot be doubted that Eudemus had before him a number of the actual works of earlier geometers, which, as before observed, seem to have vanished completely when they were superseded by the treatises of Euclid, Archimedes and Apollonius. As it is, we have to be thankful for the fragments from Eudemus which such writers as Proclus have preserved to us.

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I agree with van PeschDe Procli fontibus, pp. 73-75. that there is no sufficient reason for doubting that the work of Eudemus was accessible to Proclus at first hand. For the later writers Simplicius and Eutocius refer to it in terms such as leave no room for doubt that they had it before them. I have already quoted a passage from Simplicius' account of the lunes of Hippocrates to the effect that Eudemus must be considered the best authority since he lived nearer the timesSee above, p. 29.. In the same place Simplicius saysSimplicius, loc. cit., ed. Diels, p. 60, 27., I will set out what Eudemus says word for word (kata\ le/cin lego/mena), adding only a little explanation in the shape of reference to Euclid's Elements owing to the memorandum-like style of Eudemus (dia\ to\n u(pomnhmatiko\n tro/pon tou= *eu)dh/mou) who sets out his explanations in the abbreviated form usual with ancient writers. Now in the second book of the history of geometry he writes as followsSimplicius, loc. cit., ed. Diels, p. 60, 27.. - It is not possible to suppose that Simplicius would have written in this way about the style of Eudemus if he had merely been copying certain passages second-hand out of some other author and had not the original work itself to refer to. In like manner, Eutocius speaks of the paralogisms handed down in connexion with the attempts of Hippocrates and Antiphon to square the circleArchimedes, ed. Heiberg, vol. III. p. 228., with which I imagine that those are accurately acquainted who have examined (e)peskemme/nous) the geometrical history of Eudemus and know the Ceria Aristotelica. - How could the contemporaries of Eutocius have examined the work of Eudemus unless it was still extant in his time?

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The passages in which Proclus quotes Eudemus by name as his authority are as follows:

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(1) On 1. 26 he says that Eudemus in his history of geometry referred this theorem to Thales, inasmuch as it was necessary to Thales' method of ascertaining the distance of ships from the shoreProclus, p. 352, 14-18..

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(2) Eudemus attributed to Thales the discovery of Eucl. 1. 15ibid. p. 299, 3., and

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(3) to Oenopides the problem of 1. 23ibid. p. 333, 5..

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(4) Eudemus referred the discovery of the theorem in 1. 32 to the Pythagoreans, and gave their proof of it, which Proclus reprodueesibid. p. 379, 1-16..

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(5) On 1. 44 Proclus tells usibid. p. 419, 15-18. that Eudemus says that these things are ancient, being discoveries of the Pythagorean muse, the application (parabolh/) of areas, their exceeding (u(perbolh/) and their falling short (e)/lleiyis). - The next words about the appropriation of these terms (parabola, hyperbola and ellipse) by later writers (i.e. Apollonius) to denote the conic sections are of course not due to Eudemus.

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Coming now to notes where Eudemus is not named by Proclus, we may fairly conjecture, with van Pesch, that Eudemus was really the authority for the statements (1) that Thales first proved that a circle is bisected by its diameteribid. p. 157, 10, 11. (though the proof by reductio ad absurdum which follows in Proclus cannot be attributed to ThalesCantor (Gesch. d. Math. 13, p. 221) points out the connexion between the reductio ad absurdum and the analytical method said to have been discovered by Plato. Proclus gives the proof by reductio ad absurdum to meet an imaginary critic who desires a mathematical proof; possibly Thales may have been satisfied with the argument in the same sentence which mentions Thales, the cause of the bisection being the unswerving course of the straight line through the centre. -), (2) that Plato made over to Leodamas the analytical method, by means of which it is recorded (i(sto/rhtai) that the latter too made many discoveries in geometryProclus, p. 211, 19-23., - (3) that the theorem of 1. 5 was due to Thales, and that for equal angles he used the more archaic expression similar - anglesibid. p. 250, 20., (4) that Oenopides first investigated the problem of 1. 12, and that he called the perpendicular the gnomonic line (kata\ gnw/mona)ibid. p. 283, 7-10., (5) that the theorem that only three sorts of polygons can fill up the space round a point, viz. the equilateral triangle, the square and the regular hexagon, was Pythagoreanibid. pp. 304, 11-305, 3.. Eudemus may also be the authority for Proclus' description of the two methods, referred to Plato and Pythagoras respectively, of forming right-angled triangles in whole numbersibid. pp. 428, 7-429, 9..

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We cannot attribute to Eudemus the beginning of the note on 1. 47 where Proclus says that if we listen to those who like to recount ancient history, we may find some of them referring this theorem to Pythagoras and saying that he sacrificed an ox in honour of his discoveryibid. p. 426, 6-9.. - As such a sacrifice was contrary to the Pythagorean tenets, and Eudemus could not have been unaware of this, the story cannot rest on his authority. Moreover Proclus speaks as though he were not certain of the correctness of the tradition; indeed, so far as the story of the sacrifice is concerned, the same thing is told of Thales in connexion with his discovery that the angle in a semicircle is a right angleDiogenes Laertius, I. 24, p. 6, ed. Cobet., and Plutarch is not certain whether the ox was sacrificed on the discovery of I. 47 or of the problem about application of areasPlutarch, non posse suaviter vivi secundum Epicurum, 11; Symp. VIII, 2.. Plutarch's doubt suggests that he knew of no evidence for the story beyond the vague allusion in the distich of Apollodorus Logisticus - (the calculator -) cited by Diogenes Laertius alsoDiog. Laert. VIII. 12, p. 207, ed. Cobet: *(hni/ka *puqago/rhs to\ periklee\s eu(/reto gra/mma, kei=n) e)f) o)/w| kleinh\n h)/gage bouqusi/hn. See on this subject Tannery, La Géoinétrie grecque, p. 105.; and Proclus may have had in mind this couplet with the passages of Plutarch.

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We come now to the question of the famous historical summary given by ProclusProclus, pp. 64-70.. No one appears to maintain that Eudemus is the author of even the early part of this summary in the form in which Proclus gives it. It is, as is well known, divided into two distinct parts, between which comes the remark, Those who compiled historiesThe plural is well explained by Tannery, La Géoinétrie grecque, pp. 73, 74. No doubt the author of the summary tried to supplement Eudemus by means of any other histories which threw light on the subject. Thus e.g. the allusion (p. 64, 21) to the Nile recalls Herodotus. Cf. the expression in Proclus, p. 64, 19, para\ tw=n pollw=n i/sto/rhtai. bring the development of this science up to this point. Not much younger than these is Euclid, who put together the Elements, collecting many of the theorems of Eudoxus, perfecting many others by Theaetetus, and bringing to irrefragable demonstration the things which had only been somewhat loosely proved by his predecessors. - Since Euclid was later than Eudemus, it is impossible that Eudemus can have written this. Yet the style of the summary after this point does not show any such change from that of the former portion as to suggest different authorship. The author of the earlier portion recurs frequently to the question of the origin of the elements of geometry in a way in which no one would be likely to do who was not later than Euclid; and it must be the same hand which in the second portion connects Euclid's Elements with the work of Eudoxus and TheaetetusTannery, La Géométrie grecque, p. 75..

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If then the summary is the work of one author, and that author not Eudemus, who is it likely to have been? Tannery answers that it is Geminusibid. pp. 66-75.; but I think, with van Pesch, that he has failed to show why it should be Geminus rather than another. And certainly the extracts which we have from Geminus' work suggest that the sort of topics which it dealt with was quite different; they seem rather to have been general questions of the content of mathematics, and even Tannery admits that historical details could only have come incidentally into the workibid. p. 19..

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Could the author have been Proclus himself? Circumstances which seem to suggest this possibility are (1) that, as already stated, the question of the origin of the Elements is kept prominent, (2) that there is no mention of Democritus, whom Eudemus would not be likely to have ignored, while a follower of Plato would be likely enough to do him the injustice, following the example of Plato who was an opponent of Democritus, never once mentions him, and is said to have wished to burn all his writingsDiog. Laertius, IX. 40, p. 237, ed. Cobet., and (3) the allusion at the beginning to the inspired Aristotle - (o( daimonios *)aristote/lhs)Proclus, p. 64, 8., though this may easily have been inserted by Proclus in a quotation made by him from someone else. On the other hand there are considerations which suggest that Proclus himself was not the writer. (1) The style of the whole passage is not such as to point to him as the author. (2) If he wrote it, it is hardly conceivable that he would have passed over in silence the discovery of the analytical method, the invention of Plato to which he attached so much importanceProclus, p. 211, 19 sqq.; the passage is quoted above, p. 36..

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There is nothing improbable in the conjecture that Proclus quoted the summary from a compendium of Eudemus' history made by some later writer: but as yet the question has not been definitely settled. All that is certain is that the early part of the summary must have been made up from scattered notices found in the great work of Eudemus.

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Proclus refers to another work of Eudemus besides the history, viz. a book on The Angle (bibli/on peri\ gwni/as)ibid. p. 125, 8.. Tannery assumes that this must have been part of the history, and uses this assumption to confirm his idea that the history was arranged according to subjects, not according to chronological orderTannery, La Géométrie grecque, p. 26.. The phraseology of Proclus however unmistakably suggests a separate work; and that the history was chronologically arranged seems to be clearly indicated by the remark of Simplicius that Eudemus also counted Hippocrates among the more ancient writers - (e)n toi=s palaiote/rois)Simplicius, ed. Diels, p. 69, 23..

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The passage of Simplicius about the lunes of Hippocrates throws considerable light on the style of Eudemus' history. Eudemus wrote in a memorandum-like or summary manner (to\n u(pomnhmatiko\n tro/pon tou= *eu)dh/mou)ibid. p. 60, 29. when reproducing what he found in the ancient writers; sometimes it is clear that he left out altogether proofs or constructions of things by no means easyCf. Simplicius, p. 63, 19 sqq.; p. 64, 25 sqq.; also Usener's note de supplendis Hippocratis quas omisit Eudemus constructionibus - added to Diels' preface, pp. XXIII—XXVI..

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- Geminus. -

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The discussions about the date and birthplace of Geminus form a whole literature, as to which I must refer the reader to Manitius and TittelManitius, Gemini elementa astronomiae (Teubner, 1898), pp. 237-252; Tittel, art. Geminos - in Pauly-Wissowa's Real-Encyclopädie der classischen Altertumswissenschaft, vol. VII.. 1910.. Though the name looks like a Latin name (Gemǐnus), Manitius concluded that, since it appears as *gemi=nos in all Greek MSS. and as *gemei=nos in some inscriptions, it is Greek and possibly formed from gem as *)ergi=nos is from e)rg and *)aleci=nos from a)lec (cf. also *)ikti=nos, *krati=nos). Tittel is equally positive that it is Gemǐnus and suggests that *gemi=nos is due to a false analogy with *)aleci=nos etc. and *gemei=nos wrongly formed on the model of *)antwnei=nos, *)agrippei=na. Geminus, a Stoic philosopher, born probably in the island of Rhodes, was the author of a comprehensive work on the classification of mathematics, and also wrote, about 73-67 B.C., a not less comprehensive commentary on the meteorological textbook of his teacher Posidonius of Rhodes.

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It is the former work in which we are specially interested here. Though Proclus made great use of it, he does not mention its title, unless we may suppose that, in the passage (p. 177, 24) where, after quoting from Geminus a classification of lines which never meet, he says, these remarks I have selected from the filokali/a of Geminus, - filokali/a is a title or an alternative title. Pappus however quotes a work of Geminus on the classification of the mathematics - (e)n tw=| peri\ th=s tw=n maqhma/twn ta/cews)Pappus, ed. Hultsch, p. 1026, 9., while Eutocius quotes from the sixth book of the doctrine of the mathematics - (e)n tw=| e(/ktw| th=s tw=n maqhma/twn qewri/as)Apollonius, ed. Heiberg, vol. II. p. 170.. TanneryTannery, La Géométrie grecque, pp. 18, 19. pointed out that the former title corresponds well enough to the long extractProclus, pp. 38, 1-42, 8. which Proclus gives in his first prologue, and also to the fragments contained in the Anonymi variae collectiones published by Hultsch at the end of his edition of HeronHeron, ed. Hultsch, pp. 246, 16-249, 12.; but it does not suit most of the óther passages borrowed by Proclus. The correct title was therefore probably that given by Eutocius, The Doctrine, or Theory, of the Mathematics; and Pappus probably refers to one particular portion of the work, say the first Book. If the sixth Book treated of conics, as we may conclude from Eutocius, there must have been more Books to follow, because Proclus has preserved us details about higher curves, which must have come later. If again Geminus finished his work and wrote with the same fulness about the other branches of mathematics as he did about geometry, there must have been a considerable number of Books altogether. At all events it seems to have been designed to give a complete view of the whole science of mathematics, and in fact to be a sort of encyclopaedia of the subject.

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I shall now indicate first the certain, and secondly the probable, obligations of Proclus to Geminus, in which task I have only to follow van Pesch, who has embodied the results of Tittel's similar inquiry alsoVan Pesch, De Procli fontibus, pp. 97-113. The dissertation of Tittel is entitled De Gemini Stoici studiis mathematicis (1895).. I shall only omit the passages as regards which a case for attributing them to Geminus does not seem to me to have been made out.

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First come the following passages which must be attributed to Geminus, because Proclus mentions his name:

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(1) (In the first prologue of ProclusProclus, pp. 38, 1-42, 8, except the allusion in p. 41, 8-10, to Ctesibius and Heron and their pneumatic devices (qanmatopoii+kh/), as regards which Proclus' authority may be Pappus (VIII. p. 1024, 24-27) who uses very similar expressions. Heron, even if not later than Geminus, could hardly have been included in a historical work by him. Perhaps Geminus may have referred to Ctesibius only, and Proclus may have inserted and Heron - himself.) on the division of mathematical sciences into arithmetic, geometry, mechanics, astronomy, optics, geodesy, canonic (science of musical harmony), and logistic (apparently arithmetical problems);

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(2) (in the note on the definition of a straight line) on the classification of lines (including curves) as simple (straight or circular) and mixed, composite and incomposite, uniform (o(moiomerei=s) and non-uniform (a)nomoiomerei=s), lines about solids - and lines produced by cutting solids, including conic and spiric sectionsProclus, pp. 103, 21-107, 10; pp. 111, 1-113, 3.;

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(3) (in the note on the definition of a plane surface) on similar distinctions extended to surfaces and solidsibid. pp. 117, 14-120, 12, where perhaps in the passage pp. 117, 22-118, 23 we may have Geminus' own words.;

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(4) (in the note on the definition of parallels) on lines which do not meet (a)su/mptwtoi) but which are not on that account parallel, e.g. a curve and its asymptote, showing that the property of not meeting does not make lines parallel—a favourite observation of Geminus—and, incidentally, on bounded lines or those which enclose a figure and those which do notibid. pp. 176, 18-177, 25; perhaps also p. 175. The note ends with the words These things too we have selected from Geminus' *filokali/a for the elucidation of the matters in question. - Tannery (p. 27) takes these words coming at the end of the commentary on the definitions as referring to the whole of the portion of the commentary dealing with the definitions. Van Pesch properly regards them as only applying to the note on parallels. This seems to me clear from the use of the word too (tosau=ta kai/).;

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(5) (in the same note) the definition of parallels given by PosidoniusProclus, p. 176, 5-17.;

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(6) on the distinction between postulates and axioms, the futility of trying to prove axioms, as Apollonius tried to prove Axiom 1, and the equal incorrectness of assuming what really requires proof, as Euclid did in the fourth postulate [equality of right angles] and in the fifth postulate [the parallel-postulate]ibid. pp. 178-182, 4; pp. 183, 14-184, 10; cf. p. 188, 3-11. -;

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(7) on Postulates 1, 2, 3, which Geminus makes depend on the idea of a straight line being described by the motion of a pointibid. p. 185, 6-25.;

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(8) (in the note on Postulate 5) on the inadmissibility in geometry of an argument which is merely plausible, and the danger in this particular case owing to the existence of lines which do converge ad infinitum and yet never meetibid. p. 192, 5-29.;

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(9) (in the note on I. 1) on the subject-matter of geometry, theorems, problems and diorismoi/ (conditions of possibility) for problemsibid. pp. 200, 21-202, 25.;

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(10) (in the note on 1. 5) on a generalisation of 1. 5 by Geminus through the substitution for the rectilineal base of one uniform line (curve), - by means of which he proved that the only uniform lines - (alike in all their parts) are a straight line, a circle, and a cylindrical helixProclus, p. 251, 2-11.;

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(11) (in the note on I. 10) on the question whether a line is made up of indivisible parts (a)merh=), as affecting the problem of bisecting a given straight lineibid. pp. 277, 25-279, 11.;

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(12) (in the note on I. 35) on topical, or locus-theoremsibid. pp. 394, 11-395, 2 and p. 395, 13-21., where the illustration of the equal parallelograms described between a hyperbola and its asymptotes may also be due to Geminusibid. p. 395, 8-12..

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Other passages which may fairly be attributed to Geminus, though his name is not mentioned, are the following:

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(1) in the prologue, where there is the same allusion as in the passage (8) above to a remark of Aristotle that it is equally absurd to expect scientific proofs from a rhetorician and to accept mere plausibilities from a geometeribid. pp. 33, 21-34, 1.;

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(2) a passage in the prologue about the subject-matter, methods, and bases of geometry, the latter including axioms and postulatesibid. pp. 57, 9-58, 3.;

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(3) another on the definition and nature of elemenisibid. pp. 72, 3-75, 4.;

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(4) a remark on the Stoic use of the term axiom for every simple statement (a)po/fansis a(plh=)ibid. p. 77, 3-6.;

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(5) another discussion on theorems and problemsibid. pp. 77, 7-78, 13, and 79, 3-81, 4. in the middle of which however there are some sentences by Proclus himselfibid. pp. 78, 13-79, 2..

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(6) another passage, in connexion with Def. 3, on lines including or not including a figure (with which cf. part of the passage (4) above)ibid. pp. 102, 22-103, 18.;

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(7) a classification of different sorts of angles according as they are contained by simple or mixed lines (or curves)ibid. pp. 126, 7-127, 16.;

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(8) a similar classification of figuresibid. pp. 159, 12-160, 9., and of plane figuresibid. pp. 162, 27-164, 6.;

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(9) Posidonius' definition of a figureibid. p. 143, 5-11.;

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(10) a classification of triangles into seven kindsibid. p. 168, 4-12.;

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(11) a note distinguishing lines (or curves) producible indefinitely or not so producible, whether forming a figure or not forming a figure (like the single-turn spiral -)ibid. p. 187, 19-27.;

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(12) passages distinguishing different sorts of problemsibid. pp. 220, 7-222, 14; also p. 330, 6-9., different sorts of theoremsibid. pp. 244, 14-246, 12., and two sorts of converses (complete and partial)ibid. pp. 252, 5-254, 20.;

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(13) the definition of the term porism - as used in the title of Euclid's Porisms, as distinct from the other meaning of corollary -ibid. pp. 301, 21-302, 13.;

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(14) a note on the Epicurean objection to I. 20 as being obvious even to an assibid. pp. 322, 4-323, 3.;

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(15) a passage on the properties of parallels, with allusions to Apollonius' Conics, and the curves invented by Nicomedes, Hippias and PerseusPróclus, pp. 355, 20-356, 16.;

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(16) a passage on the parallel-postulate regarded as the converse of I. 17ibid. p. 364, 9-12; pp. 364, 20-365, 4..

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Of the authors to whom Proclus was indebted in a less degree the most important is Apollonius of Perga. Two passages allude to his Conicsibid. p. 71, 19; p. 356, 8, 6., one to a work on irrationalsibid. p. 74, 23, 24., and two to a treatise On the cochlias (apparently the cylindrical helix) by Apolloniusibid. pp. 105, 5, 6, 14, 15.. But more important for our purpose are six references to Apollonius in connexion with elementary geometry.

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(1) He appears as the author of an attempt to explain the idea of a line (possessing length but no breadth) by reference to daily experience, e.g. when we tell someone to measure, merely, the length of a road or of a wallibid. p. 100, 5-19.; and doubtless the similar passage showing how we may in like manner get a notion of a surface (without depth) is his alsoibid. p. 114, 20-25..

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(2) He gave a new general definition of an angleibid. p. 123, 15-19 (cf. p. 124, 17, p. 125, 17)..

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(3) He tried to prove certain axiomsibid. p. 183, 13, 14., and Proclus gives his attempt to prove Axiom I, word for wordibid. pp. 194, 25-195, 5..

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Proclus further quotes:

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(4) Apollonius' solution of the problem in Eucl. I. 10, avoiding Euclid's use of I. 9ibid. pp. 279, 16-280, 4.,

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(5) his solution of the problem in I. 11, differing only slightly from Euclid'sibid. p. 282, 8-19., and

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(6) his solution of the problem in I. 23ibid. pp. 335, 16-336, 5..

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HeibergPhilologus, vol. XLIII. p. 489. conjectures that Apollonius departed from Euclid's method in these propositions because he objected to solving problems of a more general, by means of problems of a more particular, character. Proclus however considers all three solutions inferior to Euclid's; and his remarks on Apollonius' handling of these elementary matters generally suggest that he was nettled by criticisms of Euclid in the work containing the things which he quotes from Apollonius, just as we conclude that Pappus was offended by the remarks of Apollonius about Euclid's incomplete treatment of the three- and four-line locusSee above, pp. 2, 3.. - If this was the case, Proclus can hardly have got his information about these things at second-hand; and there seems to be no reason to doubt that he had the actual work of Apollonius before him. This work may have been the treatise mentioned by Marinus in the words Apollonius in his general treatise - (*)apollw/nios e)n th=| kaqo/lou pragmatei/a|)Marinus in Euclidis Data, ed. Menge, p. 234, 16.. If the notice in the FihristFihrist, tr. Suter, p. 19. stating, on the authority of Thābit b. Qurra, that Apollonius wrote a tract on the parallel-postulate be correct, it may have been included in the same work. We may conclude generally that, in it, Apollonius tried to remodel the beginnings of geometry, reducing the number of axioms, appealing, in his definitions of lines, surfaces etc., more to experience than to abstract reason, and substituting for certain proofs others of a more general character.

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The probabilities are that, in quoting from the tract of Ptolemy in which he tried to prove the parallel-postulate, Proclus had the actual work before him. For, after an allusion to it as a certain bookProclus, p. 191, 23. - he gives two long extractsibid. pp. 362, 14-363, 18; pp. 365, 7-367, 27., and at the beginning of the second indicates the title of the tract, in the (book) about the meeting of straight lines produced from (angles) less than two right angles, - as he has very rarely done in other cases.

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Certain things from Posidonius are evidently quoted at secondhand, the authority being Geminus (e.g. the definitions of figure and parallels); but besides these we have quotations from a separate work which he wrote to controvert Zeno of Sidon, an Epicurean who had sought to destroy the whole of geometryibid. p. 200, 1-3.. We are told that Zeno had argued that, even if we admit the fundamental principles (a)rxai/) of geometry, the deductions from them cannot be proved without the admission of something else as well, which has not been included in the said principlesibid. pp. 199, 11-200, 1.. On I. 1 Proclus gives at some length the arguments of Zeno and the reply of Posidonius as regards this propositionibid. pp. 214, 18-215, 13; pp. 216, 10-218, 11.. In this case Zeno's something else - which he considers to be assumed is the fact that two straight lines cannot have a common segment, and then, as regards the proof - of it by means of the bisection of a circle by its diameter, he objects that it has been assumed that two circumferences (arcs) of circles cannot have a common part. Lastly, he makes up, for the purpose of attacking it, another supposed proof - of the fact that two straight lines cannot have a common part. Proclus appears, more than once, to be quoting the actual words of Zeno and Posidonius; in particular, two expressions used by Posidonius about the acrid Epicurean - (to\n drimu\n *)epikou/reion)ibid. p. 216, 21. and his misrepresentations - (*poseidw/nio/s fhsi to\n *zh/nwna sukofantei=n)ibid. p. 218, 1.. It is not necessary to suppose that Proclus had the original work of Zeno before him, because Zeno's arguments may easily have been got from Posidonius' reply; but he would appear to have quoted direct from the latter at all events.

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The work of Carpus - mechanicus (a treatise on astronomy) quoted from by Proclusibid. pp. 241, 19-243, 11. must have been accessible to him at first-hand, because a portion of the extract from it about the relation of theorems and problemsibid. pp. 242, 22-243, 11. is reproduced word for word. Moreover, if he were not using the book itself, Proclus would hardly be in a position to question whether the introduction of the subject of theorems and problems was opportune in the place where it was found (ei) me\n kata\ kairo\n h(\ mh/, parei/sqw pro\s to\ paro/n)Proclus, p. 241, 21, 22..

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It is of course evident that Proclus had before him the original works of Plato, Aristotle, Archimedes and Plotinus, as well as the *summikta/ of Porphyry and the works of his master Syrianus (o( h(me/teros kaqhgemw/n)ibid. p. 123, 19., from whom he quotes in his note on the definition of an angle. Tannery also points out that he must have had before him a group of works representing the Pythagorean tradition on its mystic, as distinct from its mathematical, side, from Philolaus downwards, and comprising the more or less apocryphal i(ero\s lo/gos of Pythagoras, the Oracles (lo/gia), and Orphic versesTannery, La Géométrie grecque, pp. 25, 26..

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Besides quotations from writers whom we can identify with more or less certainty, there are many other passages which are doubtless quoted from other commentators whose names we do not know. A list of such passages is given by van PeschVan Pesch, De Procli fontibus, p. 139., and there is no need to cite them here.

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Van Pesch also gives at the end of his workibid. p. 155. a convenient list of the books which, as the result of his investigation, he deems to have been accessible to and directly used by Proclus. The list is worth giving here, on the same ground of convenience. It is as follows: Eudemus: history of geometry. Geminus: the theory of the mathematical sciences. Heron: commentary on the Elements of Euclid. Porphyry: commentary on the Elements of Euclid. Pappus: commentary on the Elements of Euclid. Apollonius of Perga: a work relating to elementary geometry. Ptolemy: on the parallel-postulate. Posidonius: a book controverting Zeno of Sidon. Carpus: astronomy. Syrianus: a discussion on the angle. Pythagorean philosophical tradition. Plato's works. Aristotle's works. Archimedes' works. Plotinus: Enneades.

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Lastly we come to the question what passages, if any, in the commentary of Proclus represent his own contributions to the subject. As we have seen, the onus probandi must be held to rest upon him who shall maintain that a particular note is original on the part of Proclus. Hence it is not enough that it should be impossible to point to another writer as the probable source of a note; we must have a positive reason for attributing it to Proclus. The criterion must therefore be found either (1) in the general terms in which Proclus points out the deficiencies in previous commentaries and indicates the respects in which his own will differ from them, or (2) in specific expressions used by him in introducing particular notes which may indicate that he is giving his own views. Besides indicating that he paid more attention than his predecessors to questions requiring deeper study (to\ pragmateiw=des) and pursued clear distinctions - (to\ eu)diai/reton metadiw/kontas)Proclus, p. 84, 13, p. 432, 14, 15.— by which he appears to imply that his predecessors had confused the different departments of their commentaries, viz. lemmas, cases, and objections (e)nsta/seis)Cf. ibid. p. 289, 11-15; p. 432, 15-17.—Proclus complains that the earlier commentators had failed to indicate the ultimate grounds or causes of propositionsibid. p. 432, 17.. Although it is from Geminus that he borrowed a passage maintaining that it is one of the proper functions of geometry to inquire into causes (th\n ai)ti/an kai\ to\ dia\ ti/)ibid. p. 202, 9-25., yet it is not likely that Geminus dealt with Euclid's propositions one by one; and consequently, when we find Proclus, on I. 8, 16, 17, 18, 32, and 47See Proclus, p. 270, 5-24 (I. 8); pp. 309, 3-310, 8 (I. 16); pp. 310, 19-311, 23 (I. 17); pp. 316, 14-318, 2 (I. 18); p. 384, 13-21 (I. 32); pp. 426, 22-427, 8 (I. 47)., endeavouring to explain causes, we have good reason to suppose that the explanations are his own.

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Again, his remarks on certain things which he quotes from Pappus can scarcely be due to anyone else, since Pappus is the latest of the commentators whose works he appears to have used. Under this head come

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(1) his objections to certain new axioms introduced by PappusProclus, p. 198, 5-15.,

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(2) his conjecture as to how Pappus came to think of his alternative proof of I. 5ibid. p. 250, 12-19.,

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(3) an addition to Pappus' remarks about the curvilineal angle which is equal to a right angle without being oneibid. p. 190, 9-23..

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The defence of Geminus against Carpus, who combated his view of theorems and problems, is also probably due to Proclusibid. p. 243, 12-29., as well as an observation on I. 38 to the effect that I. 35-38 are really comprehended in VI. 1 as particular casesibid. pp. 405, 6-406, 9..

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Lastly, we can have no hesitation in attributing to Proclus himself (1) the criticism of Ptolemy's attempt to prove the parallel-postulateibid. p. 368, 1-23., and (2) the other attempted proof given in the same noteibid. pp. 371, 11-373, 2. (on I. 29) and assuming as an axiom that if from one point two straight lines forming an angle be produced ad infinitum the distance between them when so produced ad infinitum exceeds any finite magnitude (i.e. length), - an assumption which purports to be the equivalent of a statement in AristotleAristotle, de caelo, I. 5 (271 b 28-30).. It is introduced by words in which the writer appears to claim originality for his proof: To him who desires to see this proved (kataskeuazo/menon) let it be said by us (lege/sqw par) h(mw=n) - etc.Proclus, p. 371, 10. Moreover, Philoponus, in a note on Aristotle's Anal. post. I. 10, says that the geometer (Euclid) assumes this as an axiom, but it wants a great deal of proof, insomuch that both Ptolemy and Proclus wrote a whole book upon itBerlin Aristotle, vol. IV. p. 214 a 9-12.. -

- - - - CHAPTER V. - THE TEXTThe material for the whole of this chapter is taken from Heiberg's edition of the Elements, introduction to vol. v., and from the same scholar's Litterargeschichtliche Studien über Euklid, p. 174 sqq. and Paralipomena zu Euklid in Hermes, XXXVIII., 1903.. -

It is well known that the title of Simson's edition of Euclid (first brought out in Latin and English in 1756) claims that, in it, the errors by which Theon, or others, have long ago vitiated these books are corrected, and some of Euclid's demonstrations are restored -; and readers of Simson's notes are familiar with the phrases used, where anything in the text does not seem to him satisfactory, to the effect that the demonstration has been spoiled, or things have been interpolated or omitted, by Theon or some other unskilful editor. - Now most of the MSS. of the Greek text prove by their titles that they proceed from the recension of the Elements by Theon; they purport to be either from the edition of Theon - (e)k th=s *qe/wnos e)kdo/sews) or from the lectures of Theon - (a)po\ sunousiw=n tou= *qe/wnos). This was Theon of Alexandria (4th c. A.D.) who also wrote a commentary on Ptolemy, in which there occurs a passage of the greatest importance in this connexionI. p. 201 ed. Halma=p. 50 ed. Basel.: But that sectors in equal circles are to one another as the angles on which they stand has been proved by me in my edition of the Elements at the end of the sixth book. - Thus Theon himself says that he edited the Elements and also that the second part of VI. 33, found in nearly all the MSS., is his addition.

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This passage is the key to the whole question of Theon's changes in the text of Euclid; for, when Peyrard found in the Vatican the MS. 190 which contained neither the words from the titles of the other MSS. quoted above nor the interpolated second part of VI. 33, he was justified in concluding, as he did, that in the Vatican MS. we have an edition more ancient than Theon's. It is also clear that the copyist of P, or rather of its archetype, had before him the two recensions and systematically gave the preference to the earlier one; for at XIII. 6 in P the first hand has added a note in the margin: This theorem is not given in most copies of the new edition, but is found in those of the old. - Thus we are more fortunate than Simson, since our judgment of Theon's recension can be formed on the basis, not of mere conjecture, but of the documentary evidence afforded by a comparison of the Vatican MS. just mentioned with what we may conveniently call, after Heiberg, the Theonine MSS.

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The MSS. used for Heiberg's edition of the Elements are the following:

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(1) P = Vatican MS. numbered 190, 4to, in two volumes (doubtless one originally); 10th c.

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This is the MS. which Peyrard was able to use; it was sent from Rome to Paris for his use and bears the stamp of the Paris Imperial Library on the last page. It is well and carefully written. There are corrections some of which are by the original hand, but generally in paler ink, others, still pretty old, by several different hands, or by one hand with different ink in different places (P m. 2), and others again by the latest hand (P m. rec.). It contains, first, the Elements I.—XIII. with scholia, then Marinus' commentary on the Data (without the name of the author), followed by the Data itself and scholia, then the Elements XIV., XV. (so called), and lastly three books and a part of a fourth of a commentary by Theon ei)s tou\s proxei/rous kano/nas *ptolemai/ou.

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The other MSS. are Theonine. -

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(2) F = MS. XXVIII, 3, in the Laurentian Library at Florence, 4to; 10th c.

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This MS. is written in a beautiful and scholarly hand and contains the Elements I.—XV., the Optics and the Phaenomena, but is not well preserved. Not only is the original writing renewed in many places, where it had become faint, by a later hand of the 16th c., but the same hand has filled certain smaller lacunae by gumming on to torn pages new pieces of parchment, and has replaced bodily certain portions of the MS., which had doubtless become illegible, by fresh leaves. The larger gaps so made good extend from Eucl. VII. 12 to IX. 15, and from XII. 3 to the end; so that, besides the conclusion of the Elements, the Optics and Phaenomena are also in the later hand, and we cannot even tell what in addition to the Elements I.—XIII. the original MS. contained. Heiberg denotes the later hand by f and observes that, while in restoring words which had become faint and filling up minor lacunae the writer used no other MS., yet in the two larger restorations he used the Laurentian MS. XXVIII, 6, belonging to the 13th—14th c. The latter MS. (which Heiberg denotes by f) was copied from the Viennese MS. (V) to be described below.

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(3) B = Bodleian MS., D'Orville X. 1 inf. 2, 30, 4to; A.D. 888.

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This MS. contains the Elements I.—XV. with many scholia. Leaves 15-118 contain I. 14 (from about the middle of the proposition) to the end of Book VI., and leaves 123-387 (wrongly numbered 397) Books VII.—XV. in one and the same elegant hand (9th c.). The leaves preceding leaf 15 seem to have been lost at some time, leaves 6 to 14 (containing Elem. I. to the place in I. 14 above referred to) being carelessly written by a later hand on thick and common parchment (13th c.). On leaves 2 to 4 and 122 are certain notes in the hand of Arethas, who also wrote a two-line epigram on leaf 5, the greater part of the scholia in uncial letters, a few notes and corrections, and two sentences on the last leaf, the first of which states that the MS. was written by one Stephen clericus in the year of the world 6397 (= 888 A.D.), while the second records Arethas' own acquisition of it. Arethas lived from, say, 865 to 939 A.D. He was Archbishop of Caesarea and wrote a commentary on the Apocalypse. The portions of his library which survive are of the greatest interest to palaeography on account of his exact notes of dates, names of copyists, prices of parchment etc. It is to him also that we owe the famous Plato MS. from Patmos (Cod. Clarkianus) which was written for him in November 895See Pauly-Wissowa, Real-Encyclopädie der class. Altertumswissenschaft, vol. II., 1896, p. 675..

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(4) V = Viennese MS. Philos. Gr. No. 103; probably 12th c.

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This MS. contains 292 leaves, Eucl. Elements I.—XV. occupying leaves 1 to 254, after which come the Optics (to leaf 271), the Phaenomena (mutilated at the end) from leaf 272 to leaf 282, and lastly scholia, on leaves 283 to 292, also imperfect at the end. The different material used for different parts and the varieties of handwriting make it necessary for Heiberg to discuss this MS. at some lengthHeiberg, vol. v. pp. xxix—xxxiii.. The handwriting on leaves 1 to 183 (Book I. to the middle of X. 105) and on leaves 203 to 234 (from XI. 31, towards the end of the proposition, to XIII. 7, a few lines down) is the same; between leaves 184 and 202 there are two varieties of handwriting, that of leaves 184 to 189 and that of leaves 200 (verso) to 202 being the same. Leaf 235 begins in the same handwriting, changes first gradually into that of leaves 184 to 189 and then (verso) into a third more rapid cursive writing which is the same as that of the greater part of the scholia, and also as that of leaves 243 and 282, although, as these leaves are of different material, the look of the writing and of the ink seems altered. There are corrections both by the first and a second hand, and scholia by many hands. On the whole, in spite of the apparent diversity of handwriting in the MS., it is probable that the whole of it was written at about the same time, and it may (allowing for changes of material, ink etc.) even have been written by the same man. It is at least certain that, when the Laurentian MS. XXVIII, 6 was copied from it, the whole MS. was in the condition in which it is now, except as regards the later scholia and leaves 283 to 292 which are not in the Laurentian MS., that MS. coming to an end where the Phaenomena breaks off abruptly in V. Hence Heiberg attributes the whole MS. to the 12th c.

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But it was apparently in two volumes originally, the first consisting of leaves 1 to 183; and it is certain that it was not all copied at the same time or from one and the same original. For leaves 184 to 202 were evidently copied from two MSS. different both from one another and from that from which the rest was copied. Leaves 184 to the middle of leaf 189 (recto) must have been copied from a MS. similar to P, as is proved by similarity of readings, though not from P itself. The rest, up to leaf 202, were copied from the Bologna MS. (b) to be mentioned below. It seems clear that the content of leaves 184 to 202 was supplied from other MSS. because there was a lacuna in the original from which the rest of V was copied.

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Heiberg sums up his conclusions thus. The copyist of V first copied leaves 1 to 183 from an original in which two quaterniones were missing (covering from the middle of Eucl. X. 105 to near the end of XI. 31). Noticing the lacuna he put aside one quaternio of the parchment used up to that point. Then he copied onwards from the end of the lacuna in the original to the end of the Phaenomena. After this he looked about him for another MS. from which to fill up the lacuna; finding one, he copied from it as far as the middle of leaf 189 (recto). Then, noticing that the MS. from which he was copying was of a different class, he had recourse to yet another MS. from which he copied up to leaf 202. At the same time, finding that the lacuna was longer than he had reckoned for, he had to use twelve more leaves of a different parchment in addition to the quaternio which he had put aside. The whole MS. at first formed two volumes (the first containing leaves 1 to 183 and the second leaves 184 to 282); then, after the last leaf had perished, the two volumes were made into one to which two more quaterniones were also added. A few leaves of the latter of these two have since perished.

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(5) b = MS. numbered 18-19 in the Communal Library at Bologna, in two volumes, 4to; 11th c.

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This MS. has scholia in the margin written both by the first hand and by two or three later hands; some are written by the latest hand, Theodorus Cabasilas (a descendant apparently of Nicolaus Cabasilas, 14th c.) who owned the MS. at one time. It contains (a) in 14 quaterniones the definitions and the enunciations (without proofs) of the Elements I.—XIII. and of the Data, (b) in the remainder of the volumes the Proem to Geometry (published among the Variae Collectiones in Hultsch's edition of Heron, pp. 252, 24 to 274, 14) followed by the Elements I.—XIII. (part of XIII. 18 to the end being missing), and then by part of the Data (from the last three words of the enunciation of Prop. 38 to the end of the penultimate clause in Prop. 87, ed. Menge). From XI. 36 inclusive to the end of XII. this MS. appears to represent an entirely different recension. Heiberg is compelled to give this portion of b separately in an appendix. He conjectures that it is due to a Byzantine mathematician who thought Euclid's proofs too long and tiresome and consequently contented himself with indicating the course followedZeitschrift fiir Math. u. Physik, XXIX., hist.-litt. Abtheilung, p. 13.. At the same time this Byzantine must have had an excellent MS. before him, probably of the ante-Theonine variety of which the Vatican MS. 190 (P) is the sole representative.

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(6) p = Paris MS. 2466, 4to; 12th c.

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This manuscript is written in two hands, the finer hand occupying leaves 1 to 53 (recto), and a more careless hand leaves 53 (verso) to 64, which are of the same parchment as the earlier leaves, and leaves 65 to 239, which are of a thinner and rougher parchment showing traces of writing of the 8th—9th c. (a Greek version of the Old Testament). The MS. contains the Elements I.—XIII. and some scholia after Books XI., XII. and XIII.

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(7) q = Paris MS. 2344, folio; 12th c.

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It is written by one hand but includes scholia by many hands. On leaves 1 to 16 (recto) are scholia with the same title as that found by Wachsmuth in a Vatican MS. and relied upon by him to prove that Proclus continued his commentaries beyond Book I.[ei)s t]a\ tou= *eu)klei/dou stoixei=a prolambano/mena e)k tw=n *pro/klou spora/dhn kai\ kat) e)pitomh/n. Cf. p. 32, note 8, above. Leaves 17 to 357 contain the Elements I.—XIII. (except that there is a lacuna from the middle of VIII. 25 to the e)/kqesis of IX. 14); before Books VII. and X. there are some leaves filled with scholia only, and leaves 358 to 366 contain nothing but scholia.

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(8) Heiberg also used a palimpsest in the British Museum (Add. 17211). Five pages are of the 7th—8th c. and are contained (leaves 49-53) in the second volume of the Syrian MS. Brit. Mus. 687 of the 9th c.; half of leaf 50 has perished. The leaves contain various fragments from Book X. enumerated by Heiberg, Vol. III., p. v, and nearly the whole of XIII. 14.

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Since his edition of the Elements was published, Heiberg has collected further material bearing on the history of the textHeiberg, Paralipomena zu Euklid in Hermes, XXXVIII., 1903, pp. 46-74, 161-201, 321-356.. Besides giving the results of further or new examination of MSS., he has collected the fresh evidence contained in an-Nairīzī's commentary, and particularly in the quotations from Heron's commentary given in it (often word for word), which enable us in several cases to trace differences between our text and the text as Heron had it, and to identify some interpolations which actually found their way into the text from Heron's commentary itself; and lastly he has dealt with some valuable fragments of ancient papyri which have recently come to light, and which are especially important in that the evidence drawn from them necessitates some modification in the views expressed in the preface to Vol. V. as to the nature of the changes made in Theon's recension, and in the principles laid down for differentiating between Theon's recension and the original text, on the basis of a comparison between P and the Theonine MSS. alone.

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The fragments of ancient papyri referred to are the following.

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1. - Papyrus Herculanensis No. 1061.Described by Heiberg in Oversigt over det kngl. danske Videnskabernes Selskabs Forhandlinger, 1900, p. 161.

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This fragment quotes Def. 15 of Book I. in Greek, and omits the words h(\ kalei=tai perife/reia, which is called the circumference, - found in all our MSS., and the further addition pro\s th\n tou= ku/klou perife/reian also found in practically all the MSS. Thus Heiberg's assumption that both expressions are interpolations is now confirmed by this oldest of all sources.

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2. - The Oxyrhynchus Papyri I. p. 58, No. XXIX. of the 3rd or 4th c. This fragment contains the enunciation of Eucl. II. 5 (with figure, apparently without letters, immediately following, and not, as usual in our MSS., at the end of the proof) and before it the part of a word periexome belonging to II. 4 (with room for -nw| o)rqogwni/w|: o(/per e)/dei dei=cai and a stroke to mark the end), showing that the fragment had not the Porism which appears in all the Theonine MSS. and (in a later hand) in P, and thereby confirming Heiberg's assumption that the Porism was due to Theon.

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3. A fragment in Fayum towns and their papyri, p. 96, No. IX. of 2nd or 3rd c.

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This contains I. 39 and I. 41 following one another and almost complete, showing that I. 40 was wanting, whereas it is found in all the MSS. and is recognised by Proclus. Moreover the text of the beginning of I. 39 is better than ours, since it has no double diorismo/s but omits the first (I say that they are also in the same parallels -) and has and - instead of for let AD be joined - in the next sentence. It is clear that I. 40 was interpolated by someone who thought there ought to be a proposition following I. 39 and related to it as I. 38 is related to I. 37 and I. 36 to I. 35, although Euclid nowhere uses I. 40, and therefore was not likely to include it. The same interpolator failed to realise that the words let AD be joined - were part of the e)/kqesis or setting-out, and took them for the kataskeuh/ or construction - which generally follows the diorismo/s or particular statement - of the conclusion to be proved, and consequently thought it necessary to insert a diorismo/s - before the words.

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The conclusions drawn by Heiberg from a consideration of particular readings in this papyrus along with those of our MSS. will be referred to below.

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We now come to the principles which Heiberg followed, when preparing his edition, in differentiating the original text from the Theonine recension by means of a comparison of the readings of P and of the Theonine MSS. The rules which he gives are subject to a certain number of exceptions (mostly in cases where one MS. or the other shows readings due to copyists' errors), but in general they may be relied upon to give conclusive results.

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The possible alternatives which the comparison of P with the Theonine MSS. may give in particular passages are as follows:

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I. There may be agreement in three different degrees.

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(1) P and all the Theonine MSS. may agree.

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In this case the reading common to all, even if it is corrupt or interpolated, is more ancient than Theon, i.e. than the 4th c.

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(2) P may agree with some (only) of the Theonine MSS.

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In this case Heiberg considered that the latter give the true reading of Theon's recension, and the other Theonine MSS. have departed from it.

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(3) P and one only of the Theonine MSS may agree.

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In this case too Heiberg assumed that the one Theonine MS. which agrees with P gives the true Theonine reading, and that this rule even supplies a sort of measure of the quality and faithfulness of the Theonine MSS. Now none of them agrees alone with P in preserving the true reading so often as F. Hence F must be held to have preserved Theon's recension more faithfully than the other Theonine MSS.; and it would follow that in those portions where F fails us P must carry rather more weight even though it may differ from the Theonine MSS. BVpq. (Heiberg gives many examples in proof of this, as of his main rules generally, for which reference must be made to his Prolegomena in Vol. v.) The specially close relation of F and P is also illustrated by passages in which they have the same errors; the explanation of these common errors (where not due to accident) is found by Heiberg in the supposition that they existed, but were not noticed by Theon, in the original copy in which he made his changes.

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Although however F is by far the best of the Theonine MSS., there are a considerable number of passages where one of the others (B, V, p or q) alone with P gives the genuine reading of Theon's recension.

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As the result of the discovery of the papyrus fragment containing I. 39, 41, the principles above enunciated under (2) and (3) are found by Heiberg to require some qualification. For there is in some cases a remarkable agreement between the papyrus and the Theonine MSS. (some or all) as against P. This shows that Theon took more trouble to follow older MSS., and made fewer arbitrary changes of his own, than has hitherto been supposed. Next, when the papyrus agrees with some of the Theonine MSS. against P, it must now be held that these MSS. (and not, as formerly supposed, those which agree with P) give the true reading of Theon. If it were otherwise, the agreement between the papyrus and the Theonine MSS. would be accidental: but it happens too often for this. It is clear also that there must have been contamination between the two recensions; otherwise, whence could the Theonine MSS. which agree with P and not with the papyrus have got their readings? The influence of the P class on the Theonine F is especially marked.

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II. There may be disagreement between P and all the Theonine MSS.

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The following possibilities arise.

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(1) The Theonine MSS. differ also among themselves.

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In this case Heiberg considered that P nearly always has the true reading, and the Theonine MSS. have suffered interpolation in different ways after Theon's time.

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(2) The Theonine MSS. all combine against P.

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In this case the explanation was assumed by Heiberg to be one or other of the following. -

(a) The common reading is due to an error which cannot be imputed to Theon (though it may have escaped him when putting together the archetype of his edition); such error may either have arisen accidentally in all alike, or (more frequently) may be referred to a common archetype of all the MSS.

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(b) There may be an accidental error in P; e.g. something has dropped out of P in a good many places, generally through o(moiote/leuton

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(g) There may be words interpolated in P.

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(d) Lastly, we may have in the Theonine MSS. a change made by Theon himself.

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(The discovery of the ancient papyrus showing readings agreeing with some, or with all, of the Theonine MSS. against P now makes it necessary to be very cautious in applying these criteria.)

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It is of course the last class (d) of changes which we have to investigate in order to get a proper idea of Theon's recension.

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Heiberg first observes, as regards these, that we shall find that Theon, in editing the Elements, altered hardly anything without some reason, often inadequate according to our ideas, but still some reason which seemed to him sufficient. Hence, in cases of very slight differences where both the Theonine MSS. and P have readings good and probable in themselves, Heiberg is not prepared to put the differences down to Theon. In those passages where we cannot see the least reason why Theon, if he had the reading of P before him, should have altered it, Heiberg would not at once assume the superiority of P unless there was such a consistency in the differences as would indicate that they were due not to accident but to design. In the absence of such indications, he thinks that the ordinary principles of criticism should be followed and that proper weight should be attached to the antiquity of the sources. And it cannot be denied that the sources of the Theonine version are the more ancient. For not only is the British Museum palimpsest (L), which is intimately connected with the rest of our MSS., at least two centuries older than P, but the other Theonine MSS. are so nearly allied that they must be held to have had a common archetype intermediate between them and the actual edition of Theon; and, since they themselves are as old as, or older than P, their archetype must have been much older. Heiberg gives (pp. xlvi, xlvii) a list of passages where, for this reason, he has followed the Theonine MSS. in preference to P.

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It has been mentioned above that the copyist of P or rather of its archetype wished to give an ancient recension. Therefore (apart from clerical errors and interpolations) the first hand in P may be relied upon as giving a genuine reading even where a correction by the first hand has been made at the same time. But in many places the first hand has made corrections afterwards; on these occasions he must have used new sources, e.g. when inserting the scholia to the first Book which P alone has, and in a number of passages he has made additions from Theonine MSS.

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We cannot make out any family tree - for the different Theonine MSS. Although they all proceeded from a common archetype later than the edition of Theon itself, they cannot have been copied one from the other; for, if they had been, how could it have come about that in one place or other each of them agrees alone with P in preserving the genuine reading? Moreover the great variety in their agreements and disagreements indicates that they have all diverged to about the same extent from their archetype. As we have seen that P contains corrections from the Theonine family, so they show corrections from P or other MSS. of the same family. Thus V has part of the lacuna in the MS. from which it was copied filled up from a MS. similar to P, and has corrections apparently derived from the same; the copyist, however, in correcting V, also used another MS. to which he alludes in the additions to IX. 19 and 30 (and also on X. 23 Por.): in the book of the Ephesian (this) is not found. - Who this Ephesian of the 12th c. was, we do not know.

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We now come to the alterations made by Theon in his edition of the Elements. I shall indicate classes into which these alterations may be divided but without details (except in cases where they affect the mathematical content as distinct from form or language pure and simple).Exhaustive details under all the different heads are given by Heiberg (Vol. v. pp. lii—lxxv)..

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I. - Alterations made by Theon where he found, or thought he found, mistakes in the original.

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1. Real blots in the original which Theon saw and tried to remove.

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(a) Euclid has a porism (corollary) to VI. 19, the enunciation of which speaks of similar and similarly described figures though the proposition itself refers only to triangles, and therefore the porism should have come after VI. 20. Theon substitutes triangle for figure and proves the more general porism after VI. 20.

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(b) In IX. 19 there is a statement which is obviously incorrect. Theon saw this and altered the proof by reducing four alternatives to two, with the result that it fails to correspond to the enunciation even with Theon's substitution of if - for when - in the enunciation.

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(c) Theon omits a porism to IX. II, although it is necessary for the proof of the succeeding proposition, apparently because, owing to an error in the text (kata\ to\n corrected by Heiberg into e)pi\ to\), he could not get out of it the right sense.

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(d) I should also put into this category a case which Heiberg classifies among those in which Theon merely fancied that he found mistakes, viz. the porism to V. 7 stating that, if four magnitudes are proportional, they are proportional inversely. Theon puts this after V. 4 with a proof, which however has no necessary connexion with V. 4 but is obvious from the definition of proportion.

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(e) I should also put under this head XI. 1, where Euclid's argument to prove that two straight lines cannot have a common segment is altered.

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2. Passages which seemed to Theon to contain blots, and which he therefore set himself to correct, though more careful consideration would have shown that Euclid's words are right or at least may be excused and offer no difficulty to an intelligent reader. Under this head come:

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(a) an alteration in III. 24.

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(b) a perfectly unnecessary alteration, in Vi. 14, of equiangular parallelograms - into parallelograms having one angle equal to one angle, - where Theon followed the false analogy of VI. 15.

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(c) an omission of words in V. 26, owing to his having been misled by a wrong figure.

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(d) an alteration of the order of XI. Deff. 27, 28.

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(e) the substitution of parallelepipedal solid - for cube - in XI. 38, because Theon observed, correctly enough, that it was true of the parallelepipedal solid in general as well as of the cube, but failed to give weight to the fact that Euclid must have given the particular case of the cube for the simple reason that that was all he wanted for use in XIII. 17.

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(f) the substitution of the letter *f for *w (V for Z in my figure) because he saw that the perpendicular from *k to *b*f would fall on *f itself, so that *f, *w coincide. But, if the substitution is made, it should be proved that *f, *w coincide. Euclid can hardly have failed to notice the fact, but it may be that he deliberately ignored it as unnecessary for his purpose, because he did not want to lengthen his proposition by giving the proof.

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II. - Emendations intended to improve the form or diction of Euclid.

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Some of these emendations of Theon affect passages of appreciable length. Heiberg notes about ten such passages; the longest is in Eucl. XII. 4 where a whole page of Heiberg's text is affected and Theon's version is put in the Appendix. The kind of alteration may be illustrated by that in IX. 15 where Euclid uses successively the propositions VII. 24, 25, quoting the enunciation of the former but not of the latter; Theon does exactly the reverse. In a few of the cases here quoted by Heiberg, Theon shortened the original somewhat.

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But, as a rule, the emendations affect only a few words in each sentence. Sometimes they are considerable enough to alter the conformation of the sentence, sometimes they are trifling alterations more magistellorum ineptorum - and unworthy of Theon. Generally speaking, they were prompted by a desire to change anything which was out of the common in expression or in form, in order to reduce the language to one and the same standard or norm. Thus Theon changed the order of words, substituted one word for another where the latter was used in a sense unusual with Euclid (e.g. e)peidh/per, since, - for o(/ti in the sense of because -), or one expression for another in like circumstances (e.g. where, finding that which was enjoined would be done - in a theorem, VII. 31, and deeming the phrase more appropriate to a problem, he substituted for it that which is sought would be manifest -; probably also and for similar reasons he made certain variations between the two expressions usual at the end of propositions o(/per e)/dei dei=cai and o(/per e)/dei poih=sai, quod erat demonstrandum and quod erat faciendum). Sometimes his alterations show carelessness in the use of technical terms, as when he uses a(/ptesqai (to meet) for e)fa/ptesqai (to touch) although the ancients carefully distinguished the two words. The desire of keeping to a standard phraseology also led Theon to omit or add words in a number of cases, and also, sometimes, to change the lettering of figures.

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But Theon seems, in editing the Elements, to have bestowed the most attention upon

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III. - Additions designed to supplement or explain Euclid.

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First, he did not hesitate to interpolate whole propositions where he thought there was room or use for them. We have already mentioned the addition to VI. 33 of the second part relating to sectors, for which Theon himself takes credit in his commentary on Ptolemy. Again, he interpolated the proposition commonly known as VII. 22 (ex aequo in proportione perturbata for numbers, corresponding to V. 23), and perhaps also VII. 20, a particular case of VII. 19 as VI. 17 is of VI. 16. He added a second case to VI. 27, a porism to II. 4, a second porism to III. 16, and a lemma after X. 12; perhaps also the porism to V. 19 and the first porism to VI. 20. He also inserted alternative proofs here and there, e.g. in II. 4 (where the alternative differs little from the original) and in VII. 31; perhaps also in X. 1, 6, and 9.

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Secondly, he sometimes repeats an argument where Euclid had said For the same reason, - adds specific references to points, straight lines etc. in the figures in order to exclude the possibility of mistake arising from Euclid's reference to them in general terms, or inserts words to make the meaning of Euclid more plain, e.g. componendo and alternately, where Euclid had left them out. Sometimes he thought to increase by his additions the mathematical precision of Euclid's language in enunciations or elsewhere, sometimes to make smoother and clearer things which Euclid had expressed with unusual brevity and harshness or carelessness, in reliance on the intelligence of his readers.

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Thirdly, he supplied intermediate steps where Euclid's argument seemed too rapid and not easy enough to follow. The form of these additions varies; they are sometimes placed as a definite intermediate step with therefore - or so that, - sometimes they are additions to the statement of premisses, sometimes phrases introduced by since, - for - and the like, after the inference.

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Lastly, there is a very large class of additions of a word, or one or two words, for the sake of clearness or consistency. Heiberg gives a number of examples of the addition of such nouns as triangle, - square, - rectangle, - magnitude, - number, - point, - side, - circle, - straight line, - area - and the like, of adjectives such as remaining, - right, - whole, - proportional, - and of other parts of speech, even down to words like is - (e)sti/) which is added 600 times, dh/, a)/ra, me/n, ga/r, kai/ and the like.

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IV. - Omissions by Theon.

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Heiberg remarks that, Theon's object having been, as above shown, to amplify and explain Euclid, we should not natuially have expected to find him doing much in the contrary process of compression, and it is only owing to the recurrence of a certain sort of omissions so frequently (especially in the first Books) as to exclude the hypothesis of their being all due to chance that we are bound to credit him with alterations making for greater brevity. We have seen, it is true, that he made omissions as well as additions for the purpose of reducing the language to a certain standard form. But there are also a good number of cases where in the enunciation of propositions, and in the exposition (the re-statement of them with reference to the figure), he has left out words because, apparently, he regarded Euclid's language as being too careful and precise. Again, he is apparently responsible for the frequent omission of the words o(/per e)/dei dei=cai (or poih=sai), Q.E.D. (or F.), at the end of propositions. This is often the case at the end of porisms, where, in omitting the words, Theon seems to have deliberately departed from Euclid's practice. The MS. P seems to show clearly that, where Euclid put a porism at the end of a proposition, he omitted the Q.E.D. at the end of the proposition but inserted it at the end of the porism, as if he regarded the latter as being actually a part of the proposition itself. As in the Theonine MSS. the Q.E.D. is generally omitted, the omission would seem to have been due to Theon. Sometimes in these cases the Q.E.D. is interpolated at the end of the proposition.

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Heiberg summed up the discussion of Theon's edition by the remark that Theon evidently took no pains to discover and restore from MSS. the actual words which Euclid had written, but aimed much more at removing difficulties that might be felt by learners in studying the book. His edition is therefore not to be compared with the editions of the Alexandrine grammarians, but rather with the work done by Eutocius in editing Apollonius and with an interpolated recension of some of the works of Archimedes by a certain Byzantine, Theon occupying a position midway between these two editors, being superior to the latter in mathematical knowledge but behind Eutocius in industry (these views now require to be somewhat modified, as above stated). But however little Theon's object may be approved by those of us who would rather know the ipsissima verba of Euclid, there is no doubt that his work was approved by his pupils at Alexandria for whom it was written; and his edition was almost exclusively used by later Greeks, with the result that the more ancient text is only preserved to us in one MS.

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As the result of the above investigation, we may feel satisfied that, where P and the Theonine MSS. agree, they give us (except in a few accidental instances) Euclid as he was read by the Greeks of the 4th c. But even at that time the text had been passed from hand to hand through more than six centuries, so that it is certain that it had already suffered changes, due partly to the fault of copyists and partly to the interpolations of mathematicians. Some errors of copyists escaped Theon and were corrected in some MSS. by later hands. Others appear in all our MSS. and, as they cannot have arisen accidentally in all, we must put them down to a common source more ancient than Theon. A somewhat serious instance is to be found in III. 8; and the use of a(pte/sqw for e)fapte/sqw in the sense of touch - may also be mentioned, the proper distinction between the words having been ignored as it was by Theon also. But there are a number of imperfections in the ante-Theonine text which it would be unsafe to put down to the errors of copyists, those namely where the good MSS. agree and it is not possible to see any motive that a copyist could have had for altering a correct reading. In these cases it is possible that the imperfections are due to a certain degree of carelessness on the part of Euclid himself; for it is not possible Euclidem ab omni naevo vindicare, - to use the words of SaccheriEuclides ab omni naevo vindicatus, Mediolani, 1733., and consequently Simson is not right in attributing to Theon and other editors all the things in Euclid to which mathematical objection can be taken. Thus, when Euclid speaks of the ratio compounded of the sides - for the ratio compounded of the ratios of the sides, - there is no reason for doubting that Euclid himself is responsible for the more slip-shod expression. Again, in the Books XI.—XIII. relating to solid geometry there are blots neither few nor altogether unimportant which can only be attributed to Euclid himselfCf. especially the assumption, without proof or definition, of the criterion for equal solid angles, and the incomplete proof of XII. 17.; and there is the less reason for hesitation in so attributing them because solid geometry was then being treated in a thoroughly systematic manner for the first time. Sometimes the conclusion (sumpe/rasma) of a proposition does not correspond exactly to the enunciation, often it is cut short with the words kai\ ta\ e(ch=s and the rest - (especially from Book X. onwards), and very often in Books VIII., IX. it is omitted. Where all the MSS. agree, there is no ground for hesitating to attribute the abbreviation or omission to Euclid; though, of course, where one or more MSS. have the longer form, it must be retained because this is one of the cases where a copyist has a temptation to abbreviate.

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Where the true reading is preserved in one of the Theonine MSS. alone, Heiberg attributes the wrong reading to a mistake which arose before Theon's time, and the right reading of the single MS. to a successful correction.

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We now come to the most important question of the Interpolations introduced before Theon's time.

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I. Alternative proofs or additional cases.

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It is not in itself probable that Euclid would have given two proofs of the same proposition; and the doubt as to the genuineness of the alternatives is increased when we consider the character of some of them and the way in which they are introduced. First of all, we have those of VI. 20 and XII. 17 introduced by we shall prove this otherwise more readily (proxeiro/teron) - or that of X. 90 it is possible to prove more shortly (suntomw/teron). - Now it is impossible to suppose that Euclid would have given one proof as that definitely accepted by him and then added another with the express comment that the latter has certain advantages over the former. Had he considered the two proofs and come to this conclusion, he would have inserted the latter in the received text instead of the former. These alternative proofs must therefore have been interpolated. The same argument applies to alternatives introduced with the words or even thus - (h)\ kai\ ou(/tws), or even otherwise - (h)\ kai\ a)/llws). Under this head come the alternatives for the last portions of III. 7, 8; and Heiberg also compares the alternatives for parts of III. 31 (that the angle in a semicircle is a right angle) and XIII. 18, and the alternative proof of the lemma after X. 32. The alternatives to X. 105 and 106, again, are condemned by the place in which they occur, namely after an alternative proof to X. 115. The above alternatives being all admitted to be spurious, suspicion must necessarily attach to the few others which are in themselves unobjectionable. Heiberg instances the alternative proofs to III. 9, III. 10, VI. 30, VI. 31 and XI. 22, observing that it is quite comprehensible that any of these might have occurred to a teacher or editor and seemed to him, rightly or wrongly, to be better than the corresponding proofs in Euclid. Curiously enough, Simson adopted the alternatives to III. 9, 10 in preference to the genuine proofs. Since Heiberg's preface was written, his suspicion has been amply confirmed as regards III. 10 by the commentary of an-Nairīzī (ed. Curtze) which shows not only that this alternative is Heron's, but also that the substantive proposition III. 12 in Euclid is also Heron's, having been given by him to supplement III. II which must originally have been enunciated of circles touching one another - simply, i.e. so as to include the case of external as well as internal contact, though the proof covered the case of internal contact only. Euclid, in the 11th proposition, - says Heron, supposed two circles touching one another internally and wrote the proposition on this case, proving what it was required to prove in it. But I will show how it is to be proved if the contact be external.An-Nairīzī, ed. Curtze, p. 121.. - This additional proposition of Heron's is by way of adding another case, which brings us to that class of interpolation. It was the practice of Euclid and the ancients to give only one case (generally the most difficult one) and to leave the others to be investigated by the reader for himself. One interpolation of a second case (VI. 27) is due, as we have seen, to Theon. The two extra cases of XI. 23 were manifestly interpolated before Theon's time, for the preliminary distinction of three cases, (the centre) will either be within the triangle LMN, or on one of the sides, or outside. First let it be within, - is a spurious addition (B and V only). Similarly an unnecessary case is interpolated in III. 11.

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II. Lemmas.

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Heiberg has unhesitatingly placed in his Appendix to Vol. III. certain lemmas interpolated either by Theon (on X. 13) or later writers (on X. 27, 29, 31, 32, 33, 34, where V only has the lemmas). But we are here concerned with the lemmas found in all the MSS., which however are, for different reasons, necessarily suspected. We will deal with the Book X. lemmas last.

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(1) There is an a priori ground of objection to those lemmas which come after the propositions to which they relate and prove properties used in those propositions; for, if genuine, they would be a sign of faulty arrangement such as would not be likely in a systematic work so carefully ordered as the Elements. The lemma to VI. 22 is one of this class, and there is the further objection to it that in VI. 28 Euclid makes an assumption which would equally require a lemma though none is found. The lemma after XII. 4 is open to the further objections that certain altitudes are used but are not drawn in the figure (which is not in the manner of Euclid), and that a peculiar expression parallelepipedal solids described on (a)nagrafo/mena a)po/) prisms - betrays a hand other than Euclid's. There is an objection on the score of language to the lemma after XIII. 2. The lemmas on XI. 23, XIII. 13, XIII. 18, besides coming after the propositions to which they relate, are not very necessary in themselves and, as regards the lemma to XIII. 13, it is to be noticed that the writer of a gloss in the proposition could not have had it, and the words as will be proved afterwards - in the text are rightly suspected owing to differences between the MS. readings. The lemma to XII. 2 also, to which Simson raised objection, comes after the proposition; but, if it is rejected, the words as was proved before - used in XII. 5 and 18, and referring to this lemma, must be struck out.

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(2) Reasons of substance are fatal to the lemma before X. 60, which is really assumed in X. 44 and therefore should have appeared there if anywhere, and to the lemma on X. 20, which tries to prove what is already stated in X. Def. 4.

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We now come to the remaining lemmas in Book X., eleven in number, which come before the propositions to which they relate and remove difficulties in the way of their demonstration. That before X. 42 introduces a set of propositions with the words that the said irrational straight lines are uniquely divided ... we will prove after premising the following lemma, - and it is not possible to suppose that these words are due to an interpolator; nor are there any objections to the lemmas before X. 14, 17, 22, 33, 54, except perhaps that they are rather easy. The lemma before X. 10 and X. 10 itself should probably be removed from the Elements; for X. 10 really uses the following proposition X. 11, which is moreover numbered 10 by the first hand in P, and the words in X. 10 referring to the lemma for we learnt (how to do this) - betray the interpolator. Heiberg gives reason also for rejecting the lemmas before X. 19 and 24 with the words in any of the aforesaid ways - (omitted in the Theonine MSS.) in the enunciations of X. 19, 24 and in the exposition of X. 20. Lastly, the lemmas before X. 29 may be genuine, though there is an addition to the second of them which is spurious.

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Heiberg includes under this heading of interpolated lemmas two which purport to be substantive propositions, XI. 38 and XIII. 6. These must be rejected as spurious for reasons which will be found in detail in my notes on XI. 37 and XIII. 6 respectively. The latter proposition is only quoted once (in XIII. 17); probably the words quoting it (with grammh/ instead of eu)qei=a) are themselves interpolated, and Euclid thought the fact stated a sufficiently obvious inference from XIII. 1.

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III. Porisms (or corollaries).

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Most of the porisms in the text are both genuine and necessary; but some are shown by differences in the MSS. not to be so, e.g. those to I. 15 (though Proclus has it), III. 31 and VI. 20 (Por. 2). Sometimes parts of porisms are interpolated. Such are the last few lines in the porisms to IV. 5, VI. 8; the latter addition is proved later by means of VI. 4, 8, so that the writer of these proofs could not have had the addition to VI. 8 Por. before him. Lastly, interpolators have added a sort of proof to some porisms, as though they were not quite obvious enough; but to add a demonstration is inconsistent with the idea of a porism, which, according to Proclus, is a by-product of a proposition appearing without our seeking it.

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IV. Scholia.

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Several interpolated scholia betray themselves by their wording, e.g. those given by Heiberg in the Appendix to Book X. and containing the words kalei=, e)ka/lese (he calls - or called -); these scholia were apparently written as marginal notes before Theon's time, and, being adopted as such by Theon, found their way into the text in P and some of the Theonine MSS. The same thing no doubt accounts for the interpolated analyses and syntheses to XIII. 1-5, as to which see my note on XIII. 1.

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V. Interpolations in Book X.

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First comes the proposition Let it be proposed to us to show that in square figures the diameter is incommensurable in length with the side, - which, with a scholium after it, ends the tenth Book. The form of the enunciation is suspicious enough and the proposition, the proof of which is indicated by Aristotle and perhaps was Pythagorean, is perfectly unnecessary when X. 9 has preceded. The scholium ends with remarks about commensurable and incommensurable solids, which are of course out of place before the Books on solids. The scholiast on Book X. alludes to this particular scholium as being due to Theon and some others. - But it is doubtless much more ancient, and may, as Heiberg conjectures have been the beginning of Apollonius' more advanced treatise on incommensurables. Not only is everything in Book X. after X. 115 interpolated, but Heiberg doubts the genuineness even of X. 112-115, on the ground that X. 111 rounds off the theory of incommensurables as we want it in the Books on solid geometry, while X. 112-115 are not really connected with what precedes, nor wanted for the later Books, but seem to form the starting-point of a new and more elaborate theory of irrationals.

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VI. Other minor interpolations are found of the same character as those above attributed to Theon. First there are two places (XI. 35 and XI. 26) where, after similarly we shall prove - and for the same reason, - an actual proof is nevertheless given. Clearly the proofs are interpolated; and there are other similar interpolations. There are also interpolations of intermediate steps in proofs, unnecessary explanations and so on, as to which I need not enter into details.

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Lastly, following Heiberg's order, I come to

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VII. Interpolated definitions, axioms etc.

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Apart from VI. Def. 5 (which may have been interpolated by Theon although it is found written in the margin of P by the first hand), the definition of a segment of a circle in Book I. is interpolated, as is clear from the fact that it occurs in a more appropriate place in Book III. and Proclus omits it. VI. Def. 2 (reciprocal figures) is rightly condemned by Simson—perhaps it was taken from Heron—and Heiberg would reject VII. Def. 10, as to which see my note on that definition. Lastly the double definition of a solid angle (XI. Def. 11) constitutes a difficulty. The use of the word e)pifa/neia suggests that the first definition may have been older than Euclid, and he may have quoted it from older elements, especially as his own definition which follows only includes solid angles contained by planes, whereas the other includes other sorts (cf. the words grammw=n, grammai=s) which are also distinguished by Heron (Def. 22). If the first definition had come last, it could have been rejected without hesitation: but it is not so easy to reject the first part up to and including otherwise - (a)/llws). No difficulty need be felt about the definitions of oblong, - rhombus, - and rhomboid, - which are not actually used in the Elements; they were no doubt taken from earlier elements and given for the sake of completeness.

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As regards the axioms or, as they are called in the text, common notions (koinai\ e)/nnoiai), it is to be observed that Proclus saysProclus, pp. 194, 10 sqq. that Apollonius tried to prove the axioms, - and he gives Apollonius' attempt to prove Axiom I. This shows at all events that Apollonius had some of the axioms now appearing in the text. But how could Apollonius have taken a controversial line against Euclid on the subject of axioms if these axioms had not been Euclid's to his knowledge? And, if they had been interpolated between Euclid's time and his own, how could Apollonius, living so comparatively short a time after Euclid, have been ignorant of the fact? Therefore some of the axioms are Euclid's (whether he called them common notions, or axioms, as is perhaps more likely since Proclus calls them axioms): and we need not hesitate to accept as genuine the first three discussed by Proclus, viz. (1) things equal to the same equal to one another, (2) if equals be added to equals, wholes equal, (3) if equals be subtracted from equals, remainders equal. The other two mentioned by Proclus (whole greater than part, and congruent figures equal) are more doubtful, since they are omitted by Heron, Martianus Capella, and others. The axiom that two lines cannot enclose a space - is however clearly an interpolation due to the fact that I. 4 appeared to require it. The others about equals added to unequals, doubles of the same thing, and halves of the same thing are also interpolated; they are connected with other interpolations, and Proclus clearly used some source which did not contain them.

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Euclid evidently limited his formal axioms to those which seemed to him most essential and of the widest application; for he not unfrequently assumes other things as axiomatic, e.g. in VII. 28 that, if a number measures two numbers, it measures their difference.

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The differences of reading appearing in Proclus suggest the question of the comparative purity of the sources used by Proclus, Heron and others, and of our text. The omission of the definition of a segment in Book I. and of the old gloss which is called the circumference - in I. Def. 15 (also omitted by Heron, Taurus, Sextus Empiricus and others) indicates that Proclus had better sources than we have; and Heiberg gives other cases where Proclus omits words which are in all our MSS. and where Proclus' reading should perhaps be preferred. But, except in these instances (where Proclus may have drawn from some ancient source such as one of the older commentaries), Proclus' MS. does not seem to have been among the best. Often it agrees with our worst MSS., sometimes it agrees with F where F alone has a certain reading in the text, so that (e.g. in I. 15 Por.) the common reading of Proclus and F must be rejected, thrice only does it agree with P alone, sometimes it agrees with P and some Theonine MSS., and once it agrees with the Theonine MSS. against P and other sources.

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Of the other external sources, those which are older than Theon generally agree with our best MSS., e.g. Heron, allowing for the difference in the plan of his definitions and the somewhat free adaptation to his purpose of the Euclidean definitions in Books X., XI.

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Heiberg concludes that the Elements were most spoiled by interpolations about the 3rd c., for Sextus Empiricus had a correct text, while Iamblichus had an interpolated one; but doubtless the purer text continued for a long time in circulation, as we conclude from the fact that our MSS. are free from interpolations already found in Iamblichus' MS.

- - - - CHAPTER VI. - THE SCHOLIA. -

Heiberg has collected scholia, to the number of about 1500, in Vol. v. of his edition of Euclid, and has also discussed and classified them in a separate short treatise, in which he added a few others.Heiberg, Om Scholierne til Euklids Elementer, Kjøbenhavn, 1888. The tract is written in Danish, but, fortunately for those who do not read Danish easily, the author has appended (pp. 70-78) a résumé in French.

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These scholia cannot be regarded as doing much to facilitate the reading of the Elements. As a rule, they contain only such observations as any intelligent reader could make for himself. Among the few exceptions are XI. Nos. 33, 35 (where XI. 22, 23 are extended to solid angles formed by any number of plane angles), XII. No. 85 (where an assumption tacitly made by Euclid in XII. 17 is proved), IX. Nos. 28, 29 (where the scholiast has pointed out the error in the text of IX. 19).

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Nor are they very rich in historical information; they cannot be compared in this respect with Proclus' commentary on Book I. or with those of Eutocius on Archimedes and Apollonius. But even under this head they contain some things of interest, e.g. II. No. 11 explaining that the gnomon was invented by geometers for the sake of brevity, and that its name was suggested by an incidental characteristic, namely that from it the whole is known (gnwri/zetai), either of the whole area or of the remainder, when it (the gnw/mwn) is either placed round or taken away -; II. No. 13, also on the gnomon; IV. No. 2 stating that Book IV. was the discovery of the Pythagoreans; V. No. 1 attributing the content of Book V. to Eudoxus; X. No. 1 with its allusion to the discovery of incommensurability by the Pythagoreans and to Apollonius' work on irrationals; X. No. 62 definitely attributing X. 9 to Theaetetus; XIII. No. 1 about the Platonic - figures, which attributes the cube, the pyramid, and the dodecahedron to the Pythagoreans, and the octahedron and icosahedron to Theaetetus.

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Sometimes the scholia are useful in connexion with the settlement of the text, (1) directly, e.g. III. No. 16 on the interpolation of the word within - (e)nto/s) in the enunciation of III. 6, and X. No. 1 alluding to the discussion by Theon and some others - of irrational surfaces - and solids, - as well as lines, - from which we may conclude that the scholium at the end of Book X. is not genuine; (2) indirectly in that they sometimes throw light on the connexion of certain MSS.

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Lastly, they have their historical importance as enabling us to judge of the state of mathematical science at the times when they were written.

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Before passing to the classification of the scholia, Heiberg remarks that we must separate from them a number of additions in the nature of scholia which are found in the text of our MSS. but which can, in one way or another, be proved to be spurious. As they are found both in P and in the Theonine MSS., they must have been in the MSS. anterior to Theon (4th c.). But they are, in great part, only found in the margin of P and the Theonine MSS.; in V they are half in the text and half in the margin. This can hardly be explained except on the supposition that these additions were originally (in the MSS. before Theon) in the margin, and that Theon kept them there in his edition, but that they afterwards found their way gradually into the text of P as well as of the Theonine MSS., or were omitted altogether, while particular MSS. have in certain places preserved the old arrangement. Of such spurious additions Heiberg enumerates the following: the axiom about equals subtracted from unequals, the last lines of the porism to VI. 8, second porisms to V. 19 and to VI. 20, the porism to III. 31, VI. Def. 5, various additions in Book X., the analyses and syntheses of XIII. 1-5, and the proposition XIII. 6.

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The two first classes of scholia distinguished by Heiberg are denoted by the convenient abbreviations Schol. Vat. - and Schol. Vind. -

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- I. Schol. Vat. -

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It is first necessary to set out the letters by which Heiberg denotes certain collections of scholia.

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P=Scholia in P written by the first hand.

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B=Scholia in B by a hand of the same date as the MS. itself, generally that of Arethas.

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F=Scholia in F by the first hand.

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Vat.=Scholia of the Vatican MS. 204 of the 10th c., which has these scholia on leaves 198-205 (the end is missing) as an independent collection. It does not contain the text of the Elements.

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Vc=Scholia found on leaves 283-292 of V and written in the same hand as that part of the MS. itself which begins at leaf 235.

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Vat. 192=a Vatican MS. of the 14th c. which contains, after (1) the Elements I.—XIII. (without scholia), (2) the Data with scholia, (3) Marinus on the Data, the Schol. Vat. as an independent collection and in their entirety, beginning with 1. No. 88 and ending with XIII. No. 44.

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The Schol. Vat., the most ancient and important collection of scholia, comprise those which are found in PBF Vat. and, from VII. 12 to IX. 15, in PB Vat. only, since in that portion of the Elements F was restored by a later hand without scholia; they also include 1. No. 88 which only happens to be erased in F, and IX. Nos. 28, 29 which may be left out because F. here has a different text. In F and Vat. the collection ends with Book X.; but it must also include Schol. PB of Books XI.—XIII., since these are found along with Schol. Vat. to Books I.—X. in several MSS. (of which Vat. 192 is one) as a separate collection. The Schol. Vat. to Books X.—XIII. are also found in the collection Vc (where, curiously enough, XIII. Nos. 43, 44 are at the beginning). The Schol. Vat. accordingly include Schol. PBVc Vat. 192, and doubtless also those which are found in two of these sources. The total number of scholia classified by Heiberg as Schol. Vat. is 138.

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As regards the contents of Schol. Vat. Heiberg has the following observations. The thirteen scholia to Book I. are extracts made from Proclus by a writer thoroughly conversant with the subject, and cleverly recast (with some additions). Their author does not seem to have had the two lacunae which our text of Proclus has (at the end of the note on I. 36 and the beginning of the next note, and at the beginning of the note on I. 43), for the scholia I. Nos. 125 and 137 seem to fill the gaps appropriately, at least in part. In some passages he had better readings than our MSS. have. The rest of Schol. Vat. (on Books II.—XIII.) are essentially of the same character as those on Book I., containing prolegomena, remarks on the object of the propositions, critical remarks on the text, converses, lemmas; they are, in general, exact and true to tradition. The reason of the resemblance between them and Proclus appears to be due to the fact that they have their origin in the commentary of Pappus, of which we know that Proclus also made use. In support of the view that Pappus is the source, heiberg places some of the Schol. Vat. to Book X. side by side with passages from the commentary of Pappus in the Arabic translation discovered by Woepcke;Om Scholierne til Euklids Elementer, pp. 11, 12: cf. Euklid-Stulien, pp. 170, 171; Woepcke, Mémoires présent. à l' Acad. des Sciences, 1856, XIV. p. 658 sqq.; he also refers to the striking confirmation afforded by the fact that XII. No. 2 contains the solution of the problem of inscribing in a given circle a polygon similar to a polygon inscribed in another circle, which problem Eutocius saysArchimedes, ed. Heiberg, III. P. 28, 19-22. that Pappus gave in his commentary on the Elements.

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But, on the other hand, Schol. Vat. contain some things which cannot have come from Pappus, e.g. the allusion in X. No. 1 to Theon and irrational surfaces and solids, Theon being later than Pappus; III. No. 10 about porisms is more like Proclus' treatment of the subject than Pappus', though one expression recalls that of Pappus about forming (sxhmati/zesqai) the enunciations of porisms like those of either theorems or problems.

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The Schol. Vat. give us important indications as regards the text of the Elements as Pappus had it. In particular, they show that he could not have had in his text certain of the lemmas in Book X. For example, three of these are identical with what we find in Schol. Vat. (the lemma to X. 17=Schol. X. No. 106, and the lemmas to X. 54, 60 come in Schol. X. No. 328); and it is not possible to suppose. that these lemmas, if they were already in the text, would also be given as scholia. Of these three lemmas, that before X. 60 has already been condemned for other reasons; the other two, unobjectionable in themselves, must be rejected on the ground now stated. There were four others against which Heiberg found nothing to urge when writing his prolegomena to Vol. v., viz. the lemmas before X. 42, X. 14, X. 22 and X. 33. Of these, the lemma to X. 22 is not reconcilable with Schol. X. No. 161, which takes up the assumption in the text of Eucl. X. 22 as if no lemma had gone before. The lemma to X. 42, which, on account of the words introducing it (see p. 60 above), Heiberg at first hesitated to regard as an interpolation, is identical with Schol. X. No. 270. It is true that in Schol. X. No. 269 we find the words this lemma has been proved before (e)n toi=s e)/mprosqen), but it shall also be proved now for convenience' sake (tou= e(toi/mou e(/neka), - and it is possible to suppose that before - may mean in Euclid's text before X. 42; but a proof in that place would surely have been as convenient - as could be desired, and it is therefore more probable that the proof had been given by Pappus in some earlier place. (It may be added that the lemma to X. 14, which is identical with the lemma to XI. 23, condemned on other grounds, is for that reason open to suspicion.)

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Heiberg's conclusion is that all the lemmas are spurious, and that most or all of them have found their way into the text from Pappus' commentary, though at a time anterior to Theon's edition, since they are found in all our MSS. This enables us to fix a date for these interpolations, namely the first half of the 4th c.

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Of course Pappus had not in his text the interpolations which, from the fact of their appearing only in some of our MSS., are seen to be later than those above-mentioned. Such are the lemmas which are found in the text of V only after X. 29 and X. 31 respectively and are given in Heiberg's Appendix to Book X. (numbered 10 and 11). On the other hand it appears from Woepcke's tractWoepcke, op. cit. p. 702. that Pappus already had X. 115 in his text: though it does not follow from this that the proposition is genuine but only that interpolations began very early.

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Theon interpolated a proposition (or lemma) between X. 12 and X. 13 (No. 5 in Heiberg's Appendix). Schol. Vat. has the same thing (X. No. 125). The writer of the scholia therefore did not find this lemma in the text. Schol. Vat. IX. Nos. 28, 29 show that neither did he find in his text the alterations which Theon made in Eucl. IX. 19; the scholia in fact only agree with the text of P, not with Theon's. This suggests that Schol. Vat. were written for use with a MS. of the ante-Theonine recension such as P is. This probability is further confirmed by a certain independence which P shows in several places when compared with the Theonine MSS. Not only has P better readings in some passages, but more substantial divergences; and, in particular, the absence in P of three notes of a historical character which are added, wholly or partly from Proclus, in the Theonine MSS. attests an independent and more primitive point of view in P.

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In view of the distinctive character of P, it is possible that some of the scholia found in it in the first hand, but not in the other sources of Schol. Vat., also belong to that collection; and several circumstances confirm this. Schol. XIII. No. 45, found in P only, which relates to a passage in Eucl. XIII. 13, shows that certain words in the text, though older than Theon, are interpolated; and, as the scholium is itself older than Theon, is headed third lemma, - and follows a second lemma - relating to a passage in the text immediately preceding, which second lemma - belongs to Schol. Vat. and is taken from Pappus, the third - in all probability came from Pappus also. The same is true of Schol. XII. No. 72 and XIII. No. 69, which are respectively identical with the propositions vulgo XI. 38 (Heiberg, App. to Book XI., No. 3) and XIII. 6; for both of these interpolations are older than Theon. Moreover most of the scholia which P in the first hand alone has are of the same character as Schol. Vat. Thus VII. No. 7 and XIII. No. I introducing Books VII. and XIII. respectively are of the same historical character as several of Schol. Vat.; that VII. No. 7 appears in the text of P at the beginning of Book VII. constitutes no difficulty. There are a number of converses, remarks on the relation of propositions to one another, explanations such as XII. No. 89 in which it is remarked that *f, *w in Euclid's figure to XII. 17 (Z, V in my figure) are really the same point but that this makes no difference in the proof. Two other Schol. P on XII. 17 are connected by their headings with XII. No. 72 mentioned above. XI. No. 10 (P) is only another form of XI. No. 11 (B); and B often, alone with P, has preserved Schol. Vat. On the whole Heiberg considers some 40 scholia found in P alone to belong to Schol. Vat.

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The history of Schol. Vat. appears to have been, in its main outlines, the following. They were put together after 500 A.D., since they contain extracts from Proclus, to which we ought not to assign a date too near to that of Proclus' work itself; and they must at least be earlier than the latter half of the 9th c., in which B was written. As there must evidently have been several intermediate links between the archetype and B, we must assign them rather to the first half of the period between the two dates, and it is not improbable that they were a new product of the great development of mathematical studies at the end of the 6th c. (Isidorus of Miletus). The author extracted what he found of interest in the commentary of Proclus on Book I. and in that of Pappus on the rest of the work, and put these extracts in the margin of a MS. of the class of P. As there are no scholia to I. 1-22, the first leaves of the archetype or of one of the earliest copies must have been lost at an early date, and it was from that mutilated copy that partly P and partly a MS. of the Theonine class were taken, the scholia being put in the margin in both. Then the collection spread through the Theonine MSS., gradually losing some scholia which could not be read or understood, or which were accidentally or deliberately omitted. Next it was extracted from one of these MSS. and made into a separate work which has been preserved, in part, in its entirety (Vat. 192 etc.) and, in part, divided into sections, so that the scholia to Books X.—XIII. were detached (Vc). It had the same fate in the MSS. which kept the original arrangement (in the margin), and in consequence there are some MSS. where the scholia to the stereometric Books are missing, those Books having come to be less read in the period of decadence. It is from one of these MSS. that the collection was extracted as a separate work such as we find it in Vat. (10th c.).

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II. The second great division of the scholia is Schol. Vind.

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This title is taken from the Viennese MS. (V), and the letters used by Heiberg to indicate the sources here in question are as follows.

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Va=scholia in V written by the same hand that copied the MS. itself from fol. 235 onward.

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q=scholia of the Paris MS. 2344 (q) written by the first hand.

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l=scholia of the Florence MS. Laurent. XXVIII, 2 written in the 13th—14th c., mostly in the first hand, but partly in two later hands.

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Vb=scholia in V written by the same hand as the first part (leaves 1-183) of the MS. itself; Vb wrote his scholia after Va.

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q1=scholia of the Paris MS. (q) found here and there in another hand of early date.

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Schol. Vind. include scholia found in Vaq. 1 is nearly related to q; and in fact the three MSS. which, so far as Euclid's text is concerned, show no direct interdependence, are, as regards their scholia, derived from one original. Heiberg proves this by reference to the readings of the three in two passages (found in Schol. 1. No. 109 and X. No. 39 respectively). The common source must have contained, besides the scholia found in the three MSS. Vaql, those also which are contained in two of them, for it is more unlikely that two of the three should contain common interpolations than that a particular scholium should drop out of one of them. Besides Va and q, the scholia Vb and q1 must equally be referred to Schol. Vind., since the greater part of their scholia are found in 1. There is a lacuna in q from Eucl. VIII. 25 to IX. 14, so that for this portion of the Elements Schol. Vind. are represented by Vl only. Heiberg gives about 450 numbers in all as belonging to this collection.

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Schol. Vind. did not all come from one source; this is shown by differences of substance, e.g. between X. Nos. 36 and 39, and by differences of time of writing: e.g. VI. No. 52 refers at the beginning to No. 55 with the words as the scholium has it - and is therefore later than that scholium; X. No. 247 is also later than X. No. 246.

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The scholia to Book I. are here also extracts from Proclus, but more copious and more verbatim than in Schol. Vat. The author has not always understood Proclus; and he had a text as bad as that of our MSS., with the same lacunae. The scholia to the other Books are partly drawn (1) from Schol. Vat., the MSS. representing Schol. Vind. and Schol. Vat. in these cases showing nearly all possible combinations; but there is no certain trace in Schol. Vind. of the scholia peculiar to P. The author used a copy of Schol. Vat. in the form in which they were attached to the Theonine text; thus Schol. Vind. correspond to BF Vat., where these diverge from P, and especially closely to B. Besides Schol. Vat., the editors of Schol. Vind. used (2) other old collections of scholia of which we find traces in B and F; Schol. Vind. have also some scholia common with b. The scholia which Schol. Vind. have in common with BF come from two different sources, and were apparently afterwards introduced into the other MSS.; one result of this is that several scholia are reproduced twice.

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But, besides the scholia derived from these sources, Schol. Vind. contain a large number of others of late date, characterised by incorrect language or by triviality of content (there are many examples in numbers, citations of propositions used, absurd a)pori/ai, and the like). Unlike Schol. Vat., these scholia often quote words from Euclid as a heading (in one case a heading is inserted in Schol. Vind. where a scholium without the heading is quoted from Schol. Vat., see V. No. 14). The explanations given often presuppose very little knowledge on the part of the reader and frequently contain obscurities and gross errors.

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Schol. Vind. were collected for use with a MS. of the Theonine class; this follows from the fact that they contain a note on the proposition vulgo VII. 22 interpolated by Theon (given in Heiberg's App. to Vol. 11. p. 430). Since the scholium to VII. 39 given in V and p in the text after the title of Book VIII. quotes the proposition as VII. 39, it follows that this scholium must have been written before the interpolation of the two propositions vulgo VII. 20, 22; Schol. Vind. contain (VII. No. 80) the first sentence of it, but without the heading referring to VII. 39. Schol. VII. No. 97 quotes VII. 33 as VII. 34, so that the proposition vulgo VII. 22 may have stood in the scholiast's text but not the later interpolation vulgo VII. 20 (later because only found in B in the margin by the first hand). Of course the scholiast had also the interpolations earlier than Theon.

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For the date of the collection we have a lower limit in the date (12th c.) of MSS. in which the scholia appear. That it was not much earlier than the 12th c. is indicated (1) by the poverty of its contents, (2) by the quality of the MS. of Proclus which was used in the compilation of it (the Munich MS. used by Friedlein with which the scholiast's excerpts are essentially in agreement belongs to the 11th— 12th c.), (3) by the fact that Schol. Vind. appear only in MSS. of the 12th c. and no trace of them is found in our MSS. belonging to the 9th—10th c. in which Schol. Vat. are found. The collection may therefore probably be assigned to the 11th c. Perhaps it may be in part due to Psellus who lived towards the end of that century: for in a Florence MS. (Magliabecch. XI, 53 of the 15th c.) containing a mathematical compendium intended for use in the reading of Aristotle the scholia 1. Nos. 40 and 49 appear with the name of Psellus attached.

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Schol. Vind. are not found without the admixture of foreign elements in any of our three sources. In 1 there are only very few such in the first hand. In q there are several new scholia in the first hand, for the most part due to the copyist himself. The collection of scholia on Book X. in q (Heiberg's qc) is also in the first hand; it is not original, and it may perhaps be due to Psellus (Maglb. has some definitions of Book X. with a heading scholia of...Michael Psellus on the definitions of Euclid's 10th Element - and Schol. X. No. 9), whose name must have been attached to it in the common source of Maglb. and q; to a great extent it consists of extracts from Schol. Vind. taken from the same source as Vl. The scholia q1 (in an ancient hand in q), confined to Book II., partly belong to Schol. Vind. and partly correspond to b1 (Bologna MS.). qa and qb are in one hand (Theodorus Antiochita), the nearest to the first hand of q; they are doubtless due to an early possessor of the MS. of whom we know nothing more.

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Va has, besides Schol. Vind., a number of scholia which also appear in other MSS., one in BFb, some others in P, and some in v (Codex Vat. 1038, 13th c.); these scholia were taken from a source in which many abbreviations were used, as they were often misunderstood by Va. Other scholia in Va which are not found in the older sources—some appearing in Va alone—are also not original, as is proved by mistakes or corruptions which they contain; some others may be due to the copyist himself.

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Vb seldom has scholia common with the other older sources; for the most part they either appear in Vb alone or only in the later sources as v or F2 (later scholia in F), some being original, others not. In Book X. Vb has three series of numerical examples, (1) with Greek numerals, (2) alternatives added later, also mostly with Greek numerals, (3) with Arabic numerals. The last class were probably the work of the copyist himself. These examples (cf. p. 74 below) show the facility with which the Byzantines made calculations at the date of the MS. (12th c.). They prove also that the use of the Arabic numerals (in the East-Arabian form) was thoroughly established in the 12th c.; they were actually known to the Byzantines a century earlier, since they appear, in the first hand, in an Escurial MS. of the 11th c.

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Of collections in other hands in V distinguished by Heiberg (see preface to Vol. v.), V1 has very few scholia which are found in other sources, the greater part being original; V2, V3 are the work of the copyist himself; V4 are so in part only, and contain several scholia from Schol. Vat. and other sources. V3 and V4 are later than 13th—14th c., since they are not found in f (cod. Laurent. XXVIII, 6) which was copied from V and contains, besides Va Vb, the greater part of V1 and VI. No. 20 of V2 (in the text).

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In P there are, besides P3 (a quite late hand, probably one of the old Scriptores Graeci at the Vatican), two late hands (P2), one of which has some new and independent scholia, while the other has added the greater part of Schol. Vind., partly in the margin and partly on pieces of leaves stitched on.

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Our sources for Schol. Vat. also contain other elements. In P there were introduced a certain number of extracts from Proclus, to supplement Schol. Vat. to Book I.; they are all written with a different ink from that used for the oldest part of the MS., and the text is inferior. There are additions in the other sources of Schol. Vat. (F and B) which point to a common source for FB and which are nearly all found in other MSS., and, in particular, in Schol. Vind., which also used the same source; that they are not assignable to Schol. Vat. results only from their not being found in Vat. Of other additions in F, some are peculiar to F and some common to it and b; but they are not original. F2 (scholia in a later hand in F) contains three original scholia; the rest come from V. B contains, besides scholia common to it and F, b or other sources, several scholia which seem to have been put together by Arethas, who wrote at least a part of them with his own hand.

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Heiberg has satisfied himself, by a closer study of b, that the scholia which he denotes by b, b and b1 are by one hand; they are mostly to be found in other sources as well, though some are original. By the same hand (Theodorus Cabasilas, 15th c.) are also the scholia denoted by b2, B2, b3 and B3. These scholia come in great part from Schol. Vind., and in making these extracts Theodorus probably used one of our sources, l, mistakes in which often correspond to those of Theodorus. To one scholium is attached the name of Demetrius (who must be Demetrius Cydonius, a friend of Nicolaus Cabasilas, 14th c.); but it could not have been written by him, since it appears in B and Schol. Vind. Nor are all the scholia which bear the name of Theodorus due to Theodorus himself, though some are so.

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As B3 (a late hand in B) contains several of the original scholia of b2, B3 must have used b itself as his source, and, as all the scholia in B3 are in b, the latter is also the source of the scholia in B3 which are found in other MSS. B and b were therefore, in the 15th c., in the hands of the same person; this explains, too, the fact that b in a late hand has some scholia which can only come from B. We arrive then at the conclusion that Theodorus Cabasilas, in the 15th c., owned both the MSS. B and b, and that he transferred to B scholia which he had before written in b, either independently or after other sources, and inversely transferred some scholia from B to b. Further, B2 are earlier than Theodorus Cabasilas, who certainly himself wrote B3 as well as b2 and b3.

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An author's name is also attached to the scholia VI. No. 6 and X. No. 223, which are attributed to Maximus Planudes (end of 13th c.) along with scholia on I. 31, X. 14 and X. 18 found in 1 in a quite late hand and published on pp. 46, 47 of Heiberg's dissertation. These seem to have been taken from lectures of Planudes on the Elements by a pupil who used l as his copy.

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There are also in l two other Byzantine scholia, written by a late hand, and bearing the names Ioannes and Pediasimus respectively; these must in like manner have been written by a pupil after lectures of Ioannes Pediasimus (first half of 14th c.), and this pupil must also have used l.

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Before these scholia were edited by Heiberg, very few of them had been published in the original Greek. The Basel editio princeps has a few (V. No. I, VI. Nos. 3, 4 and some in Book X.) which are taken, some from the Paris MS. (Paris. Gr. 2343) used by Grynaeus, others probably from the Venice MS. (Marc. 301) also used by him; one published by Heiberg, not in his edition of Euclid but in his paper on the scholia, may also be from Venet. 301, but appears also in Paris. Gr. 2342. The scholia in the Basel edition passed into the Oxford edition in the text, and were also given by August in the Appendix to his Vol. II.

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Several specimens of the two series of scholia (Vat. and Vind.) were published by C. Wachsmuth (Rhein. Mus. XVIII. p. 132 sqq.) and by Knoche (Untersuchungen über die neu aufgefundenen Scholien des Proklus, Herford, 1865).

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The scholia published in Latin were much more numerous. G. Valla (De expetendis et fugiendis rebus, 1501) reproduced apparently some 200 of the scholia included in Heiberg's edition. Several of these he obtained from two Modena MSS. which at one time were in his possession (Mutin. III B, 4 and II E, 9, both of the 15th c.); but he must have used another source as well, containing extracts from other series of scholia, notably Schol. Vind. with which he has some 87 scholia in common. He has also several that are new.

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Commandinus included in his translation under the title Scholia antiqua - the greater part of the Schol. Vat. which he certainly obtained from a MS. of the class of Vat. 192; on the whole he adhered closely to the Greek text. Besides these scholia Commandinus has the scholia and lemmas which he found in the Basel editio princeps, and also three other scholia not belonging to Schol. Vat., as well as one new scholium (to XII. 13) not included in Heiberg's edition, which are distinguished by different type and were doubtless taken from the Greek MS. used by him along with the Basel edition.

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In Conrad Dasypodius' Lexicon mathematicum published in 1573 there is (on fol. 42-44) Graecum scholion in definitiones Euclidis libri quinti elementorum appendicis loco propter pagellas vacantes annexum. - This contains four scholia, and part of two others, published in Heiberg's edition, with some variations of readings, and with some new matter added (for which see pp. 64-6 of Heiberg's pamphlet). The source of these scholia is revealed to us by another work of Dasypodius, Isaaci Monachi Scholia in Euclidis elementorum geometriae sex priores libros per C. Dasypodium in latinum sermonem translata et in lucem edita (1579). This work contains, besides excerpts from Proclus on Book I. (in part closely related to Schol. Vind.), some 30 scholia included in Heiberg's edition, several new scholia, and the above-mentioned scholia to the definitions of Book V. published in Greek in 1573. After the scholia follow Isaaci Monachi prolegomena in Euclidis Elementorum geometriae libros - (two definitions of geometry) and Varia miscellanea ad geometriae cognitionem necessaria ab Isaaco Monacho collecta - (mostly the same as pp. 252, 24-272, 27 in the Variae Collectiones included in Hultsch's Heron); lastly, a note of Dasypodius to the reader says that these scholia were taken ex clarissimi viri Joannis Sambuci antiquo codice manu propria Isaaci Monachi scripto. - Isaak Monachus is doubtless Isaak Argyrus, 14th c.; and Dasypodius used a MS. in which, besides the passage in Hultsch's Variae Collectiones, there were a number of scholia marked in the margin with the name of Isaak (cf. those in b under the name of Theodorus Cabasilas). Whether the new scholia are original cannot be decided until they are published in Greek; but it is not improbable that they are at all events independent arrangements of older scholia. All but five of the others, and all but one of the Greek scholia to Book v., are taken from Schol. Vat.; three of the excepted ones are from Schol. Vind., and the other three seem to come from F (where some words of them are illegible, but can be supplied by means of Mut. III B, 4, which has these three scholia and generally shows a certain likeness to Isaak's scholia).

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Dasypodius also published in 1564 the arithmetical commentary of Barlaam the monk (14th c.) on Eucl. Book II., which finds a place in Appendix IV. to the Scholia in Heiberg's edition.

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Hultsch has some remarks on the origin of the scholiaArt. Eukleides - in Pauly-Wissowa's Real-Encyclopädie.. He observes that the scholia to Book I. contain a considerable portion of Geminus' commentary on the definitions and are specially valuable because they contain extracts from Geminus only, whereas Proclus, though drawing mainly upon him, quotes from others as well. On the postulates and axioms the scholia give more than is found in Proclus. Hultsch conjectures that the scholium on Book v., No. 3, attributing the discovery of the theorems to Eudoxus but their arrangement to Euclid, represents the tradition going back to Geminus, and that the scholium XIII., No. 1, has the same origin.

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A word should be added about the numerical illustrations of Euclid's propositions in the scholia to Book X. They contain a large number of calculations with sexagesimal fractionsHultsch has written upon these in Bibliotheca Mathematica, V_{3}, 1904, pp. 225-233.; the fractions go as far as fourth-sixtieths (1/604). Numbers expressed in these fractions are handled with skill and include some results of surprising accuracyThus \sgrt{(27)} is given (allowing for a slight correction by means of the context) as 5 II' 46'' 10''', which gives for \sgrt{3} the value 1 43' 55'' 23''', being the same value as that given by Hipparchus in his Table of Chords, and correct to the seventh decimal place. Similarly \sgrt{8} is given as 2 49' 42'' 20''' 10'''', which is equivalent to\sgrt{2}=1.41421335. Hultsch gives instances of the various operations, addition, subtraction, etc., carried out in these fractions, and shows how the extraction of the square root was effected. Cf. T. L. Heath, Históry of Greek Mathematics, 1., pp. 59-63.

- - - - CHAPTER VII. - EUCLID IN ARABIA. -

We are told by [Hnull ]ājī KhalfaLexicoṅ bibliogr. et encyclop. ed. Flügel, III. pp. 91, 92. that the Caliph al-Mansūr (754-775) sent a mission to the Byzantine Emperor as the result of which he obtained from him a copy of Euclid among other Greek books, and again that the Caliph al-Ma'mūn (813-833) obtained manuscripts of Euclid, among others, from the Byzantines. The version of the Elements by al-[Hnull ]ajjāj b. Yūsuf b. Matar is, if not the very first, at least one of the first books translated from the Greek into ArabicKlamroth, Zeitschrift der Deutschen Morgenländischen Gesellschaft, XXXV. p. 303.. According to the FihristFihrist (tr. Suter), p. 16. it was translated by al-[Hnull ]ajjāj twice; the first translation was known as Hārūni - (for Hārūn -), the second bore the name Ma'mūni - (for al-Ma'mūn -) and was the more trustworthy. Six Books of the second of these versions survive in a Leiden MS. (Codex Leidensis 399, 1) now in part published by Besthorn and HeibergCodex Leidensis 399, 1. Euclidis Elementa ex interpretatione al-Hadschdschadschii cum commentariis al-Narizii, Hauniae, part 1. i. 1893, part I. ii. 1897, part II. i. 1900, part II. ii. 1905, part III. i. 1910.. In the preface to this MS. it is stated that, in the reign of Hārūn ar-Rashīd (786-809), al-[Hnull ]ajjāj was commanded by Ya[hnull ]yā b. Khālid b. Barmak to translate the book into Arabic. Then, when al-Ma'mūn became Caliph, as he was devoted to learning, al-[Hnull ]ajjāj saw that he would secure the favour of al-Ma'mūn if he illustrated and expounded this book and reduced it to smaller dimensions. He accordingly left out the superfluities, filled up the gaps, corrected or removed the errors, until he had gone through the book and reduced it, when corrected and explained, to smaller dimensions, as in this copy, but without altering the substance, for the use of men endowed with ability and devoted to learning, the earlier edition being left in the hands of readers. -

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The Fihrist goes on to say that the work was next translated by Ishāq b. Hunain, and that this translation was improved by Thābit b. Qurra. This Abū Ya`qūb Is[hnull ]āq b. [Hnull ]unain b. Is[hnull ]āq al-`Ibādī (d. 910) was the son of the most famous of Arabic translators, Hunain b. Ishāq al-`Ibādī (809-873), a Christian and physician to the Caliph alMutawakkil (847-861). There seems to be no doubt that Is[hnull ]āq, who must have known Greek as well as his father, made his translation direct from the Greek. The revision must apparently have been the subject of an arrangement between Is[hnull ]āq and Thābit as the latter died in 901 or nine years before Is[hnull ]āq. Thābit undoubtedly consulted Greek MSS. for the purposes of his revision. This is expressly stated in a marginal note to a Hebrew version of the Elements, made from Ishāq's, attributed to one of two scholars belonging to the same family, viz. either to Moses b. Tibbon (about 1244-1274) or to Jakob b. Machir (who died soon after 1306)Steinschneider, Zeitschrift für Math. u. Physik, XXXI., hist.-litt. Abtheilung, pp. 85, 86, 99.. Moreover Thābit observes, on the proposition which he gives as IX. 31, that he had not found this proposition and the one before it in the Greek but only in the Arabic; from which statement Klamroth draws two conclusions, (1) that the Arabs had already begun to interest themselves in the authenticity of the text and (2) that Thābit did not alter the numbers of the propositions in Ishāq's translationKlamroth, p. 279.. The Fihrist also says that Yu[hnull ]annā al-Qass (i.e. the Priest -) had seen in the Greek copy in his possession the proposition in Book I. which Thābit took credit for, and that this was confirmed by Naẓīf, the physician, to whom Yuhannā had shown it. This proposition may have been wanting in Ishāq, and Thābit may have added it, but without claiming it as his own discoverySteinschneider, p. 88.. As a fact, I. 45 is missing in the translation by al-[Hnull ]ajjāj.

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The original version of Is[hnull ]āq without the improvements by Thābit has probably not survived any more than the first of the two versions by al-Hajjāj; the divergences between the MSS. are apparently due to the voluntary or involuntary changes of copyists, the former class varying according to the degree of mathematical knowledge possessed by the copyists and the extent to which they were influenced by considerations of practical utility for teaching purposesKlamroth, p. 306.. Two MSS. of the Ishāq-Thābit version exist in the Bodleian Library (No. 279 belonging to the year 1238, and No. 280 written in 1260-1)These MSS. are described by Nicoll and Pusey, Catalogus cod. mss. orient. bibl. Bodleianae, pt. II. 1835 (pp. 257-262).; Books I.—XIII. are in the Is[hnull ]āq-Thābit version, the non-Euclidean Books XIV., XV. in the translation of Qustā b. L'ūqā al-Ba`labakkī (d. about 912). The first of these MSS. (No. 279) is that (O) used by Klamroth for the purpose of his paper on the Arabian Euclid. The other MS. used by Klamroth is (K) Kjobenhavn LXXXI, undated but probably of the 13th c., containing Books V.—XV., Books V.—X. being in the Is[hnull ]āq-Thābit version, Books XI.—XIII. purporting to be in al-Hajjāj's translation, and Books XIV., XV. in the version of Qusṭā b. Lūqā. In not a few propositions K and O show not the slightest difference, and, even where the proofs show considerable differences, they are generally such that, by a careful comparison, it is possible to reconstruct the common archetype, so that it is fairly clear that we have in these cases, not two recensions of one translation, but arbitrarily altered and shortened copies of one and the same recensionKlamroth, pp. 306-8.. The Bodleian MS. No. 280 contains a preface, translated by Nîcoll, which cannot be by Thābit himself because it mentions Avicenna (980-1037) and other later authors. The MS. was written at Marāġa in the year 1260-1 and has in the margin readings and emendations from the edition of Naṣĩraddĩn aṭ-[Tnull ]ĩsī (shortly to be mentioned) who was living at Marāġa at the time. Is it possible that aṭ-[Tnull ]ūsī himself is the author of the prefaceSteinschneider, p. 98. Heiberg has quoted the whole of this preface in the Zeitschrift fűr Math. u. Physik, XXIX., hist.-litt. Abth. p. 16.? Be this as it may, the preface is interesting because it throws light on the liberties which the Arabians allowed themselves to take with the text. After the observation that the book (in spite of the labours of many editors) is not free from errors, obscurities, redundancies, omissions etc., and is without certain definitions necessary for the proofs, it goes on to say that the man has not yet been found who could make it perfect, and next proceeds to explain (1) that Avicenna cut out postulates and many definitions - and attempted to clear up difficult and obscure passages, (2) that Abū'l Wafā al-Būzjānĩ (939-997) introduced unnecessary additions and left out many things of great importance and entirely necessary, - inasmuch as he was too long in various places in Book VI. and too short in Book X. where he left out entirely the proofs of the apotomae, while he made an unsuccessful attempt to emend XII. 14, (3) that Abū Ja`far al-Khāzin (d. between 961 and 971) arranged the postulates excellently but disturbed the number and order of the propositions, reduced several propositions to one - etc. Next the preface describes the editor's own claimsThis seems to include a rearrangement of the contents of Books XIV., XV. added to the Elements. and then ends with the sentences, But we have kept to the order of the books and propositions in the work itself (i.e. Euclid's) except in the twelfth and thirteenth books. For we have dealt in Book XIII. with the (solid) bodies and in Book XII. with the surfaces by themselves. -

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After Thābit the Fihrist mentions Abū ’Uthmān ad-Dimashqī as having translated some Books of the Elements including Book X. (It is Abū ’Uthmān's translation of Pappus’ commentary on Book X. which Woepcke discovered at Paris.) The Fihrist adds also that Naẓīf the physician told me that he had seen the tenth Book of Euclid in Greek, that it had 40 propositions more than the version in common circulation which had 109 propositions, and that he had determined to translate it into Arabic. -

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But the third form of the Arabian Euclid actually accessible to us is the edition of Abū Ja`far Mu[hnull ]. b. Mu[hnull ]. b. al-[Hnull ]asan Naṣraddĩn aţ-Ţĩsī (whom we shall call aţ-Ţĩsī for short), born at Tĩs (in Khurāsān) in 1201 (d. 1274). This edition appeared in two forms, a larger and a smaller. The larger is said to survive in Florence only (Pal. 272 and 313, the latter MS. containing only six Books); this was published at Rome in 1594, and, remarkably enough, some copies of this edition are to be found with 12 and some with 13 Books, some with a Latin title and some withoutSuter, Die Mathematiker und Astronomen der Araber, p. 151. The Latin title is Euclidis elementorum geometricorum libri tredecim. Ex traditione doctissimi Nasiridini Tusini nunc primum arabice impressi. Romae in typographia Medicea MDXCIV. Cum licentia superiorum.. But the book was printed in Arabic, so that Kästner remarks that he will say as much about it as can be said about a book which one cannot readKästner, Geschichte der Mathematik, I. p. 367.. The shorter form, which however, in most MSS., is in 15 Books, survives at Berlin, Munich, Oxford, British Museum (974, 1334Suter has a note that this MS. is very old, having been copied from the original in the author's lifetime., 1335), Paris (2465, 2466), India Office, and Constantinople; it was printed at Constantinople in 1801, and the first six Books at Calcutta in 1824Suter, p. 151..

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Aṭ-[Tnull ]ũsĩ's work is however not a translation of Euclid's text, but a re-written Euclid based on the older Arabic translations. In this respect it seems to be like the Latin version of the Elements by Campanus (Campano), which was first published by Erhard Ratdolt at Venice in 1482 (the first printed edition of EuclidDescribed by Kästner, Geschichte der Mathematik, I. pp. 289-299, and by Weissenborn, Die Ubersetzungen des Euklid durch Campano und Zamberti, Halle a. S., 1882, pp. 1-7. See also infra, Chapter VIII, p. 97.). Campanus (13th c.) was a mathematician, and it is likely enough that he allowed himself the same liberty as aṭ-[Tnull ]ĩsī in reproducing Euclid. Whatever may be the relation between Campanus’ version and that of Athelhard of Bath (about 1120), and whether, as Curtze thinksSonderabdruck des Jahresberichtes über die Fortschritte der klassischen Alterthumswissenschaft vom Okt. 1879-1882, Berlin, 1884., they both used one and the same Latin version of 10th—11th c., or whether Campanus used Athelhard's version in the same way as aṭ-[Tnull ]ĩsī used those of his predecessorsKlamroth, p. 271.,it is certain that both versions came from an Arabian source, as is evident from the occurrence of Arabic words in themCurtze, op. cit. p. 20; Heiberg, Euklid-Studien, p. 178.. Campanus’ version is not of much service for the purpose of forming a judgment on the relative authenticity of the Greek and Arabian tradition ; but it sometimes preserves traces of the purer source, as when it omits Theon's addition to VI. 33Heiberg's Euclid, vol. v. p. ci.. A curious circumstance is that, while Campanus’ version agrees with aţ-Ţĩsī's in the number of the propositions in all the genuine Euclidean Books except V. and IX., it agrees with Athelhard's in having 34 propositions in Book V. (as against 25 in other versions), which confirms the view that the two are not independent, and also leads, as Klamroth says, to this dilemma: either the additions to Book V. are Athelhard's own, or he used an Arabian Euclid which is not known to usKlamroth, pp. 273-4.. Heiberg also notes that Campanus’ Books XIV., XV. show a certain agreement with the preface to the Thābit-Is[hnull ]āq version, in which the author claims to have (1) given a method of inscribing spheres in the five regular solids, (2) carried further the solution of the problem how to inscribe any one of the solids in any other and (3) noted the cases where this could not be done.Heiberg, Zeitschrift für Matk. u. Physik, XXIX., hist.-litt. Abtheilung, p. 21.

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With a view to arriving at what may be called a common measure of the Arabian tradition, it is necessary to compare, in the first place, the numbers of propositions in the various Books. [Hnull ]ājĩ Khalfa says that al-[Hnull ]ajjāj's translation contained 468 propositions, and Thābit's 478; this is stated on the authority of aṭ-[Tnull ]ĩsĩ, whose own edition contained 468Klamroth, p. 274; Steinschneider, Zeitschrift für Math. u. Physik, XXXI., hist.-litt. Abth. p. 98.. The fact that Thābit's version had 478 propositions is confirmed by an index in the Bodleian MS. 279 (called O by Klamroth). A register at the beginning of the Codex Leidensis 399, 1 which gives Is[hnull ]aq's numbers (although the translation is that of al-[Hnull ]ajjāj) apparently makes the total 479 propositions (the number in Book XIV. being apparently 11, instead of the 10 of OBesthorn-Heiberg read 11? - as the number, Klamroth had read it as 21 (p. 273).). I subjoin a table of relative numbers taken from Klamroth, to which I have added the corresponding numbers in August's and Heiberg's editions of the Greek text.

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The numbers in the case of Heiberg include all propositions which he has printed in the text; they include therefore XIII. 6 and III. 12 now to be regarded as spurious, and X. 112-115 which he brackets as doubtful. He does not number the propositions in Books XIV., XV., but I conclude that the numbers in P reach at least 9 in XIV., and 9 in XV.

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The Fihrist confirms the number 109 for Book X., from which Klamroth concludes that Is[hnull ]āq's version was considered as by far the most authoritative.

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In the text of O, Book IV. consists of 17 propositions and Book XIV. of 12, differing in this respect from its own table of contents; IV. 15, 16 in O are really two proofs of the same proposition.

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In al-[Hnull ]ajjāj's version Book I. consists of 47 propositions only, I. 45 being omitted. It has also one proposition fewer in Book III., the Heronic proposition III. 12 being no doubt omitted.

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In speaking of particular propositions, I shall use Heiberg's numbering, except where otherwise stated.

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The difference of 10 propositions between Thābit-Is[hnull ]āq and aṭ-[Tnull ]ĩsī is accounted for thus:

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(1) The three propositions VI. 12 and X. 28, 29 which both Is[hnull ]āq and the Greek text have are omitted in aṭ-[Tnull ]ĩsī.

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(2) Is[hnull ]āq divides each of the propositions XIII. 1-3 into two, making six instead of three in aṭ-[Tnull ]ĩsī and in the Greek.

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(3) Is[hnull ]āq has four propositions (numbered by him VIII. 24, 25, IX. 30, 31) which are neither in the Greek Euclid nor in aţ-Ţĩsī.

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Apart from the above differences al-[Hnull ]ajjāj (so far as we know), Ishāq and aṭ-[Tnull ]ĩsī agree; but their Euclid shows many differences from our Greek text. These differences we will classify as followsSee Klamroth, pp. 275-6, 280, 282-4, 314-15, 326;Heiberg, vol. v. pp. xcvi, xcvii..

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1. - Propositions.

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The Arabian Euclid omits VII. 20, 22 of Gregory's and August's editions (Heiberg, App. to Vol. II. pp. 428-32); VIII. 16, 17; X. 7, 8, 13, 16, 24, 112, 113, 114, besides a lemma vulgo X. 13, the proposition X. 117 of Gregory's edition, and the scholium at the end of the Book (see for these Heiberg's Appendix to Vol. III. pp. 382, 408-416); XI. 38 in Gregory and August (Heiberg, App. to Vol. IV. p. 354); XII. 6, 13, 14; (also all but the first third of Book XV.).

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The Arabian Euclid makes III. 11, 12 into one proposition, and divides some propositions (X. 31, 32; XI. 31, 34; XIII. 1-3) into two each.

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The order is also changed in the Arabic to the following extent. V. 12, 13 are interchanged and the order in Books VI., VII., IX.— XIII. is:

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VI. 1-8, 13, 11, 12, 9, 10, 14-17, 19, 20, 18, 21, 22, 24, 26, 23, 25, 27-30, 32, 31, 33.

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VII. 1-20, 22, 21, 23-28, 31, 32, 29, 30, 33-39.

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IX. 1-13, 20, 14-19, 21-25, 27, 26, 28-36, with two new propositions coming before prop. 30.

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X. 1-6, 9-12, 15, 14, 17-23, 26-28, 25, 29-30, 31, 32, 33— 111, 115.

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XI. 1-30, 31, 32, 34, 33, 35-39.

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XII. 1-5, 7, 9, 8, 10, 12, 11, 15, 16-18.

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XIII. 1-3, 5, 4, 6, 7, 12, 9, 10, 8, 11, 13, 15, 14, 16-18.

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2. - Definitions.

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The Arabic omits the following definitions : IV. Deff. 3-7, VII. Def. 9 (or 10), XI. Deff. 5-7, 15, 17, 23, 25-28; but it has the spurious definitions VI. Deff. 2, 5, and those of proportion and ordered proportion in Book V. (Deff. 8, 19 August), and wrongly interchanges V. Deff. 11, 12 and also VI. Deff. 3, 4.

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The order of the definitions is also different in Book VII. where, after Def. 11, the order is 12, 14, 13, 15, 16, 19, 20, 17, 18, 21, 22, 23, and in Book XI. where the order is 1, 2, 3, 4, 8, 10, 9, 13, 14, 16, 12, 21, 22, 18, 19, 20, 11, 24.

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3. - Lemmas and porisms.

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All are omitted in the Arabic except the porisms to VI. 8, VIII. 2, X. 3; but there are slight additions here and there, not found in the Greek, e.g. in VIII. 14, 15 (in K).

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4. - Alternative proofs.

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These are all omitted in the Arabic, except that in X. 105, 106 they are substituted for the genuine proofs; but one or two alternative proofs are peculiar to the Arabic (VI. 32 and VIII. 4, 6).

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The analyses and syntheses to XIII. 1-5 are also omitted in the Arabic.

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Klamroth is inclined, on a consideration of all these differences, to give preference to the Arabian tradition over the Greek (1) on historical grounds, - subject to the proviso that no Greek MS. as ancient as the 8th c. is found to contradict his conclusions, which are based generally (2) on the improbability that the Arabs would have omitted so much if they had found it in their Greek MSS., it being clear from the Fihrist that the Arabs had already shown an anxiety for a pure text, and that the old translators were subjected in this matter to the check of public criticism. Against the historical grounds, - Heiberg is able to bring a considerable amount of evidenceHeiberg in Zeitschrift für Math. u. Physik, XXIX., hist.-litt. Abth. p. 3 sqq.. First of all there is the British Museum palimpsest (L) of the 7th or the beginning of the 8th c. This has fragments of propositions in Book X. which are omitted in the Arabic; the numbering of one proposition, which agrees with the numbering in other Greek MS., is not comprehensible on the assumption that eight preceding propositions were omitted in it, as they are in the Arabic; and lastly, the readings in L are tolerably like those of our MSS., and surprisingly like those of B. It is also to be noted that, although P dates from the 10th c. only, it contains, according to all appearance, an ante-Theonine recension.

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Moreover there is positive evidence against certain omissions by the Arabians. Aṭ-[Tnull ]ĩsī omits VI. 12, but it is scarcely possible that, if Eutocius had not had it, he would have quoted VI. 23 by that numberApollonius, ed. Heiberg, vol. 11. p. 218, 3-5.. This quotation of VI. 23 by Eutocius also tells against Is[hnull ]āq who has the proposition as VI. 25. Again, Simplicius quotes VI. 10 by that number, whereas it is VI. 13 in Is[hnull ]āq; and Pappus quotes, by number, XIII. 2 (Is[hnull ]āq 3, 4), XIII. 4 (Is[hnull ]āq 8), XIII. 16 (Is[hnull ]āq 19). On the other hand the contraction of III. 11, 12 into one proposition in the Arabic tells in favour of the Arabic.

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Further, the omission of certain porisms in the Arabic cannot be supported; for Pappus quotes the porism to XIII. 17Pappus, V. p. 436, 5., Proclus those to II. 4, III. 1, VII. 2Proclus, pp. 303-4., and Simplicius that to IV. 15.

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Lastly, some propositions omitted in the Arabic are required in later propositions. Thus X. 13 is used in X. 18, 22, 23, 26 etc.; X. 17 is wanted in X. 18, 26, 36; XII. 6, 13 are required for XII. 11 and XII. 15 respectively.

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It must also be remembered that some of the things which were properly omitted by the Arabians are omitted or marked as doubtful in Greek MSS. also, especially in P, and others are rightly suspected for other reasons (e.g. a number of alternative proofs, lemmas, and porisms, as well as the analyses and syntheses of XIII. 1-5). On the other hand, the Arabic has certain interpolations peculiar to our inferior MSS. (cf. the definition VI. Def. 2 and those of proportion and ordered proportion).

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Heiberg comes to the general conclusion that, not only is the Arabic tradition not to be preferred offhand to that of the Greek MSS., but it must be regarded as inferior in authority. It is a question how far the differences shown in the Arabic are due to the use of Greek MSS. differing from those which have been most used as the basis of our text, and how far to the arbitrary changes made by the Arabians themselves. Changes of order and arbitrary omissions could not surprise us, in view of the preface above quoted from the Oxford MS. of Thābit-Is[hnull ]āq, with its allusion to the many important and necessary things left out by Abĩ'l Wafā and to the author's own rearrangement of Books XII., XIII. But there is evidence of differences due to the use by the Arabs of other Greek MSS. HeibergZeitschrift für Math. u. Physik, XXIX., hist.-litt. Abth. p. 6 sqq. is able to show considerable resemblances between the Arabic text and the Bologna MS. b in that part of the MS. where it diverges so remarkably from our other MSS. (see the short description of it above, p.49); in illustration he gives a comparison of the proofs of XII. 7 in b and in the Arabic respectively, and points to the omission in both of the proposition given in Gregory's edition as XI. 38, and to a remarkable agreement between them as regards the order of the propositions of Book XII. As above stated, the remarkable divergence of b only affects Books XI. (at end) and XII.; and Book XIII. in b shows none of the transpositions and other peculiarities of the Arabic. There are many differences between b and the Arabic, especially in the definitions of Book XI., as well as in Book XIII. It is therefore a question whether the Arabians made arbitrary changes, or the Arabic form is the more ancient, and b has been altered through contact with other MSS. Heiberg points out that the Arabians must be alone responsible for their definition of a prism, which only covers a prism with a triangular base. This could not have been Euclid's own, for the word prism already has the wider meaning in Archimedes, and Euclid himself speaks of prisms with parallelograms and polygons as bases (XI. 39; XII. 10). Moreover, a Greek would not have been likely to leave out the definitions of the Platonic - regular solids.

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Heiberg considers that the Arabian translator had before him a MS. which was related to b, but diverged still further from the rest of our MSS. He does not think that there is evidence of the existence of a redaction of Books I.—X. similar to that of Books XI., XII. in b; for Klamroth observes that it is the Books on solid geometry (XI.—XIII.) which are more remarkable than the others for omissions and shorter proofs, and it is a noteworthy coincidence that it is just in these Books that we have a divergent text in b.

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An advantage in the Arabic version is the omission of VII. Def. 10, although, as Iamblichus had it, it may have been deliberately omitted by the Arabic translator. Another advantage is the omission of the analyses and syntheses of XIII. 1-5; but again these may have been omitted purposely, as were evidently a number of porisms which are really necessary.

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One or two remarks may be added about the Arabic versions as compared with one another. Al-[Hnull ]ajjāj's object seems to have been less to give a faithful reflection of the original than to write a useful and convenient mathematical text-book. One characteristic of it is the careful references to earlier propositions when their results are used. Such specific quotations of earlier propositions are rare in Euclid; but in al-[Hnull ]ajjāj we find not only such phrases as by prop. so and so, - which was proved - or which we showed how to do in prop. so and so, - but also still longer phrases. Sometimes he kepeats a construction, as in I. 44 where, instead of constructing the parallelogram BEFG equal to the triangle C in the angle EBG which is equal to the angle D - and placing it in a certain position, he produces AB to G, making BG equal to half DE (the base of the triangle CDE in his figure), and on GB so constructs the parallelogram BHKG by I. 42 that it is equal to the triangle CDE, and its angle GBH is equal to the given angle.

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Secondly, al-[hnull ]ajjāj, in the arithmetical books, in the theory of proportion, in the applications of the Pythagorean I. 47, and generally where possible, illustrates the proofs by numerical examples. It is true, observes Klamroth, that these examples are not apparently separated from the commentary of an-Nairīzī, and might not therefore have been due to al-[Hnull ]ajjāj himself; but the marginal notes to the Hebrew translation in Municn MS. 36 show that these additions were in the copy of al-[Hnull ]ajjāj used by the translator, for they expressly give these proofs in numbers as variants taken from al-[Hnull ]ajjājKlamroth, p. 310; Steinschneider, pp. 85-6..

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These characteristics, together with al-[Hnull ]ájjāj's freer formulation of the propositions and expansion of the proofs, constitute an intelligible reason why Is[hnull ]āq should have undertaken a fresh translation from the Greek. Klamroth calls Is[hnull ]āq's version a model of a good translation of a mathematical text; the introductory and transitional phrases are stereotyped and few in number, the technical terms are simply and consistently rendered, and the less formal expressions connect themselves as closely with the Greek as is consistent with intelligibility and the character of the Arabic language. Only in isolated cases does the formulation of definitions and enunciations differ to any considerable extent from the original. In general, his object seems to have been to get rid of difficulties and unevennesses in the Greek text by next devices, while at the same time giving a faithful reproduction of it.Klamroth, p. 290, illustrates is[hnull ]āq's method by his way of distinguishing e)farmo/zein (to be congruent with) and e)farmo/zesqai (to be applied to), the confusion of which by translators was animadverted on by Savile. Is[hnull ]āq avoided the confusion by using two entirely different words..

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There are curious points of contact between the versions of al-[Hnull ]ajjāj and T[hnull ]amacr;bit-Is[hnull ]āq. For example, the definitions and enunciations of propositions are often word for word the same. Presumably this is owing to the fact that Is[hnull ]āq found these definitions and enunciations already established in the schools in his time, where they would no doubt be learnt by heart, and refrained from translating them afresh, merely adopting the older version with some changesKlamroth, pp. 310-1.. Secondly, there is remarkable agreement between the Arabic versions as regards the figures, which show considerable variations from the figures of the Greek text, especially as regards the letters; this is also probably to be explained in the same way, all the later translators having most likely borrowed al-[hnull ]ajjāj's adaptation of the Greek figuresibid. p. 287.. Lastly, it is remarkable that the version of Books XI.—XIII. in the Kjfbenhavn MS. (K), purporting to be by al-[Hnull ]ajjāj, is almost exactly the same as the Thābit-Is[hnull ]āq version of the same Books in O. Klamroth conjectures that Is[hnull ]āq may not have translated the Books on solid geometry at all, and that Thābit took them from al-[Hnull ]ajjāj, only making some changes in order to fit them to the translation of Is[hnull ]āqibid. pp. 304-5..

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From the facts (1) that aṭ-[Tnull ]ĩsī's edition had the same number of propositions (468) as al-[hnull ]ajjāj's version, while Thābit-Is[hnull ]āq's had 478, and (2) that aṭ-[Tnull ]ĩsī has the same careful references to earlier propositions, Klamroth concludes that aṭ-[Tnull ]ĩsī deliberately preferred al-[hnull ]ajjāj's version to that of Is[hnull ]āqibid. p. 274.. Heiberg, however, points out (1) that aṭ-[Tnull ]ĩsī left out VI. 12 which, if we may judge by Klamroth's silence, al-[hnull ]ajjāj had, and (2) al-[hnull ]ajjāj's version had one proposition less in Books I. and III. than aṭ-[Tnull ]ĩsī has. Besides, in a passage quoted by [hnull ]ājī Khalfa[hnull ]ājī Khalfa, I. p. 383. from aṭ-[Tnull ]ĩsī, the latter says that he separated the things which, in the approved editions, were taken from the archetype from the things which had been added thereto, - indicating that he had compiled his edition from both the earlier translationsHeiberg, Zeitschrift für Math. u. Physik, XXIX., hist.-litt. Abth. pp. 2, 3..

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There were a large number of Arabian commentaries on, or reproductions of, the Elements or portions thereof, which will be found fully noticed by SteinschneiderSteinschneider, Zeitschrift für Math. u. Physik, XXXI., hist.-litt. Abth. pp. 86 sqq.. I shall mention here the commentators etc. referred to in the Fihrist, with a few others.

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1. Abĩ 'l `Abbās al-Fa[dnull ]l b. [hnull ]ātim an-Nairīzī (born at Nairīz, died about 922) has already been mentionedSteinschneider, p. 86, Fihrist (tr. Suter), pp. 16, 67; Suter, Die Mathematiker und Astronomen der Araber (1900), p. 45.. His commentary survives, as regards Books I.—VI., in the Codex Leidensis 399, I, now edited, as to four Books, by Besthorn and Heiberg, and as regards Books I.—X. in the Latin translation made by Gherard of Cremona in the 12th c. and now published by Curtze from a Cracow MSSupplementum ad Euclidis opera omnia, ed. Heiberg and Menge, Leipzig, 1899.. Its importance lies mainly in the quotations from Heron and Simplicius.

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2. Ahmad b. `Umar al-Karābīsī (date uncertain, probably 9th— 10th c.), who was among the most distinguished geometers and arithmeticiansFihrist, pp. 16, 38; Steinschneider, p. 87; Suter, p. 65.. -

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3. Al-'Abbās b. Sa`īd al-Jauharī (fl. 830) was one of the astronomical observers under al-Ma'mĩn, but devoted himself mostly to geometry. He wrote a commentary to the whole of the Elements, from the beginning to the end; also the Book of the propositions which he added to the first book of EuclidFihrist, pp. 16, 25; Steinschneider, p. 88; Suter, p. 12.. -

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4. Muh. b. `Īsā Abĩ `Abdallāh al-Māhānī (d. between 874 and 884) wrote, according to the Fihrist, (I) a commentary on Eucl. Book v., (2) On proportion, - (3) On the 26 propositions of the first Book of Euclid which are proved without reductio ad absurdumFihrist, pp. 16, 25, 58.. - The work On proportion - survives and is probably identical with, or part of, the commentary on Book v.Suter, p. 26, note, quotes the Paris MS. 2467, 16o containing the work on proportion - as the authority for this conjecture. He also wrote, what is not mentioned by the Fihrist, a commentary on Eucl. Book X., a fragment of which survives in a Paris MS.MS. 2457, 39o (cf. Woepcke in Me/m. pre/s. agrave; l'acad. des sciences, XIV., 1856, p. 669).

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5. Abĩ Ja`far al-Khāzin (i.e. the treasurer - or librarian -), one of the first mathematicians and astronomers of his time, was born in Khurāsān and died between the years 961 and 971. The Fihrist speaks of him as having written a commentary on the whole of the ElementsFihrist, p. 17., but only the commentary on the beginning of Book X. survives (in Leiden, Berlin and Paris); therefore either the notes on the rest of the Books have perished, or the Fihrist is in errorSuter, p. 58, note b.. The latter would seem more probable, for, at the end of his commentary, al-Khāzin remarks that the rest had already been commented on by Sulaimān b. 'Usma (Leiden MS.)Steinschneider, p. 89. or 'Oqba (Suter), to be mentioned below. Al-Khāzin's method is criticised unfavourably in the preface to the Oxford MS. quoted by Nicoll (see p. 77 above).

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6. Abĩ'l Wafā al-Bĩzjānī (940-997), one of the greatest Arabian mathematicians, wrote a commentary on the Elements, but did not complete itFihrist, p. 17.. His method is also unfavourably regarded in the same preface to the Oxford MS. 280. According to Hājī Khalfa, he also wrote a book on geometrical constructions, in thirteen chapters. Apparently a book answering to this description was compiled by a gifted pupil from lectures by Abĩ 'l Wafā, and a Paris MS. (Anc. fonds 169) contains a Persian translation of this work, not that of Abĩ 'l Wafā himself. An analysis of the work was given by WoepckeWoepcke, Fournal Asiatique, Sér. v. T. v. pp. 218-256 and 309-359., and some particulars will be found in CantorGesch. d. Math. vol. 13, pp. 743-6.. Abĩ 'l Wafā also wrote a commentary on Diophantus, as well as a separate book of proofs to the propositions which Diophantus used in his book and to what he (Abĩ 'l Wafā) employed in his commentaryFihrist, p. 39; Suter, p. 71.. -

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7. Ibn Rāhawaihi al-Arjānī also commented on Eucl. Book X.Fihrist, p. 17; Suter, p. 17..

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8. `Alī b. Ahmad Abĩ 'l-Qāsim al-Anṭākī (d. 987) wrote a commentary on the whole bookFihrist, p. 17.; part of it seems to survive (from the 5th Book onwards) at Oxford (Catal. MSS. orient. II. 281)Suter, p. 64..

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9. Sind b. `Alī Abĩ 'ṭ-[Tnull ]aiyib was a Jew who went over to Islam in the time of al-Ma'mĩn, and was received among his astronomical observers, whose head he becameFihrist, p. 17, 29; Suter, pp. 13, 14. (about 830); he died after 864. He wrote a commentary on the whole of the Elements; Abĩ `Alī saw nine books of it, and a part of the tenthFihrist, p. 17.. - His book On the Apotomae and the Medials, - mentioned by the Fihrist, may be the same as, or part of, his commentary on Book X.

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10. Abĩ Yĩsuf Ya`qĩb b. Muh. ar-Rāzī wrote a commentary on Book X., and that an excellent one, at the instance of Ibn al`AmīdFihrist, p. 17; Suter, p. 66.. -

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11. The Fihrist next mentions al-Kindī (Abĩ Yĩsuf Ya'qĩb b. Ishāq b. as-Sabbāh al-Kindī, d. about 873), as the author (1) of a work on the objects of Euclid's book, - in which occurs the statement that the Elements were originally written by Apollonius, the carpenter (see above, p. 5 and note), (2) of a book on the improvement of Euclid's work, - and (3) of another on the improvement of the 14th and 15th Books of Euclid. - He was the most distinguished man of his time, and stood alone in the knowledge of the old sciences collectively; he was called ’the philosopher of the Arabians’; his writings treat of the most different branches of knowledge, as logic, philosophy, geometry, calculation, arithmetic, music, astronomy and othersFihrist, p. 17, 10-15.. - Among the other geometrical works of al-Kindī mentioned by the FihristThe mere catalogue of al-Kindī's works on the various branches of science takes up four octavo pages (11-15) of Suter's translation of the Fihrist. are treatises on the closer investigation of the results of Archimedes concerning the measure of the diameter of a circle in terms of its circumference, on the construction of the figure of the two mean proportionals, on the approximate determination of the chords of the circle, on the approximate determination of the chord (side) of the nonagon, on the division of triangles and quadrilaterals and constructions for that purpose, on the manner of construction of a circle which is equal to the surface of a given cylinder, on the division of the circle, in three chapters etc.

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12. The physician Naẓīf b. Yumn (or Yaman) al-Qass (the priest -) is mentioned by the Fihrist as having seen a Greek copy of Eucl. Book X. which had 40 more propositions than that which was in general circulation (containing 109), and having determined to translate it into ArabicFihrist, pp. 16, 17.. Fragments of such a translation exist at Paris, Nos. 18 and 34 of the MS. 2457 (952, 2 Suppl. Arab. in Woepcke's tract); No. 18 contains additions to some propositions of the 10th Book, existing in the Greek languageWoepcke, Mém. prés. à l'acad. des sciences, XIV. pp. 666, 668.. - Nazĩf must have died about 990Suter, p. 68..

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13. Yũ[hnull ]annã b. Yūsuf b. al-[Hnull ]ãrith b. al-Bitrĩq al-Qass (d. about 980) lectured on the Elements and other geometrical books, made translations from the Greek, and wrote a tract on the proof - of the case of two straight lines both meeting a third and making with it, on one side, two angles together less than two right anglesFihrist, p. 38; Suter, p. 60.. Nothing of his appears to survive, except that a tract on rational and irrational magnitudes, - No. 48 in the Paris MS. just mentioned, is attributed to him.

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14. Abũ Mu[hnull ]. al-[Hnull ]asan b. 'Ubaidallãh b. Sulaimãn b. Wahb (d. 901) was a geometer of distinction, who wrote works under the two distinct titles A commentary on the difficult parts of the work of Euclid - and The Book on ProportionFihrist, p. 26, and Suter's note, p. 60.. - Suter thinks that another reading is possible in the case of the second title, and that it may refer to the Euclidean work on the divisions (of figures)Suter, p. 211, note 23.. -

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15. Qustã b. Lũqã al-Ba'labakkĩ (d. about 912), a physician, philosopher, astronomer, mathematician and translator, wrote on the difficult passages of Euclid's book - and on the solution of arithmetical problems from the third book of EuclidFihrist, p. 43. -; also an introduction to geometry, - in the form of question and answerFihrist, p. 43; Suter, p. 41..

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16. Thãbit b. Qurra (826-901), besides translating some parts of Archimedes and Books V.mdashVII. of the Conics of Apollonius, and revising Ishãq's translation of Euclid's Elements, also revised the translation of the Data by the same Ishãq and the book On divisions of figures translated by an anonymous writer. We are told also that he wrote the following works: (I) On the Premisses (Axioms, Postulates etc.) of Euclid, (2) On the Propositions of Euclid, (3) On the propositions and questions which arise when two straight lines are cut by a third (or on the proof - of Euclid's famous postulate). The last tract is extant in the MS. discovered by Woepcke (Paris 2457, 32o). He is also credited with an excellent work - in the shape of an Introduction to the Book of Euclid, - a treatise on Geometry dedicated to Ismã'il b. Bulbul, a Compendium of Geometry, and a large number of other works for the titles of which reference may be made to Suter, who also gives particulars as to which are extantSuter, pp. 34-8..

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17. Abũ Sa'ĩd Sinãn b. Thãbit b. Qurra, the son of the translator of Euclid, followed in his father's footsteps as geometer, astronomer and physician. He wrote an improvement of the book of ...... on the Elements of Geometry, in which he made various additions to the original. - It is natural to conjecture that Euclid is the name missing in this description (by Ibn abĩ Uṣaibi'a); Casiri has the name AqãtonFihrist (ed. Suter), p. 59, note 132; Suter, p. 52, note b.. The latest editor of the Ta'rĩkh al-[Hnull ]ukamã, however, makes the name to be Iflãton (=Plato), and he refers to the statement by the Fihrist and Ibn al-Qiftĩ attributing to Plato a work on the Elements of Geometry translated by Qusṭã. It is just possible, therefore, that at the time of Qusṭã the Arabs were acquainted with a book on the Elements of Geometry translated from the Greek, which they attributed to PlatoSee Suter in Bibliotheca Mathematica, IV_{3}, 1903-4, pp. 296-7, review of Julius Lippert's Ibn al-Qiftĩ. Ta'rĩch al-hukamã, Leipzig, 1903.. Sinãn died in 943.

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18. Abũ Sahl Wĩjan (or Waijan) b. Rustam al-Kũhĩ (fl. 988), born at Kũh in [Tnull ]abaristãn, a distinguished geometer and astronomer, wrote, according to the Fihrist, a Book of the Elements - after that of EuclidFihrist, p. 40.; the 1st and 2nd Books survive at Cairo, and a part of the 3rd Book at Berlin (5922)Suter, p. 75.. He wrote also a number of other geometrical works: Additions to the 2nd Book of Archimedes on the Sphere and Cylinder (extant at Paris, at Leiden, and in the India Office), On the finding of the side of a heptagon in a circle (India Office and Cairo), On two mean proportionals (India Office), which last may be only a part of the Additions to Archimedes' On the Sphere and Cylinder, etc.

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19. Abũ Naṣr Mu[hnull ]. b. Mu[hnull ]. b. [Tnull ]arkhãn b. Uzlaġ al-Fãrãbĩ (870-950) wrote a commentary on the difficulties of the introductory matter to Books I. and V.Suter, p. 55. This appears to survive in the Hebrew translation which is, with probability, attributed to Moses b. TibbonSteinschneider, p. 92..

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20. Abũ 'Alĩ al-Hasan b. al-Hasan b. al-Haitham (about 9651039), known by the name Ibn al-Haitham or Abũ 'Alĩ al-Basrĩ, was a man of great powers and knowledge, and no one of his time approached him in the field of mathematical science. He wrote several works on Euclid the titles of which, as translated by Woepcke from Usaibi'a, are as followsSteinschneider, pp. 92-3.: -

1. Commentary and abridgment of the Elements.

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2. Collection of the Elements of Geometry and Arithmetic, drawn from the treatises of Euclid and Apollonius.

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3. Collection of the Elements of the Calculus deduced from the principles laid down by Euclid in his Elements.

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4. Treatise on measure - after the manner of Euclid's Elements.

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5. Memoir on the solution of the difficulties in Book I.

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6. Memoir for the solution of a doubt about Euclid, relative to Book V.

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7. Memoir on the solution of a doubt about the stereometric portion.

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8. Memoir on the solution of a doubt about Book XII.

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9. Memoir on the division of the two magnitudes mentioned in X. I (the theorem of exhaustion).

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10. Commentary on the definitions in the work of Euclid (where Steinschneider thinks that some more general expression should be substituted for definitions -).

-

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The last-named work (which Suter calls a commentary on the Postulates of Euclid) survives in an Oxford MS. (Catal. MSS. orient. I. 908) and in Algiers (1446, 1o).

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A Leiden MS. (966) contains his Commentary on the difficult places - up to Book V. We do not know whether in this commentary, which the author intended to form, with the commentary on the Musãdarãt, a sort of complete commentary, he had collected the separate memoirs on certain doubts and difficult passages mentioned in the above list.

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A commentary on Book V. and following Books found in a Bodleian MS. (Catal. II. p. 262) with the title Commentary on Euclid and solution of his difficulties - is attributed to b. Haitham; this might be a continuation of the Leiden MS.

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The memoir on X. 1 appears to survive at St Petersburg, MS. de l'Institut des langues orient. 192, 5o (Rosen, Catal. p. 125).

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21. Ibn Sĩnã, known as Avicenna (980-1037), wrote a Compendium of Euclid, preserved in a Leiden MS. No. 1445, and forming the geometrical portion of an encyclopaedic work embracing Logic, Mathematics, Physics and MetaphysicsSteinschneider, p. 92; Suter, p. 89..

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22. Ahmad b. al-Husain al-Ahwãzĩ al-Kãtib wrote a commentary on Book X., a fragment of which (some 10 pages) is to be found at Leiden (970), Berlin (5923) and Paris (2467, 18o)Suter, p. 57..

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23. Naṣĩraddĩn aṭ-[Tnull ]ũsĩ (1201-1274) who, as we have seen, brought out a Euclid in two forms, wrote: -

1. A treatise on the postulates of Euclid (Paris, 2467, 5o).

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2. A treatise on the 5th postulate, perhaps only a part of the foregoing (Berlin, 5942, Paris, 2467, 6o).

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3. Principles of Geometry taken from Euclid, perhaps identical with No. 1 above (Florence, Pal. 298).

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4. 105 problems out of the Elements (Cairo). He also edited the Data (Berlin, Florence, Oxford etc.)Suter, pp. 150-1..

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24. Muh. b. Ashraf Shamsaddĩn as-Samarqandĩ (fl. 1276) wrote Fundamental Propositions, being elucidations of 35 selected propositions of the first Books of Euclid, - which are extant at Gotha (1496 and 1497), Oxford (Catal. I. 967, 2o), and Brit. Mus.Suter, p. 157..

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25. Mũsã b. Mu[hnull ]. b. Ma[hnull ]mũd, known as Qã[dnull ]ĩzãde ar-Rũmĩ (i.e. the son of the judge from Asia Minor), who died between 1436 and 1446, wrote a commentary on the Fundamental Propositions - just mentioned, of which many MSS. are extantibid. p. 175.. It contained biographical statements about Euclid alluded to above (p. 5, note).

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26. Abũ Dã'ũd Sulaimãn b. ’Uqba, a contemporary of al-Khãzin (see above, No. 5), wrote a commentary on the second half of Book X., which is, at least partly, extant at Leiden (974) under the title On the binomials and apotomae found in the 10th Book of Euclidibid. p. 56.. -

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27. The Codex Leidensis 399, 1 containing al-[Hnull ]ajjãj's translation of Books I.—VI. is said to contain glosses to it by Sa`ĩd b. Mas'ũd b. al-Qass, apparently identical with Abũ Nasr Gars al-Na'ma, son of the physician Mas'ũd b. al-Qass al-Bagdãdĩ, who lived in the time of the last Caliph al-Musta'sim (d. 1258)ibid. pp. 153-4, 227..

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28. Abũ Mu[hnull ]ammad b. Abdalbãqĩ al-Bag[dnull ]ãdĩ al-Fara[dnull ]ĩ (d. 1141, at the age of over 70 years) is stated in the Ta'rĩkh al-[Hnull ]ukamã to have written an excellent commentary on Book X. of the Elements, in which he gave numerical examples of the propositionsGartz, p. 14; Steinschneider, pp. 94-5.. This is published in Curtze's edition of an-Nairĩzĩ where it occupies pages 252-386Suter in Bibliotheca Mathematica, IV3, 1903, pp. 25, 295; Suter has also an article on its contents, Bibliotheca Mathematica, VII3, 1906-7, pp. 234-251..

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29. Ya[hnull ]yã b. Mu[hnull ]. b. 'Abdãn b. 'Abdalwã[hnull ]id, known by the name of Ibn al-Lubũdĩ (1210-1268), wrote a Compendium of Euclid, and a short presentation of the postulatesSteinschneider, p. 94; Suter, p. 146..

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30. Abũ 'Abdallãh Mu[hnull ]. b. Mu'ãdh al-Jayyãnĩ wrote a commentary on Eucl. Book V. which survives at Algiers (1446, 3o)Suter, Nachträge und Berichtigungen, in Abhandlungen zur Gesch. der math. Wissenschaften, XIV., 1902, p. 170..

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31. Abũ Naṣr Mansũr b. 'Aliã b.'Irãq wrote, at the instance of Mu[hnull ]. b. A[hnull ]mad Abũ ’r-Rai[hnull ]ãn al-Bĩrũnĩ (973-1048), a tract on a doubtful (difficult) passage in Eucl. Book XIII. - (Berlin, 5925). Suter, p. 81, and Nachträge, p. 172..

- - - - CHAPTER VIII. - PRINCIPAL TRANSLATIONS AND EDITIONS OF THE ELEMENTS. -

Cicero is the first Latin author to mention EuclidDe oratore III. 132.; but it is not likely that in Cicero's time Euclid had been translated into Latin or was studied to any considerable extent by the Romans; for, as Cicero says in another placeTusc. 1. 5., while geometry was held in high honour among the Greeks, so that nothing was more brilliant than their mathematicians, the Romans limited its scope by having regard only to its utility for measurements and calculations. How very little theoretical geometry satisfied the Roman agrimensores is evidenced by the work of Balbus de mensurisGromatici veteres, I. 97 sq. (ed. F. Blume, K. Lachmann and A. Rudorff. Berlin, 1848, 1852)., where some of the definitions of Eucl. Book I. are given. Again, the extracts from the Elements found in the fragment attributed to Censorinus (fl. 238 A.D)Censorinus, ed. Hultsch, pp. 60-3. are confined to the definitions, postulates, and common notions. But by degrees the Elements passed even among the Romans into the curriculum of a liberal education; for Martianus Capella speaks of the effect of the enunciation of the proposition how to construct an equilateral triangle on a given straight line - among a company of philosophers, who, recognising the first proposition of the Elements, straightway break out into encomiums on EuclidMartianus Capella, VI. 724.. But the Elements were then (c. 470 A.D.) doubtless read in Greek; for what Martianus Capella givesibid. VI. 708 sq. was drawn from a Greek source, as is shown by the occurrence of Greek words and by the wrong translation of I. def. 1 (punctum vero est cuius pars nihil est -). Martianus may, it is true, have quoted, not from Euclid himself, but from Heron or some other ancient source.

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But it is clear from a certain palimpsest at Verona that some scholar had already attempted to translate the Elements into Latin. This palimpsestCf. Cantor, I_{3}, p. 565. has part of the Moral reflections on the Book of Job - by Pope Gregory the Great written in a hand of the 9th c. above certain fragments which in the opinion of the best judges date from the 4th c. Among these are fragments of Vergil and of Livy, as well as a geometrical fragment which purports to be taken from the 14th and 15th Books of Euclid. As a matter of fact it is from Books XII. and XIII. and is of the nature of a free rendering, or rather a new arrangement, of Euclid with the propositions in different orderThe fragment was deciphered by W. Studemund, who communicated his results to Cantor.. The MS. was evidently the translator's own copy, because some words are struck out and replaced by synonyms. We do not know whether the translator completed the translation of the whole, or in what relation his version stood to our other sources.

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Magnus Aurelius Cassiodorus (b. about 475 A.D.) in the geometrical part of his encyclopaedia De artibus ac disciplinis liberalium literarum says that geometry was represented among the Greeks by Euclid, Apollonius, Archimedes, and others, of whom Euclid was given us translated into the Latin language by the same great man Boethius -; also in his collection of lettersCassiodorus, Variae, I. 45, p. 40, 12 ed. Mommsen. is a letter from Theodoric to Boethius containing the words, for in your translations...Nicomachus the arithmetician, and Euclid the geometer, are heard in the Ausonian tongue. - The so-called Geometry of Boethius which has come down to us by no means constitutes a translation of Euclid. The MSS. variously give five, four, three or two Books, but they represent only two distinct compilations, one normally in five Books and the other in two. Even the latter, which was edited by Friedlein, is not genuineSee especially Weissenborn in Abhandlungen zur Gesch. d. Math. II. p. 18_{5} sq.; Heiberg in Philologus, XLIII. p. 507 sq.; Cantor, 1_{3}, p. 580 sq.,but appears to have been put together in the 11th c., from various sources. It begins with the definitions of Eucl. I., and in these are traces of perfectly correct readings which are not found even in the MSS. of the 10th c., but which can be traced in Proclus and other ancient sources; then come the Postulates (five only), the Axioms (three only), and after these some definitions of Eucl. II., III., IV. Next come the enunciations of Eucl. I., of ten propositions of Book II., and of some from Books III., IV., but always without proofs; there follows an extraordinary passage which indicates that the author will now give something of his own in elucidation of Euclid, though what follows is a literal translation of the proofs of Eucl. I. 1-3. This latter passage, although it affords a strong argument against the genuineness of this part of the work, shows that the Pseudoboethius had a Latin translation of Euclid from which he extracted the three propositions.

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Curtze has reproduced, in the preface to his edition of the translation by Gherard of Cremona of an-Nairĩzĩ's Arabic commentary on Euclid, some interesting fragments of a translation of Euclid taken from a Munich MS. of the 10th c. They are on two leaves used for the cover of the MS. (Bibliothecae Regiae Universitatis Monacensis 2o 757) and consist of portions of Eucl. I. 37, 38 and II. 8, translated literally word for word from the Greek text. The translator seems to have been an Italian (cf. the words capitolo nono - used for the ninth prop. of Book II.) who knew very little Greek and had moreover little mathematical knowledge. For example, he translates the capital letters denoting points in figures as if they were numerals: thus ta\ *a*b*g, - *d*e*z is translated que primo secundo et tertio quarto quinto et septimo, - T becomes tricentissimo - and so on. The Greek MS. which he used was evidently written in uncials, for *d*e*z*q becomes in one place quod autem septimo nono, - showing that he mistook *d*e for the particle de/, and kai\ o( *s*t*u is rendered sicut tricentissimo et quadringentissimo, - showing that the letters must have been written KAIOCTU.

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The date of the Englishman Athelhard (Æthelhard) is approximately fixed by some remarks in his work Perdifficiles Quaestiones Naturales which, on the ground of the personal allusions they contain, must be assigned to the first thirty years of the 12th c.Cantor, Gesch. d. Math. I3, p. 906. He wrote a number of philosophical works. Little is known about his life. He is said to have studied at Tours and Laon, and to have lectured at the latter school. He travelled to Spain, Greece, Asia Minor and Egypt, and acquired a knowledge of Arabic, which enabled him to translate from the Arabic into Latin, among other works, the Elements of Euclid. The date of this translation must be put at about 1120. MSS. purporting to contain Athelhard's version are extant in the British Museum (Harleian No. 5404 and others), Oxford (Trin. Coll. 47 and Ball. Coll. 257 of 12th c.), Nürnberg (Johannes Regiomontanus' copy) and Erfurt.

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Among the very numerous works of Gherard of Cremona (1114— 1187) are mentioned translations of 15 Books of Euclid - and of the DataBoncompagni, Della vita e delle opere di Gherardo Cremonese, Rome, 1851, p. 5.. Till recently this translation of the Elements was supposed to be lost; but Axel Anthon Björnbo has succeeded (1904) in discovering a translation from the Arabic which is different from the two others known to us (those by Athelhard and Campanus respectively), and which he, on grounds apparently convincing, holds to be Gherard's. Already in 1901 Björnbo had found Books X.—XV. of this translation in a MS. at Rome (Codex Reginensis lat. 1268 of 14th c.)Described in an appendix to Studien über Menelaos' Sphärik (Abhandlungen zur Gesckichse der mathematischen Wissenschaften, XIV., 1902).; but three years later he had traced three MSS. containing the whole of the same translation at Paris (Cod. Paris. 7216, 15th c.), Boulogne-sur-Mer (Cod. Bononiens. 196, 14th c.), and Bruges (Cod. Brugens. 521, 14th c.), and another at Oxford (Cod. Digby 174, end of 12th c.) containing a fragment, XI. 2 to XIV. The occurrence of Greek words in this translation such as rombus, romboides (where Athelhard keeps the Arabic terms), ambligonius, orthogonius, gnomo, pyramis etc., show that the translation is independent of Athelhard's. Gherard appears to have had before him an old translation of Euclid from the Greek which Athelhard also often followed, especially in his terminology, using it however in a very different manner. Again, there are some Arabic terms, e.g. meguar for axis of rotation, which Athelhard did not use, but which is found in almost all the translations that are with certainty attributed to Gherard of Cremona; there occurs also the 4 See Bibliotheca Mathematica, VI3, 1905-6, PP. 242-8.expression superficies equidistantium laterum et rectorum angulorum, - found also in Gherard's translation of an-Nairĩzĩ, where Athelhard says parallelogrammum rectangulum. - The translation is much clearer than Athelhard's: it is neither abbreviated nor edited - as Athelhard's appears to have been; it is a word-for-word translation of an Arabic MS. containing a revised and critical edition of Thãbit's version. It contains several notes quoted from Thãbit himself (Thebit dixit), e.g. about alternative proofs etc. which Thãbit found in another Greek MS., - and is therefore a further testimony to Thãbit's critical treatment of the text after Greek MSS. The new editor also added critical remarks of his own, e.g. on other proofs which he found in other Arabic versions, but not in the Greek: whence it is clear that he compared the Thãbit version before him with other versions as carefully as Thãbit collated the Greek MSS. Lastly, the new editor speaks of Thebit qui transtulit hunc librum in arabicam linguam - and of translatio Thebit, - which may tend to confirm the statement of al-Qiftĩ who credited Thãbit with an independent translation, and not (as the Fihrist does) with a mere improvement of the version of Is[hnull ]ãq b. Hunain.

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Gherard's translation of the Arabic commentary of an-Nairĩzĩ on the first ten Books of the Elements was discovered by Maximilian Curtze in a MS. at Cracow and published as a supplementary volume to Heiberg and Menge's EuclidAnaritii in decem libros priores Elementorum Euclidis Commentarii ex interpretatione Gherardi Cremonensis in codice Cracoviensi 569 servata edidit Maximilianus Curtze, Leipzig (Teubner), 1899.: it will often be referred to in this work.

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Next in chronological order comes Johannes Campanus (Campano) of Novara. He is mentioned by Roger Bacon (1214-1294) as a prominent mathematician of his timeCantor, II_{1}, p. 88., and this indication of his date is confirmed by the fact that he was chaplain to Pope Urban IV, who was Pope from 1261 to 1281Tiraboschi, Storia della letteratura italiana, IV. 145mdash;160.. His most important achievement was his edition of the Elements including the two Books XIV. and XV. which are not Euclid's. The sources of Athelhard's and Campanus' translations, and the relation between them, have been the subject of much discussion, which does not seem to have led as yet to any definite conclusion. Cantor (II_{1}, p. 91) gives referencesH. Weissenborn in Zeitschrift für Math. u. Physik, XXV., Supplement, pp. 143mdash;166, and in his monograph, Die Übersetzungen des Euklid durch Campano und Zamberti (1882); Max. Curtze in Philologische Rundschau (1881), 1. pp. 943-950, and in Fahresbericht über die Fortschritte der classischen Alterthumswissenschaft, XL. (188_{4}, III.) pp. 19mdash;22; Heiberg in Zeitschrift für Math. u. Physik, XXXV., hist.-litt. Abth., pp. 48mdash;58 and pp. 81mdash;6. and some particulars. It appears that there is a MS. at Munich (Cod. lat. Mon. 13021) written by Sigboto in the 12th c. at Prüfning near Regensburg, and denoted by Curtze by the letter R, which contains the enunciations of part of Euclid. The Munich MSS. of Athelhard and Campanus' translations have many enunciations textually identical with those in R, so that the source of all three must, for these enunciations, have been the same; in others Athelhard and Campanus diverge completely from R, which in these places follows the Greek text and is therefore genuine and authoritative. In the 32nd definition occurs the word elinuam, - the Arabic term for rhombus, - and throughout the translation are a number of Arabic figures. But R was not translated from the Arabic, as is shown by (among other things) its close resemblance to the translation from Euclid given on pp. 377 sqq. of the Gromatici Veteres and to the so-called geometry of Boethius. The explanation of the Arabic figures and the word elinuam - in Def. 32 appears to be that R was a late copy of an earlier original with corruptions introduced in many places; thus in Def. 32 a part of the text was completely lost and was supplied by some intelligent copyist who inserted the word elinuam, - which was known to him, and also the Arabic figures. Thus Athelhard certainly was not the first to translate Euclid into Latin; there must have been in existence before the 11th c. a Latin translation which was the common source of R, the passage in the Gromatici, and Boethius. - As in the two latter there occur the proofs as well as the enunciations of I. 1mdash;3, it is possible that this translation originally contained the proofs also. Athelhard must have had before had before him this translation of the enunciations, as well as the Arabic source from which he obtained his proofs. That some sort of translation, or at least fragments of one, were available before Athelhard's time even in England is indicated by some old English versesQuoted by Halliwell in Rara Mathematica (p. 56 note) from MS. Bib. Reg. Mus. Brit. 17 A. 1. f. 24mdash;3.: The clerk Euclide on this wyse hit fonde Thys craft of gemetry yn Egypte londe Yn Egypte he tawghte hyt ful wyde, In dyvers londe on every syde. Mony erys afterwarde y understonde Yer that the craft com ynto thys londe. Thys craft com into England, as y yow say, Yn tyme of good kyng Adelstone's day, - which would put the introduction of Euclid into England as far back as 924-940 A.D.

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We now come to the relation between Athelhard and Campanus. That their translations were not independent, as Weissenborn would have us believe, is clear from the fact that in all MSS. and editions, apart from orthographical differences and such small differences as are bound to arise when MSS. are copied by persons with some knowledge of the subject-matter, the definitions, postulates, axioms, and the 364 enunciations are word for word identical in Athelhard and Campanus; and this is the case not only where both have the same text as R but where they diverge from it. Hence it would seem that Campanus used Athelhard's translation and only developed the proofs by means of another redaction of the Arabian Euclid. It is true that the difference between the proofs of the propositions in the two translations is considerable; Athelhard's are short and compressed, Campanus' clearer and more complete, following the Greek text more closely, though still at some distance. Further, the arrangement in the two is different; in Athelhard the proofs regularly precede the enunciations, Campanus follows the usual order. It is a question how far the differences in the proofs, and certain additions in each, are due to the two translators themselves or go back to Arabic originals. The latter supposition seems to Curtze and Cantor the more probable one. Curtze's general view of the relation of Campanus to Athelhard is to the effect that Athelhard's translation was gradually altered, from the form in which it appears in the two Erfurt MSS. described by Weissenborn, by successive copyists and commentators who had Arabic originals before them, until it took the form which Campanus gave it and in which it was published. In support of this view Curtze refers to Regiomontanus' copy of the Athelhard-Campanus translation. In Regiomontanus' own preface the title is given, and this attributes the translation to Athelhard; but, while this copy agrees almost exactly with Athelhard in Book I., yet, in places where Campanus is more lengthy, it has similar additions, and in the later Books, especially from Book III. onwards, agrees absolutely with Campanus; Regiomontanus, too, himself implies that, though the translation was Athelhard's, Campanus had revised it; for he has notes in the margin such as the following, Campani est hec, - dubito an demonstret hic Campanus - etc.

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We come now to the printed editions of the whole or of portions of the Elements. This is not the place for a complete bibliography, such as Riccardi has attempted in his valuable memoir issued in five parts between 1887 and 1893, which makes a large book in itselfSaggio di una Bibliografia Euclidea, memoria del Prof. Pietro Riccardi (Bologna, 1887, 1888, 1890, 1893).. I shall confine myself to saying something of the most noteworthy translations and editions. It will be convenient to give first the Latin translations which preceded the publication of the editio princeps of the Greek text in 1533, next the most important editions of the Greek text itself, and after them the most important translations arranged according to date of first appearance and languages, first the Latin translations after 1533, then the Italian, German, French and English translations in order.

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It may be added here that the first allusion, in the West, to the Greek text as still extant is found in Boccaccio's commentary on the Divina Commedia of DanteI. p. 404.. Next Johannes Regiomontanus, who intended to publish the Elements after the version of Campanus, but with the latter's mistakes corrected, saw in Italy (doubtless when staying with his friend Bessarion) some Greek MSS. and noticed how far they differed from the Latin version (see a letter of his written in the year 1471 to Christian Roder of Hamburg)Published in C. T. de Murr's Memorabilia Bibliothecarum Norimbergensium, Part I. p. 190 sqq..

- - - I. Latin translations prior to 1533. -

1482. In this year appeared the first printed edition of Euclid, which was also the first printed mathematical book of any importance. This was printed at Venice by Erhard Ratdolt and contained Campanus' translationCurtze (An-Nairīzī, p. xiii) reproduces the heading of the first page of the text as follows (there is no title-page): Preclariffimũ opus elementoꝝ Euclidis megarēfis [vmacr ]na cū cōmentis Campani pfpicaciffimi in artē geometriā incipit felicit', after which the definitions begin at once. Other copies have the shorter heading: Preclarissimus liber elementorum Euclidis perspicacissimi: in artem Geometrie incipit quam foelicissime. At the end stands the following: 󰋓Opus elementorū euclidis megarenfis in geometriā artē Jnid quoqꝫ Campani pfpicaciffimi Cōmentationes finiũt. Erhardus ratdolt Augustensis impreffor folertiffimus. venetijs impreffit . Anno falutis . M.cccc.lxxxij . Octauis . Cale[ntilde] . Ju[ntilde] . Lector . Vale.. Ratdolt belonged to a family of artists at Augsburg, where he was born about 1443. Having learnt the trade of printing at home, he went in 1475 to Venice, and founded there a famous printing house which he managed for II years, after which he returned to Augsburg and continued to print important books until 1516. He is said to have died in 1528. KästnerKästner, Geschichte der Mathematik, I. p. 289 sqq. See also Weissenborn, Die Übersetzungen des Euklid durch Campano und Zamberti, pp. 1-7. gives a short description of this first edition of Euclid and quotes the dedication to Prince Mocenigo of Venice which occupies the page opposite to the first page of text. The book has a margin of 2 1/2 inches, and in this margin are placed the figures of the propositions. Ratdolt says in his dedication that at that time, although books by ancient and modern authors were printed every day in Venice, little or nothing mathematical had appeared: a fact which he puts down to the difficulty involved by the figures, which no one had up to that time succeeded in printing. He adds that after much labour he had discovered a method by which figures could be produced as easily as lettersMea industria non sine maximo labore effeci vt qua facilitate litterarum elementa imprimuntur ea etiam geometrice figure conficerentur. -. Experts are in doubt as to the nature of Ratdolt's discovery. Was it a method of making figures up out of separate parts of figures, straight or curved lines, put together as letters are put together to make words? In a life of Joh. Gottlob Immanuel Breitkopf, a contemporary of Kästner's own, this member of the great house of Breitkopf is credited with this particular discovery. Experts in that same house expressed the opinion that Ratdolt's figures were woodcuts, while the letters denoting points in the figures were like the other letters in the text; yet it was with carved wooden blocks that printing began. If Ratdolt was the first to print geometrical figures, it was not long before an emulator arose; for in the very same year Mattheus Cordonis of Windischgrätz employed woodcut mathematical figures in printing Oresme's De latitudinibusCurtze in Zeitschrift für Math. u. Physik, XX., hist.-litt. Abth. p. 58.. How eagerly the opportunity of spreading geometrical knowledge was seized upon is proved by the number of editions which followed in the next few years. Even the year 1482 saw two forms of the book, though they only differ in the first sheet. Another edition came out in 1486 (Ulmae, apud Io. Regerum) and another in 1491 (Vincentiae per Leonardum de Basilea et Gulielmum de Papia), but without the dedication to Mocenigo who had died in the meantime (1485). If Campanus added anything of his own, his additions are at all events not distinguished by any difference of type or otherwise; the enunciations are in large type, and the rest is printed continuously in smaller type. There are no superscriptions to particular passages such as Euclides ex Campano, Campanus, Campani additio, or Campani annotatio, which are found for the first time in the Paris edition of 1516 giving both Campanus' version and that of Zamberti (presently to be mentioned).

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1501. G. Valla included in his encyclopaedic work De expetendis et fugiendis rebus published in this year at Venice (in aedibus Aldi Romani) a number of propositions with proofs and scholia translated from a Greek MS. which was once in his possession (cod. Mutin. III B, 4 of the 15th c.).

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1505. In this year Bartolomeo Zamberti (Zambertus) brought out at Venice the first translation, from the Greek text, of the whole of the Elements. From the titleThe title begins thus: Euclidis megaresis philosophi platonicj mathematicarum disciplinarum Janitoris: Habent in hoc volumine quicunque ad mathematicam substantiam aspirant: elementorum libros xiij cum expositione Theonis insignis mathematici. quibus multa quae deerant ex lectione graeca sumpta addita sunt nec non plurima peruersa et praepostere: voluta in Campani interpretatione: ordinata digesta et castigata sunt etc. - For a description of the book see Weissenborn, p. 12 sqq., as well as from his prefaces to the Catoptrica and Data, with their allusions to previous translators who take some things out of authors, omit some, and change some, - or to that most barbarous translator - who filled a volume purporting to be Euclid's with extraordinary scarecrows, nightmares and phantasies, - one object of Zamberti's translation is clear. His animus against Campanus appears also in a number of notes, e.g. when he condemns the terms helmuain - and helmuariphe - used by Campanus as barbarous, un-Latin etc., and when he is roused to wrath by Campanus' unfortunate mistranslation of v. Def. 5. He does not seem to have had the penetration to see that Campanus was translating from an Arabic, and not from a Greek, text. Zamberti tells us that he spent seven years over his translation of the thirteen Books of the Elements. As he seems to have been born in 1473, and the Elements were printed as early as 1500, though the complete work (including the Phaenomena, Optica, Catoptrica, Data etc.) has the date 1505 at the end, he must have translated Euclid before the age of 30. Heiberg has not been able to identify the MS. of the Elements which Zamberti used; but it is clear that it belonged to the worse class of MSS., since it contains most of the interpolations of the Theonine variety. Zamberti, as his title shows, attributed the proofs to Theon.

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1509. As a counterblast to Zamberti, Luca Paciuolo brought out an edition of Euclid, apparently at the expense of Ratdolt, at Venice (per Paganinum de Paganinis), in which he set himself to vindicate Campanus. The title-page of this now very rare editionSee Weissenborn, p. 30 sqq. begins thus: The works of Euclid of Megara, a most acute philosopher and without question the chief of all mathematicians, translated by Campanus their most faithful interpreter. - It proceeds to say that the translation had been, through the fault of copyists, so spoiled and deformed that it could scarcely be recognised as Euclid. Luca Paciuolo accordingly has polished and emended it with the most critical judgment, has corrected 129 figures wrongly drawn and added others, besides supplying short explanations of difficult passages. It is added that Scipio Vegius of Milan, distinguished for his knowledge of both languages - (i.e. of course Latin and Greek), as well as in medicine and the more sublime studies, had helped to make the edition more perfect. Though Zamberti is not once mentioned, this latter remark must have reference to Zamberti's statement that his translation was from the Greek text; and no doubt Zamberti is aimed at in the wish of Paciuolo's that others too would seek to acquire knowledge instead of merely showing off, or that they would not try to make a market of the things of which they are ignorant, as it were (selling) smoke - Atque utinam et alii cognoscere vellent non ostentare aut ea quae nesciunt veluti fumum venditare non conarentur. -. - Weissenborn observes that, while there are many trivialities in Paciuolo's notes, they contain some useful and practical hints and explanations of terms, besides some new proofs which of course are not difficult if one takes the liberty, as Paciuolo does, of divering from Euclid's order and assuming for the proof of a proposition results not arrived at till later. Two not inapt terms are used in this edition to describe the figures of III. 7, 8, the former of which is called the goose's foot (pes anseris), the second the peacock's tail (cauda pavonis) Paciuolo as the castigator of Campanus' translation, as he calls himself, failed to correct the mistranslation of V. Def. 5Campanus' translation in Ratdolt's edition is as follows: Quantitates quae dicuntur continuam habere proportionalitatem, sunt, quarum equè multiplicia aut equa sunt aut equè sibi sine interruptione addunt aut minuunt - (!), to which Campanus adds the note: Continuè proportionalia sunt quorum omnia multiplicia equalia sunt continuè proportionalia. Sed noluit ipsam diffinitionem proponere sub hac forma, quia tunc diffiniret idem per idem, aperte (? a parte) tamen rei est istud cum sua diffinitione convertibile. -. Before the fifth Book he inserted a discourse which he gave at Venice on the 15th August, 1508, in S. Bartholomew's Church, before a select audience of 500, as an introduction to his elucidation of that Book.

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1516. The first of the editions giving Campanus' and Zamberti's translations in conjunction was brought out at Paris (in officina Henrici Stephani e regione scholae Decretorum). The idea that only the enunciations were Euclid's, and that Campanus was the author of the proofs in his translation, while Theon was the author of the proofs in the Greek text, reappears in the title of this edition; and the enunciations of the added Books XIV., XV. are also attributed to Euclid, Hypsicles being credited with the proofsEuclidis Megarensis Geometricorum Elementorum Libri XV. Campani Galli transalpini in eosdem commentariorum libri XV. Theonis Alexandrini Bartholomaeo. Zamberto Veneto interprete, in tredecim priores, commentationum libri XIII. Hypsiclis Alexandrini in duos posteriores, eodem Bartholomaeo Zamberto Veneto interprete, commentariorum libri II. - On the last page (261) is a similar statement of content, but with the difference that the expression ex Campani...deinde Theonis...et Hypsiclis...traditionibus. - For description see Weissenborn, p. 56 sqq.. The date is not on the title-page nor at the end, but the letter of dedication to François Briconnet by Jacques Lefèvre is dated the day after the Epiphany, 1516. The figures are in the margin. The arrangement of the propositions is as follows: first the enunciation with the heading Euclides ex Campano, then the proof with the note Campanus, and after that, as Campani additio, any passage found in the edition of Campanus' translation but not in the Greek text; then follows the text of the enunciation translated from the Greek with the heading Euclides ex Zamberto, and lastly the proof headed Theo ex Zamberto. There are separate figures for the two proofs. This edition was reissued with few changes in 1537 and 1546 at Basel (apud Iohannem Hervagium), but with the addition of the Phaenomena, Optica, Catoptrica etc. For the edition of 1537 the Paris edition of 1516 was collated with a Greek copy - (as the preface says) by Christian Herlin, professor of mathematical studies at Strassburg, who however seems to have done no more than correct one or two passages by the help of the Basel editio princeps (1533), and add the Greek word in cases where Zamberti's translation of it seemed unsuitable or inaccurate.

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We now come to

- - - II. Editions of the Greek text. -

1533 is the date of the editio princeps, the title-page of which reads as follows: *e*u*k*l*e*i*d*o*u *s*t*o*i*x*e*i*w*n *b*i*b*l *i*ē - *e*k *t*w*n *q*e*w*n*o*s *s*u*n*o*u*s*i*w*n. - *ei)s tou= au)tou= to\ prw=ton, e)chghma/twn *pro/klou bibl. d_. Adiecta praefatiuncula in qua de disciplinis Mathematicis nonnihil. BASILEAE APVD IOAN. HERVAGIVM ANNO M.D.XXXIII. MENSE SEPTEMBRI.

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The editor was Simon Grynaeus the elder (d. 1541), who, after working at Vienna and Ofen, Heidelberg and Tübingen, taught last of all at Basel, where theology was his main subject. His praefatiuncula - is addressed to an Englishman, Cuthbert Tonstall (14741559), who, having studied first at Oxford, then at Cambridge, where he became Doctor of Laws, and afterwards at Padua, where in addition he learnt mathematics—mostly from the works of Regiomontanus and Paciuolo—wrote a book on arithmeticDe arté supputandi libri quatuor. as a farewell to the sciences, - and then, entering politics, became Bishop of London and member of the Privy Council, and afterwards (1530) Bishop of Durham. Grynaeus tells us that he used two MSS. of the text of the Elements, entrusted to friends of his, one at Venice by Lazarus Bayfius - (Lazare de Baïf, then the ambassador of the King of France at Venice), the other at Paris by Ioann. Rvellius - (Jean Ruel, a French doctor and a Greek scholar), while the commentaries of Proclus were put at the disposal of Grynaeus himself by Ioann. Claymundus - at Oxford. Heiberg has been able to identify the two MSS. used for the text; they are (1) cod. Venetus Marcianus 301 and (2) cod. Paris. gr. 2343 of the 16th c., containing Books I.—XV., with some scholia which are embodied in the text. When Grynaeus notes in the margin the readings from the other copy, - this other copy - is as a rule the Paris MS., though sometimes the reading of the Paris MS. is taken into the text and the other copy - of the margin is the Venice MS. Besides these two MSS. Grynaeus consulted Zamberti, as is shown by a number of marginal notes referring to Zampertus - or to latinum exemplar - in certain propositions of Books IX.—XI. When it is considered that the two MSS. used by Grynaeus are among the worst, it is obvious how entirely unauthoritative is the text of the editio princeps. Yet it remained the source and foundation of later editions of the Greek text for a long period, the editions which followed being designed, not for the purpose of giving, from other MSS., a text more nearly representing what Euclid himself wrote, but of supplying a handy compendium to students at a moderate price.

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1536. Orontius Finaeus (Oronce Fine) published at Paris (apud Simonem Colinaeum) demonstrations on the first six books of Euclid's elements of geometry, - in which the Greek text of Euclid himself is inserted in its proper places, with the Latin translation of Barth. Zamberti of Venice, - which seems to imply that only the enunciations were given in Greek. The preface, from which Kästner quotesKästner, I. p. 260., says that the University of Paris at that time required, from all who aspired to the laurels of philosophy, a most solemn oath that they had attended lectures on the said first six Books. Other editions of Fine's work followed in 1544 and 1551.

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1545. The enunciations of the fifteen Books were published in Greek, with an Italian translation by Angelo Caiani, at Rome (apud Antonium Bladum Asulanum). The translator claims to have corrected the books and purged them of six hundred things which did not seem to savour of the almost divine genius and the perspicuity of EuclidHeiberg, vol. V. p. cvii. -

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1549. Joachim Camerarius published the enunciations of the first six Books in Greek and Latin (Leipzig). The book, with preface, purports to be brought out by Rhaeticus (1514-1576), a pupil of Copernicus. Another edition with proofs of the propositions of the first three Books was published by Moritz Steinmetz in 1577 (Leipzig); a note by the printer attributes the preface to Camerarius himself.

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1550. Ioan. Scheubel published at Basel (also per Ioan. Hervagium) the first six Books in Greek and Latin together with true and appropriate proofs of the propositions, without the use of letters - (i.e. letters denoting points in the figures), the various straight lines and angles being described in wordsKästner, I. p. 359..

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1557 (also 1558). Stephanus Gracilis published another edition (repeated 1573, 1578, 1598) of the enunciations (alone) of Books I.—XV. in Greek and Latin at Paris (apud Gulielmum Cavellat). He remarks in the preface that for want of time he had changed scarcely anything in Books I.—VI., but in the remaining Books he had emended what seemed obscure or inelegant in the Latin translation, while he had adopted in its entirety the translation of Book X. by Pierre Mondoré (Petrus Montaureus), published separately at Paris in 1551. Gracilis also added a few scholia. -

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1564. In this year Conrad Dasypodius (Rauchfuss), the inventor and maker of the clock in Strassburg cathedral, similar to the present one, which did duty from 1571 to 1789, edited (Strassburg, Chr. Mylius) (1) Book I. of the Elements in Greek and Latin with scholia, (2) Book II. in Greek and Latin with Barlaam's arithmetical version of Book II., and (3) the enunciations of the remaining Books III.—XIII. Book I. was reissued with vocabula quaedam geometrica - of Heron, the enunciations of all the Books of the Elements, and the other works of Euclid, all in Greek and Latin. In the preface to (1) he says that it had been for twenty-six years the rule of his school that all who were promoted from the classes to public lectures should learn the first Book, and that he brought it out, because there were then no longer any copies to be had, and in order to prevent a good and fruitful regulation of his school from falling through. In the preface to the edition of 1571 he says that the first Book was generally taught in all gymnasia and that it was prescribed in his school for the first class. In the preface to (3) he tells us that he published the enunciations of Books III.—XIII. in order not to leave his work unfinished, but that, as it would be irksome to carry about the whole work of Euclid in extenso, he thought it would be more convenient to students of geometry to learn the Elements if they were compressed into a smaller book.

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1620. Henry Briggs (of Briggs' logarithms) published the first six Books in Greek with a Latin translation after Commandinus, corrected in many places - (London, G. Jones).

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1703 is the date of the Oxford edition by David Gregory which, until the issue of Heiberg and Menge's edition, was still the only edition of the complete works of Euclid*e*u*k*l*e*i*d*o*u *t*a *s*w*z*o*m*e*n*a. Euclidis quae supersunt omnia. Ex recensione Davidis Gregorii M.D. Astronomiae Professoris Saviliani et R.S.S. Oxoniae, e Theatro Sheldoniano, An. Dom. MDCCIII.. In the Latin translation attached to the Greek text Gregory says that he followed Commandinus in the main, but corrected numberless passages in it by means of the books in the Bodleian Library which belonged to Edward Bernard (1638-1696), formerly Savilian Professor of Astronomy, who had conceived the plan of publishing the complete works of the ancient mathematicians in fourteen volumes, of which the first was to contain Euclid's Elements I.—XV. As regards the Greek text, Gregory tells us that he consulted, as far as was necessary, not a few MSS. of the better sort, bequeathed by the great Savile to the University, as well as the corrections made by Savile in his own hand in the margin of the Basel edition. He had the help of John Hudson, Bodley's Librarian, who punctuated the Basel text before it went to the printer, compared the Latin version with the Greek throughout, especially in the Elements and Data, and, where they differed or where he suspected the Greek text, consulted the Greek MSS. and put their readings in the margin if they agreed with the Latin and, if they did not agree, affixed an asterisk in order that Gregory might judge which reading was geometrically preferable. Hence it is clear that no Greek MS., but the Basel edition, was the foundation of Gregory's text, and that Greek MSS. were only referred to in the special passages to which Hudson called attention.

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1814-1818. A most important step towards a good Greek text was taken by F. Peyrard, who published at Paris, between these years, in three volumes, the Elements and Data in Greek, Latin and FrenchEuclidis quae supersunt. Les Œuvres d'Euclide, en Grec, en Latin et en Français d'après un manuscrit très-ancien, qui était resté inconnu jusqu'à nos jours Par F. Peyrard. Ouvrage approuvé par l'Institut de France (Paris, chez M. Patris).. At the time (1808) when Napoleon was having valuable MSS. selected from Italian libraries and sent to Paris, Peyrard managed to get two ancient Vatican MSS. (190 and 1038) sent to Paris for his use (Vat. 204 was also at Paris at the time, but all three were restored to their owners in 1814). Peyrard noticed the excellence of Cod. Vat. 190, adopted many of its readings, and gave in an appendix a conspectus of these readings and those of Gregory's edition; he also noted here and there readings from Vat. 1038 and various Paris MSS. He therefore pointed the way towards a better text, but committed the error of correcting the Basel text instead of rejecting it altogether and starting afresh.

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1824-1825. A most valuable edition of Books I.—VI. is that of J. G. Camerer (and C. F. Hauber) in two volumes published at BerlinEuclidis elcmentorum libri sex priores graece et latine commentario e scriptis veterum ac recentiarum mathematicorum et Pfleidereri maxime illustrati (Berolini, sumptibus G. Reimeri). Tom. I. 1824; tom. II. 1825.. The Greek text is based on Peyrard, although the Basel and Oxford editions were also used. There is a Latin translation and a collection of notes far more complete than any other I have seen and well nigh inexhaustible. There is no editor or commentator of any mark who is not quoted from; to show the variety of important authorities drawn upon by Camerer, I need only mention the following names: Proclus, Pappus, Tartaglia, Commandinus, Clavius, Peletier, Barrow, Borelli, Wallis, Tacquet, Austin, Simson, Playfair. No words of praise would be too warm for this veritable encyclopaedia of information.

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1825. J. G. C. Neide edited, from Peyrard, the text of Books I.—VI., XI. and XII. (Halis Saxoniae).

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1826-9. The last edition of the Greek text before Heiberg's is that of E. F. August, who followed the Vatican MS. more closely than Peyrard did, and consulted at all events the Viennese MS. Gr. 103 (Heiberg's V). August's edition (Berlin, 1826-9) contains Books I.-XIII.

- - - - III. Latin versions or commentaries after 1533. -

1545. Petrus Ramus (Pierre de la Ramée, 1515-1572) is credited with a translation of Euclid which appeared in 1545 and again in 1549 at ParisDescribed by Boncompagni, Bullettino, II. p. 389.. Ramus, who was more rhetorician and logician than geometer, also published in his Scholae mathematicae (1559, Frankfurt; 1569, Basel) what amounts to a series of lectures on Euclid's Elements, in which he criticises Euclid's arrangement of his propositions, the definitions, postulates and axioms, all from the point of view of logic.

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1557. Demonstrations to the geometrical Elements of Euclid, six Books, by Peletarius (Jacques Peletier). The second edition (1610) contained the same with the addition of the Greek text of Euclid -; but only the enunciations of the propositions, as well as the definitions etc., are given in Greek (with a Latin translation), the rest is in Latin only. He has some acute observations, for instance about the angle - of contact.

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1559. Johannes Buteo, or Borrel (1492-1572), published in an appendix to his book De quadratura circuli some notes on the errors of Campanus, Zambertus, Orontius, Peletarius, Pena, interpreters of Euclid. - Buteo in these notes proved, by reasoned argument based on original authorities, that Euclid himself and not Theon was the author of the proofs of the propositions.

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1566. Franciscus Flussates Candalla (François de Foix, Comte de Candale, 1502-1594) restored - the fifteen Books, following, as he says, the terminology of Zamberti's translation from the Greek, but drawing, for his proofs, on both Campanus and Theon (i.e. Zamberti) except where mistakes in them made emendation necessary. Other editions followed in 1578, 1602, 1695 (in Dutch).

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1572. The most important Latin translation is that of Commandinus (1509-1575) of Urbino, since it was the foundation of most translations which followed it up to the time of Peyrard, including that of Simson and therefore of those editions, numerous in England, which give Euclid chiefly after the text of Simson. - Simson's first (Latin) edition (1756) has ex versione Latina Federici Commandini - on the title-page. Commandinus not only followed the original Greek more closely than his predecessors but added to his translation some ancient scholia as well as good notes of his own. The title of his work is -

Euclidis elementorum libri XV, una cum scholiis antiquis. A Federico Commandino Urbinate nuper in latinum conversi, commentariisque quibusdam illustrati (Pisauri, apud Camillum Francischinum).

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He remarks in his preface that Orontius Finaeus had only edited six Books without reference to any Greek MS., that Peletarius had followed Campanus' version from the Arabic rather than the Greek text, and that Candalla had diverged too far from Euclid, having rejected as inelegant the proofs given in the Greek text and substituted faulty proofs of his own. Commandinus appears to have used, in addition to the Basel editio princeps, some Greek MS., so far not identified; he also extracted his scholia antiqua - from a MS. of the class of Vat. 192 containing the scholia distinguished by Heiberg as Schol. Vat. - New editions of Commandinus' translation followed in 1575 (in Italian), 1619, 1749 (in English, by Keill and Stone), 1756 (Books I.—VI., XI., XII. in Latin and English, by Simson), 1763 (Keill). Besides these there were many editions of parts of the whole work, e.g. the first six Books.

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1574. The first edition of the Latin version by ClaviusEuclidis elemcntorum librit XV. Accessit XVI. de solidorum regularium comparatione. Omnes perspicuis demonstrationibus, accuratisque scholiis illustrati. Auctore Christophoro Clavio (Romae, apud Vincentium Accoltum), 2 vols. (Christoph Klau [?]. born at Bamberg 1537, died 1612) appeared in 1574, and new editions of it in 1589, 1591, 1603, 1607, 1612. It is not a translation, as Clavius himself states in the preface, but it contains a vast amount of notes collected from previous commentators and editors, as well as some good criticisms and elucidations of his own. Among other things, Clavius finally disposed of the error by which Euclid had been identified with Euclid of Megara. He speaks of the differences between Campanus who followed the Arabic tradition and the commentaries of Theon, - by which he appears to mean the Euclidean proofs as handed down by Theon; he complains of predecessors who have either only given the first six Books, or have rejected the ancient proofs and substituted worse proofs of their own, but makes an exception as regards Commandinus, a geometer not of the common sort, who has lately restored Euclid, in a Latin translation, to his original brilliancy. - Clavius, as already stated, did not give a translation of the Elements but rewrote the proofs, compressing them or adding to them, where he thought that he could make them clearer. Altogether his book is a most useful work.

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1621. Henry Savile's lectures (Praelectiones tresdecim in principium Elementorum Euclidis Oxoniae habitae MDC.XX., Oxonii 1621), though they do not extend beyond I. 8, are valuable because they grapple with the difficulties connected with the preliminary matter, the definitions etc., and the tacit assumptions contained in the first propositions.

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1654. André Tacquet's Elementa geometriae planae et solidae containing apparently the eight geometrical Books arranged for general use in schools. It came out in a large number of editions up to the end of the eighteenth century.

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1655. Barrow's Euclidis Elementorum Libri XV breviter demonstrati is a book of the same kind. In the preface (to the edition of 1659) he says that he would not have written it but for the fact that Tacquet gave only eight Books of Euclid. He compressed the work into a very small compass (less than 400 small pages, in the edition of 1659, for the whole of the fifteen Books and the Data) by abbreviating the proofs and using a large quantity of symbols (which, he says, are generally Oughtred's). There were several editions up to 1732 (those of 1660 and 1732 and one or two others are in English).

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1658. Giovanni Alfonso Borelli (1608-1679) published Euclides restitutus, on apparently similar lines, which went through three more editions (one in Italian, 1663).

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1660. Claude François Milliet Dechales' eight geometrical Books of Euclid's Elements made easy. Dechales' versions of the Elements had great vogue, appearing in French, Italian and English as well as Latin. Riccardi enumerates over twenty editions.

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1733. Saccheri's Euclides ab omni naevo vindicatus sive conatus geometricus quo stabiliuntur prima ipsa geometriae principia is important for his elaborate attempt to prove the parallel-postulate, forming an important stage in the history of the development of nonEuclidean geometry.

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1756. Simson's first edition, in Latin and in English. The Latin title is -

Euclidis elementorum libri priores sex, item undecimus et duodecimus, ex versione latina Federici Commandini; sublatis iis quibus olim libri hi a Theone, aliisve, vitiati sunt, et quibusdam Euclidis demonstrationibus restitutis. A Roberto Simson M.D. Glasguae, in aedibus Academicis excudebant Robertus et Andreas Foulis, Academiae typographi.

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1802. Euclidis elementorum libri priores XII ex Commandini et Gregorii versionibus latinis. In usum juventutis Academicae...by Samuel Horsley, Bishop of Rochester. (Oxford, Clarendon Press.)

- - - IV. Italian versions or commentaries. -

1543. Tartaglia's version, a second edition of which was published in 1565The title-page of the edition of 1565 is as follows: Euclide Megarense philosopho, solo introduttore delle scientie mathematice, diligentemente rassettato, et alla integrità ridotto, per il degno professore di tal scientie Nicolo Tartalea Brisciano. secondo le due tradottioni. con una ampla espositione dello istesso tradottore di nuouo aggiunta. talmente chiara, che ogni mediocre ingegno, sensa la notitia, ouer suffragio di alcun' altra scientia con facilità serà capace a poterlo intendere. In Venetia, Appresso Curtio Troiano, 1565., and a third in 1585. It does not appear that he used any Greek text, for in the edition of 1565 he mentions as available only the first translation by Campano, - the second made by Bartolomeo Zamberto Veneto who is still alive, - the editions of Paris or Germany in which they have included both the aforesaid translations, - and our own translation into the vulgar (tongue). -

-

1575. Commandinus' translation turned into Italian and revised by him.

-

1613. The first six Books reduced to practice - by Pietro Antonio Cataldi, re-issued in 1620, and followed by Books VII.—IX. (1621) and Book X. (1625).

-

1663. Borelli's Latin translation turned into Italian by Domenico Magni.

-

1680. Euclide restituto by Vitale Giordano.

-

1690. Vincenzo Viviani's Elementi piani e solidi di Euclide (Book V. in 1674).

-

1731. Elementi geometrici piani e solidi di Euclide by Guido Grandi. No translation, but an abbreviated version, of which new editions followed one another up to 1806.

-

1749. Italian translation of Dechales with Ozanam's corrections and additions, re-issued 1785, 1797.

-

1752. Leonardo Ximenes (the first six Books). Fifth edition, 1819.

-

1818. Vincenzo Flauti's Corso di geometria elementare e sublime (4 vols.) contains (Vol. I.) the first six Books, with additions and a dissertation on Postulate 5, and (Vol. II.) Books XI., XII. Flauti also published the first six Books in 1827 and the Elements of geometry of Euclid in 1843 and 1854.

- - - V. German. -

1558. The arithmetical Books VII.-IX. by ScheubelDas sibend acht und neunt buch des hochberümbten Mathematici Euclidis Megarensis... - durch Magistrum Fohann Scheybl, der löblichen universitet zu Tübingen, des Euclidis und Arithmetic Ordinarien, auss dem latein ins teutsch gebracht.... (cf. the edition of the first six Books, with enunciations in Greek and Latin, mentioned above, under date 1550).

-

1562. The version of the first six Books by Wilhelm Holtzmann (Xylander).Die sechs erste b'ücher Euclidis vom anfang oder grund der Geometrj...Auss Griechischer sprach in die Teütsch gebracht aigentlich erklärt...Demassen vormals in Teütscher sprach nie gesehen worden...Durch Wilhelm Holtzman genant Xylander von Augspurg. Getruckht zu Basel.. This work has its interest as the first edition in German, but otherwise it is not of importance. Xylander tells us that it was written for practical people such as artists, goldsmiths, builders etc., and that, as the simple amateur is of course content to know facts, without knowing how to prove them, he has often left out the proofs altogether. He has indeed taken the greatest possible liberties with Euclid, and has not grappled with any of the theoretical difficulties, such as that of the theory of parallels.

-

1651. Heinrich Hoffmann's Teutscher Euclides (2nd edition 1653), not a translation.

-

1694. Ant. Ernst Burkh. v. Pirckenstein's Teutsch Redender Euclides (eight geometrical Books), for generals, engineers etc. - proved in a new and quite easy manner. - Other editions 1699, 1744.

-

1697. Samuel Reyher's In teutscher Sprache vorgestellter Euclides (six Books), made easy, with symbols algebraical or derived from the newest art of solution. -

-

1714. Euclidis XV Bücher teutsch, treated in a special and brief manner, yet completely, - by Chr. Schessler (another edition in 1729).

-

1773. The first six Books translated from the Greek for the use of schools by J. F. Lorenz. The first attempt to reproduce Euclid in German word for word.

-

1781. Books XI., XII. by Lorenz (supplementary to the preceding). Also Euklid's Elemente fünfzehn Bücher translated from the Greek by Lorenz (second edition 1798; editions of 1809, 1818, 1824 by Mollweide, of 1840 by Dippe). The edition of 1824, and I presume those before it, are shortened by the use of symbols and the compression of the enunciation and setting-out - into one.

-

1807. Books I.—VI., XI., XII. newly translated from the Greek, - by J. K. F. Hauff.

-

1828. The same Books by Joh. Jos. Ign. Hoffmann as guide to instruction in elementary geometry, - followed in 1832 by observations on the text by the same editor.

-

1833. Die Geometrie des Euklid und das Wesen derselben by E. S. Unger; also 1838, 1851.

-

1901. Max Simon, Euclid und die sechs planimetrischen Bücher.

- - - VI. French. -

1564-1566. Nine Books translated by Pierre Forcadel, a pupil and friend of P. de la Ramée.

-

1604. The first nine Books translated and annotated by Jean Errard de Bar-le-Duc; second edition, 1605.

-

1615. Denis Henrion's translation of the 15 Books (seven editions up to 1676).

-

1639. The first six Books demonstrated by symbols, by a method very brief and intelligible, - by Pierre Hérigone, mentioned by Barrow as the only editor who, before him, had used symbols for the exposition of Euclid.

-

1672. Eight Books rendus plus faciles - by Claude FranÇis Milliet Dechales, who also brought out Les élémens d'Euclide expliqués d'une manière nouvelle et très facile, which appeared in many editions, 1672, 1677, 1683 etc. (from 1709 onwards revised by Ozanam), and was translated into Italian (1749 etc.) and English (by William Halifax, 1685).

-

1804. In this year, and therefore before his edition of the Greek text, F. Peyrard published the Elements literally translated into French. A second edition appeared in 1809 with the addition of the fifth Book. As this second edition contains Books I.—VI. XI., XII. and X. I, it would appear that the first edition contained Books I.—IV., VI., XI., XII. Peyrard used for this translation the Oxford Greek text and Simson.

- - - VII. Dutch. -

1606. Jan Pieterszoon Dou (six Books). There were many later editions. Kästner, in mentioning one of 1702, says that Dou explains in his preface that he used Xylander's translation, but, having afterwards obtained the French translation of the six Books by Errard de Bar-le-Duc (see above), the proofs in which sometimes pleased him more than those of the German edition, he made his Dutch version by the help of both.

-

1617. Frans van Schooten, The Propositions of the Books of Euclid's Elements -; the fifteen Books in this version enlarged - by Jakob van Leest in 1662.

-

1695. C. J. Vooght, fifteen Books complete, with Candalla's 16th. -

-

1702. Hendrik Coets, six Books (also in Latin, 1692); several editions up to 1752. Apparently not a translation. but an edition for school use.

-

1763. Pybo Steenstra, Books I.—VI., XI., XII., likewise an abberviated version, several times reissued until 1825.

- - - VIII. English. -

1570 saw the first and the most important translation, that of Sir Henry Billingsley. The title-page is as follows: THE ELEMENTS OF GEOMETRIE of the most auncient Philosopher EVCLIDE of Megara - -

- Faithfully (now first) translated into the Englishe toung, by H. Billingsley, Citizen of London. Whereunto are annexed certaine Scholies, Annotations, and Inuentions, of the best Mathematiciens, both of time past, and in this our age. -

-

With a very fruitfull Preface by M. I. Dee, specifying the chiefe Mathematicall Sciēces, what they are, and whereunto commodious: where, also, are disclosed certaine new Secrets Mathematicall and Mechanicall, vntill these our daies, greatly missed.

- - Imprinted at London by John Daye.

-

The Preface by the translator, after a sentence observing that without the diligent study of Euclides Elementes it is impossible to attain unto the perfect knowledge of Geometry, proceeds thus. Wherefore considering the want and lacke of such good authors hitherto in our Englishe tounge, lamenting also the negligence, and lacke of zeale to their countrey in those of our nation, to whom God hath geuen both knowledge and also abilitie to translate into our tounge, and to publishe abroad such good authors and bookes (the chiefe instrumentes of all learninges): seing moreouer that many good wittes both of gentlemen and of others of all degrees, much desirous and studious of these artes, and seeking for them as much as they can, sparing no paines, and yet frustrate of their intent, by no meanes attaining to that which they seeke: I haue for their sakes, with some charge and great trauaile, faithfully translated into our vulgare toũge, and set abroad in Print, this booke of Euclide. Whereunto I haue added easie and plaine declarations and examples by figures, of the definitions. In which booke also ye shall in due place finde manifolde additions, Scholies, Annotations, and Inuentions: which I haue gathered out of many of the most famous and chiefe Mathematiciēs, both of old time, and in our age: as by diligent reading it in course, ye shall well perceaue.... -

-

It is truly a monumental work, consisting of 464 leaves, and therefore 928 pages, of folio size, excluding the lengthy preface by Dee. The notes certainly include all the most important that had ever been written, from those of the Greek commentators, Proclus and the others whom he quotes, down to those of Dee himself on the last books. Besides the fifteen Books, Billingsley included the sixteenth - added by Candalla. The print and appearance of the book are worthy of its contents; and, in order that it may be understood how no pains were spared to represent everything in the clearest and most perfect form, I need only mention that the figures of the propositions in Book XI. are nearly all duplicated, one being the figure of Euclid, the other an arrangement of pieces of paper (triangular, rectangular etc.) pasted at the edges on to the page of the book so that the pieces can be turned up and made to show the real form of the solid figures represented.

-

Billingsley was admitted Lady Margaret Scholar of St John's College, Cambridge, in 1551, and he is also said to have studied at Oxford, but he did not take a degree at either University. He was afterwards apprenticed to a London haberdasher and rapidly became a wealthy merchant. Sheriff of London in 1584, he was elected Lord Mayor on 31st December, 1596, on the death, during his year of office, of Sir Thomas Skinner. From 1589 he was one of the Queen's four customers, - or farmers of customs, of the port of London. In 1591 he founded three scholarships at St John's College for poor students, and gave to the College for their maintenance two messuages and tenements in Tower Street and in Mark Lane, Allhallows, Barking. He died in 1606.

-

1651. Elements of Geometry. The first VI Boocks: In a compendious form contracted and demonstrated by Captain Thomas Rudd, with the mathematicall preface of John Dee (London).

-

1660. The first English edition of Barrow's Euclid (published in Latin in 1655), appeared in London. It contained the whole fifteen books compendiously demonstrated -; several editions followed, in 1705, 1722, 1732, 1751.

-

1661. Euclid's Elements of Geometry, with a supplement of divers Propositions and Corollaries. To which is added a Treatise of regular Solids by Campane and Flussat; likewise Euclid's Data and Marinus his Preface. Also a Treatise of the Divisions of Superficies, ascribed to Machomet Bagdedine, but published by Commandine at the request of J. Dee of London. Published by care and industry of John Leeke and Geo. Serle, students in the Math. (London). According to Potts this was a second edition of Billingsley's translation.

-

1685. William Halifax's version of Dechales' Elements of Euclid explained in a new but most easy method - (London and Oxford).

-

1705. The English Euclide; being the first six Elements of Geometry, translated out of the Greek, with annotations and usefull supplements by Edmund Scarburgh (Oxford). A noteworthy and useful edition.

-

1708. Books I.—VI., XI., XII., translated from Commandinus' Latin version by Dr John Keill, Savilian Professor of Astronomy at Oxford.

-

Keill complains in his preface of the omissions by such editors as Tacquet and Dechales of many necessary propositions (e.g. VI. 27-29), and of their substitution of proofs of their own for Euclid's. He praises Barrow's version on the whole, though objecting to the algebraical - form of proof adopted in Book II., and to the excessive use of notes and symbols, which (he considers) make the proofs too short and thereby obscure; his edition was therefore intended to hit a proper mean between Barrow's excessive brevity and Clavius' prolixity.

-

Keill's translation was revised by Samuel Cunn and several times reissued. 1749 saw the eighth edition, 1772 the eleventh, and 1782 the twelfth.

-

1714. W. Whiston's English version (abridged) of The Elements of Euclid with select theorems out of Archimedes by the learned Andr. Tacquet.

-

1756. Simson's first English edition appeared in the same year as his Latin version under the title: -

The Elements of Euclid, viz. the first six Books together with the eleventh and twelfth. In this Edition the Errors by which Theon or others have long ago vitiated these Books are corrected and some of Euclid's Demonstrations are restored. By Robert Simson (Glasgow).

-

-

As above stated, the Latin edition, by its title, purports to be ex versione latina Federici Commandini, - but to the Latin edition, as well as to the English editions, are appended -

Notes Critical and Geometrical; containing an Account of those things in which this Edition differs from the Greek text; and the Reasons of the Alterations which have been made. As also Observations on some of the Propositions.

-

-

Simson says in the Preface to some editions (e.g. the tenth, of 1799) that the translation is much amended by the friendly assistance of a learned gentleman. -

-

Simson's version and his notes are so well known as not to need any further description. The book went through some thirty successive editions. The first five appear to have been dated 1756, 1762, 1767, 1772 and 1775 respectively; the tenth 1799, the thirteenth 1806, the twenty-third 1830, the twenty-fourth 1834, the twenty-sixth 1844 The Data in like manner corrected - was added for the first time in the edition of 1762 (the first octavo edition).

-

1781, 1788. In these years respectively appeared the two volumes containing the complete translation of the whole thirteen Books by James Williamson, the last English translation which reproduced Euclid word for word. The title is -

The Elements of Euclid, with Disserlations intended to assist and encourage a critical examination of these Elements, as the most effectual means of establishing a juster taste upon mathematical subjects than that which at present prevails. By James Williamson.

-

-

In the first volume (Oxford, 1781) he is described as M.A. Fellow of Hertford College, - and in the second (London, printed by T. Spilsbury, 1788) as B.D. - simply. Books V., VI. with the Conclusion in the first volume are paged separately from the rest.

-

1781. An examination of the first six Books of Euclid's Elements, by William Austin (London).

-

1795. John Playfair's first edition, containing the first six Books of Euclid with two Books on the Geometry of Solids. - The book reached a fifth edition in 1819, an eighth in 1831, a ninth in 1836, and a tenth in 1846.

-

1826. Riccardi notes under this date Euclid's Elements of Geometry containing the whole twelve Books translated into English, from the edition of Peyrard, by George Phillips. The editor, who was President of Queens' College, Cambridge, 1857-1892, was born in 1804 and matriculated at Queens' in 1826, so that he must have published the book as an undergraduate.

-

1828. A very valuable edition of the first six Books is that of Dionysius Lardner, with commentary and geometrical exercises, to which he added, in place of Books XI., XII., a Treatise on Solid Geometry mostly based on Legendre. Lardner compresses the propositions by combining the enunciation and the setting-out, and he gives a vast number of riders and additional propositions in smaller print. The book had reached a ninth edition by 1846, and an eleventh by 1855. Among other things, Lardner gives an Appendix on the theory of parallel lines, - in which he gives a short history of the attempts to get over the difficulty of the parallel-postulate, down to that of Legendre.

-

1833. T. Perronet Thompson's Geometry without axioms, or the first Book of Euclid's Elements with alterations and notes; and an intercalary book in which the straight line and plane are derived from properties of the sphere, with an appendix containing notices of methods proposed for getting over the difficulty in the twelfth axiom of Euclid.

-

Thompson (1783-1869) was 7th wrangler 1802, midshipman 1803, Fellow of Queens' College, Cambridge, 1804, and afterwards general and politician. The book went through several editions, but, having been well translated into French by Van Tenac, is said to have received more recognition in France than at home.

-

1845. Robert Potts' first edition (and one of the best) entitled: -

Euclid's Elements of Geometry chiefly from the text of Dr Simson with explanatory notes...to which is prefixed an introduction containing a brief outline of the History of Geometry. Designed for the use of the higher forms in Public Schools and students in the Universities (Cambridge University Press, and London, John W. Parker), to which was added (1847) An Appendix to the larger edition of Euclid's Elements of Geometry, containing additional notes on the Elements, a short tract on transversals, and hints for the solution of the problems etc

- .

-

1862. Todhunter's edition.

-

The later English editions I will not attempt to enumerate; their name is legion and their object mostly that of adapting Euclid for school use, with all possible gradations of departure from his text and order.

- - - IX. Spanish. -

1576. The first six Books translated into Spanish by Rodrigo Çamorano.

-

1637. The first six Books translated, with notes, by L. Carduchi.

-

1689. Books I.—VI., XI., XII., translated and explained by Jacob Knesa.

- - - - X. Russian. -

1739. Ivan Astaroff (translation from Latin).

-

1789. Pr. Suvoroff and Yos. Nikitin (translation from Greek).

-

1880. Vachtchenko-Zakhartchenko.

-

(1817. A translation into Polish by Jo. Czecha.)

- - - XI. Swedish. -

1744. Mårten Strömer, the first six Books; second edition 1748. The third edition (1753) contained Books XI.—XII. as well; new editions continued to appear till 1884.

-

1836. H. Falk, the first six Books.

-

1844, 1845, 1859. P. R. Bråkenhjelm, Books I.—VI., XI., XII.

-

1850. F. A. A. Lundgren.

-

1850. H. A. Witt and M. E. Areskong, Books I.—VI., XI., XII.

- - - XII. Danish. -

1745. Ernest Gottlieb Ziegenbalg.

-

1803. H. C. Linderup, Books I.—VI.

- - - XIII. Modern Greek. -

1820. Benjamin of Lesbos.

-

I should add a reference to certain editions which have appeared in recent years.

-

A Danish translation (Euklid's Elementer oversat af Thyra Eibe) was completed in 1912; Books I.—II. were published (with an Introduction by Zeuthen) in 1897, Books III.—IV. in 1900, Books V.—VI. in 1904, Books VII.—XIII. in 1912.

-

The Italians, whose great services to elementary geometry are more than once emphasised in this work, have lately shown a noteworthy disposition to make the ipsissima verba of Euclid once more the object of study. Giovanni Vacca has edited the text of Book I. (Il primo libro degli Elementi. Testo greco, versione italiana, introduzione e note, Firenze 1916.) Federigo Enriques has begun the publication of a complete Italian translation (Gli Elementi d' Euclide e la critica antica e moderna); Books I.—IV. appeared in 1925 (Alberto Stock, Roma).

-

An edition of Book I. by the present writer was published in 1918 (Euclid in Greek, Book I., with Introduction and Notes, Camb. Univ. Press).

- - - - - CHAPTER IX. - - § 1. ON THE NATURE OF ELEMENTS. -

It would not be easy to find a more lucid explanation of the terms element and elementary, and of the distinction between them, than is found in ProclusProclus, Comm. on Eucl. I., ed. Friedlein, pp. 72 sqq., who is doubtless, here as so often, quoting from Geminus. There are, says Proclus, in the whole of geometry certain leading theorems, bearing to those which follow the relation of a principle, all-pervading, and furnishing proofs of many properties. Such theorems are called by the name of elements; and their function may be compared to that of the letters of the alphabet in relation to language, letters being indeed called by the same name in Greek (stoixei=a).

-

The term elementary, on the other hand, has a wider application: it is applicable to things which extend to greater multiplicity, and, though possessing simplicity and elegance, have no longer the same dignity as the elements, because their investigation is not of general use in the whole of the science, e.g. the proposition that in triangles the perpendiculars from the angles to the transverse sides meet in a point. -

-

“Again, the term element is used in two senses, as Menaechmus says. For that which is the means of obtaining is an element of that which is obtained, as the first proposition in Euclid is of the second, and the fourth of the fifth. In this sense many things may even be said to be elements of each other, for they are obtained from one another. Thus from the fact that the exterior angles of rectilineal figures are (together) equal to four right angles we deduce the number of right angles equal to the internal angles (taken together)to\ plh=qos tw=n e)nto\s o)rqai=s i)/swn. If the text is right, we must apparently take it as the number of the angles equal to right angles that there are inside, - i.e. that are made up by the internal angles, and vice versa. Such an element is like a lemma. But the term element is otherwise used of that into which, being more simple, the composite is divided; and in this sense we can no longer say that everything is an element of everything, but only that things which are more of the nature of principles are elements of those which stand to them in the relation of results, as postulates are elements of theorems. It is according to this signification of the term element that the elements found in Euclid were compiled, being partly those of plane geometry, and partly those of stereometry. In like manner many writers have drawn up elementary treatises in arithmetic and astronomy.

-

“Now it is difficult, in each science, both to select and arrange in due order the elements from which all the rest proceeds, and into which all the rest is resolved. And of those who have made the attempt some were able to put together more and some less; some used shorter proofs, some extended their investigation to an indefinite length; some avoided the method of reductio ad absurdum, some avoided proportion; some contrived preliminary steps directed against those who reject the principles; and, in a word, many different methods have been invented by various writers of elements.

-

It is essential that such a treatise should be rid of everything superfluous (for this is an obstacle to the acquisition of knowledge); it should select everything that embraces the subject and brings it to a point (for this is of supreme service to science); it must have great regard at once to clearness and conciseness (for their opposites trouble our understanding); it must aim at the embracing of theorems in general terms (for the piecemeal division of instruction into the more partial makes knowledge difficult to grasp). In all these ways Euclid's system of elements will be found to be superior to the rest; for its utility avails towards the investigation of the primordial figurestw=n a)rxikw=n sxhma/twn, by which Proclus probably means the regular polyhedra (Tannery, P. 143 n.)., its clearness and organic perfection are secured by the progression from the more simple to the more complex and by the foundation of the investigation upon common notions, while generality of demonstration is secured by the progression through the theorems which are primary and of the nature of principles to the things sought. As for the things which seem to be wanting, they are partly to be discovered by the same methods, like the construction of the scalene and isosceles (triangle), partly alien to the character of a selection of elements as introducing hopeless and boundless complexity, like the subject of unordered irrationals which Apollonius worked out at lengthWe have no more than the most obscure indications of the character of this work in an Arabic MS. analysed by Woepcke, Essai d'une restitution de travaux perdus d'Apollonius sur les quantités irrationelles d'après des indications tirées d'un manuscrit arabe in Mémoires présentés à l'académie des sciences, XIV. 658-720, Paris, 1856. Cf. Cantor, Gesch. d. Math. I_{3}, pp. 348-9: details are also given in my notes to Book X., and partly developed from things handed down (in the elements) as causes, like the many species of angles and of lines. These things then have been omitted in Euclid, though they have received full discussion in other works; but the knowledge of them is derived from the simple (elements). -

-

Proclus, speaking apparently on his own behalf, in another place distinguishes two objects aimed at in Euclid's Elements. The first has reference to the matter of the investigation, and here, like a good Platonist, he takes the whole subject of geometry to be concerned with the cosmic figures, - the five regular solids, which in Book XIII. are constructed, inscribed in a sphere and compared with one another. The second object is relative to the learner; and, from this standpoint, the elements may be described as a means of perfecting the learner's understanding with reference to the whole of geometry. For, starting from these (elements), we shall be able to acquire knowledge of the other parts of this science as well, while without them it is impossible for us to get a grasp of so complex a subject, and knowledge of the rest is unattainable. As it is, the theorems which are most of the nature of principles, most simple, and most akin to the first hypotheses are here collected, in their appropriate order; and the proofs of all other propositions use these theorems as thoroughly well known, and start from them. Thus Archimedes in the books on the sphere and cylinder, Apollonius, and all other geometers, clearly use the theorems proved in this very treatise as constituting admitted principlesProclus, pp. 70, 19-71, 21. -

-

Aristotle too speaks of elements of geometry in the same sense. Thus: - in geometry it is well to be thoroughly versed in the elements - Topics VIII. 14, 163 b 23. - ; in general the first of the elements are, given the definitions, e.g. of a straight line and of a circle, most easy to prove, although of course there are not many data that can be used to establish each of them because there are not many middle terms - Topics VIII. 3, 158 b 35. - ; among geometrical propositions we call those ’elements’ the proofs of which are contained in the proofs of all or most of such propositions - - Metaph. 998 a 25 - .; (as in the case of bodies), so in like manner we speak of the elements of geometrical propositions and, generally, of demonstrations; for the demonstrations which come first and are contained in a variety of other demonstrations are called elements of those demonstrations... the term element is applied by analogy to that which, being one and small, is useful for many purposes - Metaph. 1014 a 35-b 5..

- - - § 2. ELEMENTS ANTERIOR TO EUCLID'S. -

The early part of the famous summary of Proclus was no doubt drawn, at least indirectly, from the history of geometry by Eudemus; this is generally inferred from the remark, made just after the mention of Philippus of Medma, a disciple of Plato, that those who have written histories bring the development of this science up to this point. - We have therefore the best authority for the list of writers of elements given in the summary. Hippocrates of Chios (fl. in second half of 5th c.) is the first; then Leon, who also discovered diorismi, put together a more careful collection, the propositions proved in it being more numerous as well as more serviceableProclus, p. 66, 20 w)/ste to\n *le/onta kai\ ta\ stoixei=a sunqei=nai tw=| te plh(qei kai\ th= xrei/a| tw=n deiknume/nwn e)pimele/steron.. Leon was a little older than Eudoxus (about 408-355 B.C.) and a little younger than Plato (428/7-347/6 B.C.), but did not belong to the latter's school. The geometrical text-book of the Academy was written by Theudius of Magnesia, who, with Amyclas of Heraclea, Menaechmus the pupil of Eudoxus, Menaechmus' brother Dinostratus and Athenaeus of Cyzicus consorted together in the Academy and carried on their investigations in common. Theudius put together the elements admirably, making many partial (or limited) propositions more generalProclus, p. 67, 14 kai\ ga\r ta\ stoixei=a kalw=s sune/tacen kai\ polla\ tw=n merikw=n [o(rikw=n (?) Friedlein] kaqolikw/tera e)poi/hsen.. - Eudemus mentions no text-book after that of Theudius, only adding that Hermotimus of Colophon discovered many of the elementsProclus, p. 67, 22 tw=n stoixei/wn polla\ a)neu=re.. - Theudius then must be taken to be the immediate precursor of Euclid, and no doubt Euclid made full use of Theudius as well as of the discoveries of Hermotimus and all other available material. Naturally it is not in Euclid's Elements that we can find much light upon the state of the subject when he took it up; but we have another source of information in Aristotle. Fortunately for the historian of mathematics, Aristotle was fond of mathematical illustrations; he refers to a considerable number of geometrical propositions, definitions etc., in a way which shows that his pupils must have had at hand some textbook where they could find the things he mentions; and this text-book must have been that of Theudius. Heiberg has made a most valuable collection of mathematical extracts from AristotleMathematisches zu Aristoteles in Abhandlungen zur Gesch. d. math. Wissenschaften, XVIII. Heft (1904), pp. 1-49., from which much is to be gathered as to the changes which Euclid made in the methods of his predecessors; and these passages, as well as others not included in Heiberg's selection, will often be referred to in the sequel.

- - - § 3. FIRST PRINCIPLES: DEFINITIONS, POSTULATES, AND AXIOMS. -

On no part of the subject does Aristotle give more valuable information than on that of the first principles as, doubtless, generally accepted at the time when he wrote. One long passage in the Posterior Analytics is particularly full and lucid, and is worth quoting in extenso. After laying it down that every demonstrative science starts from necessary principlesAnal. post. 1. 6, 74 b 5., he proceedsibid. 1. 10, 76 a 31-77 a 4.:

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“By first principles in each genus I mean those the truth of which it is not possible to prove. What is denoted by the first (terms) and those derived from them is assumed; but, as regards their existence, this must be assumed for the principles but proved for the rest. Thus what a unit is, what the straight (line) is, or what a triangle is (must be assumed); and the existence of the unit and of magnitude must also be assumed, but the rest must be proved. Now of the premisses used in demonstrative sciences some are peculiar to each science and others common (to all), the latter being common by analogy, for of course they are actually useful in so far as they are applied to the subject-matter included under the particular science. Instances of first principles peculiar to a science are the assumptions that a line is of such and such a character, and similarly for the straight (line); whereas it is a common principle, for instance, that, if equals be subtracted from equals, the remainders are equal. But it is enough that each of the common principles is true so far as regards the particular genus (subject-matter); for (in geometry) the effect will be the same even if the common principle be assumed to be true, not of everything, but only of magnitudes, and, in arithmetic, of numbers.

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“Now the things peculiar to the science, the existence of which must be assumed, are the things with reference to which the science investigates the essential attributes, e.g. arithmetic with reference to units, and geometry with reference to points and lines. With these things it is assumed that they exist and that they are of such and such a nature. But, with regard to their essential properties, what is assumed is only the meaning of each term employed: thus arithmetic assumes the answer to the question what is (meant by) ’odd’ or ’even,’ ’a square’ or ’a cube,’ and geometry to the question what is (meant by) ’the irrational’ or ’deflection’ or (the so-called) ’verging’ (to a point); but that there are such things is proved by means of the common principles and of what has already been demonstrated. Similarly with astronomy. For every demonstrative science has to do with three things, (1) the things which are assumed to exist, namely the genus (subject-matter) in each case, the essential properties of which the science investigates, (2) the common axioms so-called, which are the primary source of demonstration, and (3) the properties with regard to which all that is assumed is the meaning of the respective terms used. There is, however, no reason why some sciences should not omit to speak of one or other of these things. Thus there need not be any supposition as to the existence of the genus, if it is manifest that it exists (for it is not equally clear that number exists and that cold and hot exist); and, with regard to the properties, there need be no assumption as to the meaning of terms if it is clear: just as in the common (axioms) there is no assumption as to what is the meaning of subtracting equals from equals, because it is well known. But none the less is it true that there are three things naturally distinct, the subject-matter of the proof, the things proved, and the (axioms) from which (the proof starts).

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Now that which is per se necessarily true, and must necessarily be thought so, is not a hypothesis nor yet a postulate. For demonstration has not to do with reasoning from outside but with the reason dwelling in the soul, just as is the case with the syllogism. It is always possible to raise objection to reasoning from outside, but to contradict the reason within us is not always possible. Now anything that the teacher assumes, though it is matter of proof, without proving it himself, is a hypothesis if the thing assumed is believed by the learner, and it is moreover a hypothesis, not absolutely, but relatively to the particular pupil; but, if the same thing is assumed when the learner either has no opinion on the subject or is of a contrary opinion, it is a postulate. This is the difference between a hypothesis and a postulate; for a postulate is that which is rather contrary than otherwise to the opinion of the learner, or whatever is assumed and used without being proved, although matter for demonstration. Now definitions are not hypotheses, for they do not assert the existence or non-existence of anything, while hypotheses are among propositions. Definitions only require to be understood: a definition is therefore not a hypothesis, unless indeed it be asserted that any audible speech is a hypothesis. A hypothesis is that from the truth of which, if assumed, a conclusion can be established. Nor are the geometer's hypotheses false, as some have said: I mean those who say that ’you should not make use of what is false, and yet the geometer falsely calls the line which he has drawn a foot long when it is not, or straight when it is not straight.’ The geometer bases no conclusion on the particular line which he has drawn being that which he has described, but (he refers to) what is illustrated by the figures. Further, the postulate and every hypothesis are either universal or particular statements; definitions are neither - (because the subject is of equal extent with what is predicated of it).

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Every demonstrative science, says Aristotle, must start from indemonstrable principles: otherwise, the steps of demonstration would be endless. Of these indemonstrable principles some are (a) common to all sciences, others are (b) particular, or peculiar to the particular science; (a) the common principles are the axioms, most commonly illustrated by the axiom that, if equals be subtracted from equals, the remainders are equal. Coming now to (b) the principles peculiar to the particular science which must be assumed, we have first the genus or subject-matter, the existence of which must be assumed, viz. magnitude in the case of geometry, the unit in the case of arithmetic. Under this we must assume definitions of manifestations or attributes of the genus, e.g. straight lines, triangles, deflection etc. The definition in itself says nothing as to the existence of the thing defined: it only requires to be understood. But in geometry, in addition to the genus and the definitions, we have to assume the existence of a few primary things which are defined, viz. points and lines only: the existence of everything else, e.g. the various figures made up of these, as triangles, squares, tangents, and their properties, e.g. incommensurability etc., has to be proved (as it is proved by construction and demonstration). In arithmetic we assume the existence of the unit: but, as regards the rest, only the definitions, e.g. those of odd, even, square, cube, are assumed, and existence has to be proved. We have then clearly distinguished, among the indemonstrable principles, axioms and definitions. A postulate is also distinguished from a hypothesis, the latter being made with the assent of the learner, the former without such assent or even in opposition to his opinion (though, strangely enough, immediately after saying this, Aristotle gives a wider meaning to postulate - which would cover hypothesis - as well, namely whatever is assumed, though it is matter for proof, and used without being proved). Heiberg remarks that there is no trace in Aristotle of Euclid's Postulates, and that postulate - in Aristotle has a different meaning. He seems to base this on the alternative description of postulate, indistinguishable from a hypothesis; but, if we take the other description in which it is distinguished from a hypothesis as being an assumption of something which is a proper subject of demonstration without the assent or against the opinion of the learner, it seems to fit Euclid's Postulates fairly well, not only the first three (postulating three constructions), but eminently also the other two, that all right angles are equal, and that two straight lines meeting a third and making the internal angles on the same side of it less than two right angles will meet on that side. Aristotle's description also seems to me to suit the postulates - with which Archimedes begins his book On the equilibrium of planes, namely that equal weights balance at equal distances, and that equal weights at unequal distances do not balance but that the weight at the longer distance will prevail.

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Aristotle's distinction also between hypothesis and definition, and between hypothesis and axiom, is clear from the following passage: Among immediate syllogistic principles, I call that a thesis which it is neither possible to prove nor essential for any one to hold who is to learn anything; but that which it is necessary for any one to hold who is to learn anything whatever is an axiom: for there are some principles of this kind, and that is the most usual name by which we speak of them. But, of theses, one kind is that which assumes one or other side of a predication, as, for instance, that something exists or does not exist, and this is a hypothesis; the other, which makes no such assumption, is a definition. For a definition is a thesis: thus the arithmetician posits (ti/qetai) that a unit is that which is indivisible in respect of quantity; but this is not a hypothesis, since what is meant by a unit and the fact that a unit exists are different thingsAnal. post. 1. 2, 72 a 14-24.. -

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Aristotle uses as an alternative term for axioms common (things), - ta\ koina/, or common opinions - (koinai\ do/cai), as in the following passages. That, when equals are taken from equals, the remainders are equal is (a) common (principle) in the case of all quantities, but mathematics takes a separate department (a)polabou=sa) and directs its investigation to some portion of its proper subject-matter, as e.g. lines or angles, numbers, or any of the other quantitiesMetaph. 1061 b 19-24.. - The common (principles), e.g. that one of two contradictories must be true, that equals taken from equals etc., and the likeAnal. post. 1. 11, 77 a 30..... - With regard to the principles of demonstration, it is questionable whether they belong to one science or to several. By principles of demonstration I mean the common opinions from which all demonstration proceeds, e.g. that one of two contradictories must be true, and that it is impossible for the same thing to be and not beMetaph. 996 b 26-30.. - Similarly every demonstrative (science) investigates, with regard to some subject-matter, the essential attributes, starting from the common opinionsMetaph. 997 a 20-22.. - We have then here, as Heiberg says, a sufficient explanation of Euclid's term for axioms, viz. common notions (koinai\ e)/nnoiai), and there is no reason to suppose it to be a substitution for the original term due to the Stoics: cf. Proclus' remark that, according to Aristotle and the geometers, axiom and common notion are the same thingProclus, p. 194, 8..

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Aristotle discusses the indemonstrable character of the axioms in the Metaphysics. Since all the demonstrative sciences use the axiomsMetaph. 997 a 10., - the question arises, to what science does their discussion belongibid. 996 b 26.? The answer is that, like that of Being (ou)si/a), it is the province of the (first) philosopheribid. 1005 a 21—b 11.. It is impossible that there should be demonstration of everything, as there would be an infinite series of demonstrations: if the axioms were the subject of a demonstrative science, there would have to be here too, as in other demonstrative sciences, a subject-genus, its attributes and corresponding axiomsibid. 997 a 5-8.; thus there would be axioms behind axioms, and so on continually. The axiom is the most firmly established of all principlesibid. 1005 b 11-17.. It is ignorance alone that could lead any one to try to prove the axiomsibid. 1006 a 5.; the supposed proof would be a petitio principiiibid. 1006 a 17.. If it is admitted that not everything can be proved, no one can point to any principle more truly indemonstrableibid. 1006 a 10.. If any one thought he could prove them, he could at once be refuted; if he did not attempt to say anything, it would be ridiculous to argue with him: he would be no better than a vegetableibid. 1006 a 11-15.. The first condition of the possibility of any argument whatever is that words should signify something both to the speaker and to the hearer: without this there can be no reasoning with any one. And, if any one admits that words can mean anything to both hearer and speaker, he admits that something can be true without demonstration. And so onibid. 1006 a 18 sqq..

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It was necessary to give some sketch of Aristotle's view of the first principles, if only in connexion with Proclus' account, which is as follows. As in the case of other sciences, so the compiler of elements in geometry must give separately the principles of the science, and after that the conclusions from those principles, not giving any account of the principles but only of their consequences. No science proves its own principles, or even discourses about them: they are treated as self-evident....Thus the first essential was to distinguish the principles from their consequences. Euclid carries out this plan practically in every book and, as a preliminary to the whole enquiry, sets out the common principles of this science. Then he divides the common principles themselves into hypotheses, postulates, and axioms. For all these are different from one another: an axiom, a postulate and a hypothesis are not the same thing, as the inspired Aristotle somewhere says. But, whenever that which is assumed and ranked as a principle is both known to the learner and convincing in itself, such a thing is an axiom, e.g. the statement that things which are equal to the same thing are also equal to one another. When, on the other hand, the pupil has not the notion of what is told him which carries conviction in itself, but nevertheless lays it down and assents to its being assumed, such an assumption is a hypothesis. Thus we do not preconceive by virtue of a common notion, and without being taught, that the circle is such and such a figure, but, when we are told so, we assent without demonstration. When again what is asserted is both unknown and assumed even without the assent of the learner, then, he says, we call this a postulate, e.g. that all right angles are equal. This view of a postulate is clearly implied by those who have made a special and systematic attempt to show, with regard to one of the postulates, that it cannot be assented to by any one straight off. According then to the teaching of Aristotle, an axiom, a postulate and a hypothesis are thus distinguishedProclus, pp. 75, 10-77, 2.. -

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We observe, first, that Proclus in this passage confuses hypotheses and definitions, although Aristotle had made the distinction quite plain. The confusion may be due to his having in his mind a passage of Plato from which he evidently got the phrase about not giving an account of - the principles. The passage isRepublic, VI. 510 c. Cf. Aristotle, Nic. Eth. 1151 a 17.: I think you know that those who treat of geometries and calculations (arithmetic) and such things take for granted (u(poqe/menoi) odd and even, figures, angles of three kinds, and other things akin to these in each subject, implying that they know these things, and, though using them as hypotheses, do not even condescend to give any account of them either to themselves or to others, but begin from these things and then go through everything else in order, arriving ultimately, by recognised methods, at the conclusion which they started in search of. - But the hypothesis is here the assumption, e.g. ’that there may be such a thing as length without breadth, henceforward called a lineH. Jackson, Journal of Philology, vol. x. p. 144.,’ and so on, without any attempt to show that there is such a thing; it is mentioned in connexion with the distinction between Plato's ’superior’ and ’inferior’ intellectual method, the former of which uses successive hypotheses as stepping-stones by which it mounts upwards to the idea of Good.

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We pass now to Proclus' account of the difference between postulates and axioms. He begins with the view of Geminus, according to which they differ from one another in the same way as theorems are also distinguished from problems. For, as in theorems we propose to see and determine what follows on the premisses, while in problems we are told to find and do something, in like manner in the axioms such things are assumed as are manifest of themselves and easily apprehended by our untaught notions, while in the postulates we assume such things as are easy to find and effect (our understanding suffering no strain in their assumption), and we require no complication of machineryProclus, pp. 178, 12-179, 8. In illustration Proclus contrasts the drawing of a straight line or a circle with the drawing of a single-turn spiral - or of an equilateral triangle, the spiral requiring more complex machinery and even the equilateral triangle needing a certain method. For the geometrical intelligence will say that by conceiving a straight line fixed at one end but, as regards the other end, moving round the fixed end, and a point moving along the straight line from the fixed end, I have described the single-turn spiral; for the end of the straight line describing a circle, and the point moving on the straight line simultaneously, when they arrive and meet at the same point, complete such a spiral. And again, if I draw equal circles, join their common point to the centres of the circles and draw a straight line from one of the centres to the other, I shall have the equilateral triangle. These things then are far from being completed by means of a single act or of a moment's thought - (p. 180, 8-21).. - ...Both must have the characteristic of being simple and readily grasped, I mean both the postulate and the axiom; but the postulate bids us contrive and find some subject-matter (u(/lh) to exhibit a property simple and easily grasped, while the axiom bids us assert some essential attribute which is self-evident to the learner, just as is the fact that fire is hot, or any of the most obvious thingsProclus, p. 181,4-11.. -

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Again, says Proclus, some claim that all these things are alike postulates, in the same way as some maintain that all things that are sought are problems. For Archimedes begins his first book on InequilibriumIt is necessary to coin a word to render a)nisorropiw=n, which is moreover in the plural. The title of the treatise as we have it is Equilibria of planes or cenires of gravity of planes in Book I and Equilibria of planes in Book II. with the remark ’I postulate that equal weights at equal distances are in equilibrium,’ though one would rather call this an axiom. Others call them all axioms in the same way as some regard as theorems everything that requires demonstrationProclus, p. 181, 16-23.. -

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Others again will say that postulates are peculiar to geometrical subject-matter, while axioms are common to all investigation which is concerned with quantity and magnitude. Thus it is the geometer who knows that all right angles are equal and how to produce in a straight line any limited straight line, whereas it is a common notion that things which are equal to the same thing are also equal to one another, and it is employed by the arithmetician and any scientific person who adapts the general statement to his own subjectibid. p. 182, 6-14.. -

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The third view of the distinction between a postulate and an axiom is that of Aristotle above describedPp. 118, 119..

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The difficulties in the way of reconciling Euclid's classification of postulates and axioms with any one of the three alternative views are next dwelt upon. If we accept the first view according to which an axiom has reference to something known, and a postulate to something done, then the 4th postulate (that all right angles are equal) is not a postulate; neither is the 5th which states that, if a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the straight lines, if produced indefinitely, will meet on that side on which are the angles less than two right angles. On the second view, the assumption that two straight lines cannot enclose a space, which even now, - says Proclus, some add as an axiom, - and which is peculiar to the subject-matter of geometry, like the fact that all right angles are equal, is not an axiom. According to the third (Aristotelian) view, everything which is confirmed (pistou=tai) by a sort of demonstration will be a postulate, and what is incapable of proof will be an axiomProclus, pp. 182, 21-183, 13.. - This last statement of Proclus is loose, as regards the axiom, because it omits Aristotle's requirement that the axiom should be a selfevident truth, and one that must be admitted by any one who is to learn anything at all, and, as regards the postulate, because Aristotle calls a postulate something assumed without proof though it is matter of demonstration - (a)podeikto\n o)/n), but says nothing of a quasi-demonstration of the postulates. On the whole I think it is from Aristotle that we get the best idea of what Euclid understood by a postulate and an axiom or common notion. Thus Aristotle's account of an axiom as a principle common to all sciences, which is self-evident, though incapable of proof, agrees sufficiently with the contents of Euclid's common notions as reduced to five in the most recent text (not omitting the fourth, that things which coincide are equal to one another -). As regards the postulates, it must be borne in mind that Aristotle says elsewhereAnal. post. 1. 25, 86 a 33-35. that, other things being equal, that proof is the better which proceeds from the fewer postulates or hypotheses or propositions. - If then we say that a geometer must lay down as principles, first certain axioms or common notions, and then an irreducible minimum of postulates in the Aristotelian sense concerned only with the subject-matter of geometry, we are not far from describing what Euclid in fact does. As regards the postulates we may imagine him saying: Besides the common notions there are a few other things which I must assume without proof, but which differ from the common notions in that they are not self-evident. The learner may or may not be disposed to agree to them; but he must accept them at the outset on the superior authority of his teacher, and must be left to convince himself of their truth in the course of the investigation which follows. In the first place certain simple constructions, the drawing and producing of a straight line, and the drawing of a circle, must be assumed to be possible, and with the constructions the existence of such things as straight lines and circles; and besides this we must lay down some postulate to form the basis of the theory of parallels. - It is true that the admission of the 4th postulate that all right angles are equal still presents a difficulty to which we shall have to recur.

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There is of course no foundation for the idea, which has found its way into many text-books, that the object of the postulates is to declare that the only instruments the use of which is permitted in geometry are the rule and compassCf. Lardner's Euclid: also Todhunter.. -

- - - § 4. THEOREMS AND PROBLEMS. -

Again the deductions from the first principles, - says Proclus, are divided into problems and theorems, the former embracing the generation, division, subtraction or addition of figures, and generally the changes which are brought about in them, the latter exhibiting the essential attributes of eachProclus, p. 77, 7-12.. -

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“Now, of the ancients, some, like Speusippus and Amphinomus, thought proper to call them all theorems, regarding the name of theorems as more appropriate than that of problems to theoretic sciences, especially as these deal with eternal objects. For there is no becoming in things eternal, so that neither could the problem have any place with them, since it promises the generation and making of what has not before existed, e.g. the construction of an equilateral triangle, or the describing of a square on a given straight line, or the placing of a straight line at a given point. Hence they say it is better to assert that all (propositions) are of the same kind, and that we regard the generation that takes place in them as referring not to actual making but to knowledge, when we treat things existing eternally as if they were subject to becoming: in other words, we may say that everything is treated by way of theorem and not by way of problemibid. pp. 77, 15-78, 8. (pa/nta qewrhmatikw=s a)ll) ou) problhmatikw=s lamba/nesqai).

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“Others on the contrary, like the mathematicians of the school of Menaechmus, thought it right to call them all problems, describing their purpose as twofold, namely in some cases to furnish (pori/sasqai) the thing sought, in others to take a determinate object and see either what it is, or of what nature, or what is its property, or in what relations it stands to something else.

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In reality both assertions are correct. Speusippus is right because the problems of geometry are not like those of mechanics, the latter being matters of sense and exhibiting becoming and change of every sort. The school of Menaechmus are right also because the discoveries even of theorems do not arise without an issuing-forth into matter, by which I mean intelligible matter. Thus forms going out into matter and giving it shape may fairly be said to be like processes of becoming. For we say that the motion of our thought and the throwing-out of the forms in it is what produces the figures in the imagination and the conditions subsisting in them. It is in the imagination that constructions, divisions, placings, applications, additions and subtractions (take place), but everything in the mind is fixed and immune from becoming and from every sort of changeibid. pp. 78, 8-79, 2.. -

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Now those who distinguish the theorem from the problem say that every problem implies the possibility, not only of that which is predicated of its subject-matter, but also of its opposite, whereas every theorem implies the possibility of the thing predicated but not of its opposite as well. By the subject-matter I mean the genus which is the subject of inquiry, for example, a triangle or a square or a circle, and by the property predicated the essential attribute, as equality, section, position, and the like. When then any one enunciates thus, To inscribe an equilateral triangle in a circle, he states a problem; for it is also possible to inscribe in it a triangle which is not equilateral. Again, if we take the enunciation On a given limited straight line to construct an equilateral triangle, this is a problem; for it is possible also to construct one which is not equilateral. But, when any one enunciates that In isosceles triangles the angles at the base are equal, we must say that he enunciates a theorem; for it is not also possible that the angles at the base of isosceles triangles should be unequal. It follows that, if any one were to use the form of a problem and say In a semicircle to describe a right angle, he would be set down as no geometer. For every angle in a semicircle is rightProclus, pp. 79, 11-80, 5.. -

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Zenodotus, who belonged to the succession of Oenopides, but was a disciple of Andron, distinguished the theorem from the problem by the fact that the theorem inquires what is the property predicated of the subject-matter in it, but the problem what is the cause of what effect (ti/nos o)/ntos ti/ e)stin). Hence too Posidonius defined the one (the problem) as a proposition in which it is inquired whether a thing exists or not (ei) e)/stin h)\ mh/), the other (the theoremIn the text we have to\ de\ pro/blhma answering to to\ me\n without substantive: pro/blhma was obviously inserted in error.) as a proposition in which it is inquired what (a thing) is or of what nature (ti/ e)stin h)\ poi=o/n ti); and he said that the theoretic proposition must be put in a declaratory form, e.g., Any triangle has two sides (together) greater than the remaining side and In any isosceles triangle the angles at the base are equal, but that we should state the problematic propositi\on as if inquiring whether it is possible to construct an equilateral triangle upon such and such a straight line. For there is a difference between inquiring absolutely and indeterminately (a(plw=s te kai\ a)ori/stws) whether there exists a straight line from such and such a point at right angles to such and such a straight line and investigating which is the straight line at right anglesProclus, pp. 80, 15-81, 4.. -

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That there is a certain difference between the problem and the theorem is clear from what has been said; and that the Elements of Euclid contain partly problems and partly theorems will be made manifest by the individual propositions, where Euclid himself adds at the end of what is proved in them, in some cases, ’that which it was required to do,’ and in others, ’that which it was required to prove,’ the latter expression being regarded as characteristic of theorems, in spite of the fact that, as we have said, demonstration is found in problems also. In problems, however, even the demonstration is for the purpose of (confirming) the construction: for wė bring in the demonstration in order to show that what was enjoined has been done; whereas in theorems the demonstration is worthy of study for its own sake as being capable of putting before us the nature of the thing sought. And you will find that Euclid sometimes interweaves theorems with problems and employs them in turn, as in the first book, while at other times he makes one or other preponderate. For the fourth book consists wholly of problems, and the fifth of theoremsProclus, p. 81, 5-22.. -

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Again, in his note on Eucl. 1. 4, Proclus says that Carpus, the writer on mechanics, raised the question of theorems and problems in his treatise on astronomy. Carpus, we are told, says that the class of problems is in order prior to theorems. For the subjects, the properties of which are sought, are discovered by means of problems. Moreover in a problem the enunciation is simple and requires no skilled intelligence; it orders you plainly to do such and such a thing, to construct an equilateral triangle, or, given two straight lines, to cut off from the greater (a straight line) equal to the lesser, and what is there obscure or elaborate in these things? But the enunciation of a theorem is a matter of labour and requires much exactness and scientific judgment in order that it may not turn out to exceed or fall short of the truth; an example is found even in this proposition (1. 4), the first of the theorems. Again, in the case of problems, one general way has been discovered, that of analysis, by following which we can always hope to succeed; it is this method by which the more obscure problems are investigated. But, in the case of theorems, the method of setting about them is hard to get hold of since ’up to our time,’ says Carpus, ’no one has been able to hand down a general method for their discovery. Hence, by reason of their easiness, the class of problems would naturally be more simple.’ After these distinctions, he proceeds: ’Hence it is that in the Elements too problems precede theorems, and the Elements begin from them; the first theorem is fourth in order, not because the fifthto\ pe/mpton. This should apparently be the fourth because in the next words it is implied that none of the first three propositions are required in proving it. is proved from the problems, but because, even if it needs for its demonstration none of the propositions which precede it, it was necessary that they should be first because they are problems, while it is a theorem. In fact, in this theorem he uses the common notions exclusively, and in some sort takes the same triangle placed in different positions; the coincidence and the equality proved thereby depend entirely upon sensible and distinct apprehension. Nevertheless, though the demonstration of the first theorem is of this character, the problems properly preceded it, because in general problems are allotted the order of precedenceProclus, pp. 241, 19-243, 11..’ -

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Proclus himself explains the position of Prop. 4 after Props. 1-3 as due to the fact that a theorem about the essential properties of triangles ought not to be introduced before we know that such a thing as a triangle can be constructed, nor a theorem about the equality of sides or straight lines until we have shown, by constructing them, that there can be two straight lines which are equal to one anotheribid. pp. 233, 21-234, 6.. It is plausible enough to argue in this way that Props. 2 and 3 at all events should precede Prop. 4. And Prop. 1 is used in Prop. 2, and must therefore precede it. But Prop. I showing how to construct an equilateral triangle on a given base is not important, in relation to Prop. 4, as dealing with the production of triangles - in general: for it is of no use to say, as Proclus does, that the construction of the equilateral triangle is common to the three species (of triangles)Proclus, p. 234, 21., - as we are not in a position to know this at such an early stage. The existence of triangles in general was doubtless assumed as following from the existence of straight lines and points in one plane and from the possibility of drawing a straight line from one point to another.

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Proclus does not however seem to reject definitely the view of Carpus, for he goes onibid. p. 243, 12-25.: And perhaps problems are in order before theorems, and especially for those who need to ascend from the arts which are concerned with things of sense to theoretical investigation. But in dignity theorems are prior to problems....It is then foolish to blame Geminus for saying that the theorem is more perfect than the problem. For Carpus himself gave the priority to problems in respect of order, and Geminus to theorems in point of more perfect dignity, - so that there was no real inconsistency between the two.

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Problems were classified according to the number of their possible solutions. Amphinomus said that those which had a unique solution (monaxw=s) were called ordered - (the word has dropped out in Proclus, but it must be tetagme/na, in contrast to the third kind, a)/takta); those which had a definite number of solutions intermediate - (me/sa); and those with an infinite variety of solutions unordered - (a)/takta)ibid. p. 220, 7-12.. Proclus gives as an example of the last the problem To divide a given straight line into three parts in continued proportionibid. pp. 220, 16-221, 6.. This is the same thing as solving the equations . Proclus' remarks upon the problem show that it was solved, like all quadratic equations, by the method of application of areas. - The straight line a was first divided into any two parts, (x+z) and y, subject to the sole limitation that (x+z) must not be less than 2y, which limitation is the diorismo/s, or condition of possibility. Then an area was applied to (x+z), or (a-y), falling short by a square figure - (e)llei=pon ei)/dei tetragw/nw|) and equal to the square on y. This determines x and z separately in terms of a and y. For, if z be the side of the square by which the area (i.e. rectangle) falls short, - we have whence And y may be chosen arbitrarily, provided that it is not greater than a/3. Hence there are an infinite number of solutions. If then, as Proclus remarks, the three parts are equal.

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Other distinctions between different kinds of problems are added by Proclus. The word problem, - he says, is used in several senses. In its widest sense it may mean anything propounded - (proteino/menon), whether for the purpose of instruction (maqh/sews) or construction (poih/sews). (In this sense, therefore, it would include a theorem.) But its special sense in mathematics is that of something propounded with a view to a theoretic constructionProclus, p. 221, 7-11.. -

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Again you may apply the term (in this restricted sense) even to something which is impossible, although it is more appropriately used of what is possible and neither asks too much nor contains too little in the shape of data. According as a problem has one or other of these defects respectively, it is called (1) a problem in excess (pleona/zon) or (2) a deficient problem (e)llipe\s pro/blhma). The problem in excess (1) is of two kinds, (a) a problem in which the properties of the figure to be found are either inconsistent (a)su/mbata) or non-existent (a)nu/parkta), in which case the problem is called impossible, or (b) a problem in which the enunciation is merely redundant: an example of this would be a problem requiring us to construct an equilateral triangle with its vertical angle equal to two-thirds of a right angle; such a problem is possible and is called more than a problem - (mei=zon h)\ pro/blhma). The deficient problem (2) is similarly called less than a problem - (e)/lasson h)\ pro/blhma), its characteristic being that something has to be added to the enunciation in order to convert it from indeterminateness (a)oristi/a) to order (ta/cis) and scientific determinateness (o(/ros e)pisthmoniko/s): such would be a problem bidding you to construct an isosceles triangle, - for the varieties of isosceles triangles are unlimited. Such problems - are not problems in the proper sense (kuri/ws lego/mena problh/mata), but only equivocallyibid. pp. 221, 13-222,14..

- - - § 5. THE FORMAL DIVISIONS OF A PROPOSITION. -

Every problem, - says Proclusibid. pp. 203, 1-204, 13; 204, 23-205, 8., and every theorem which is complete with all its parts perfect purports to contain in itself all of the following elements: enunciation (pro/tasis), setting-out (e)/kqesis), definition or specification (diorismo/s), construction or machinery (kataskeuh/), proof (a)po/deicis), conclusion (sumpe/rasma). Now of these the enunciation states what is given and what is that which is sought, the perfect enunciation consisting of both these parts. The setting-out marks off what is given, by itself, and adapts it beforehand for use in the investigation. The definition or specification states separately and makes clear what the particular thing is which is sought. The construction or machinery adds what is wanting to the datum for the purpose of finding what is sought. The proof draws the required inference by reasoning scientifically from acknowledged facts. The conclusion reverts again to the enunciation, confirming what has been demonstrated. These are all the parts of problems and theorems, but the most essential and those which are found in all are enunciation, proof, conclusion. For it is equally necessary to know beforehand what is sought, to prove this by means of the intermediate steps, and to state the proved fact as a conclusion; it is impossible to dispense with any of these three things. The remaining parts are often brought in, but are often left out as serving no purpose. Thus there is neither setting-out nor definition in the problem of constructing an isosceles triangle having each of the angles at the base double of the remaining angle, and in most theorems there is no construction because the setting-out suffices without any addition for proving the required property from the data. When then do we say that the setting-out is wanting? The answer is, when there is nothing given in the enunciation; for, though the enunciation is in general divided into what is given and what is sought, this is not always the case, but sometimes it states only what is sought, i.e. what must be known or found, as in the case of the problem just mentioned. That problem does not, in fact, state beforehand with what datum we are to construct the isosceles triangle having each of the equal angles double of the remaining angle, but (simply) that we are to find such a triangle.... When, then, the enunciation contains both (what is given and what is sought), in that case we find both definition and setting-out, but, whenever the datum is wanting, they too are wanting. For not only is the setting-out concerned with the datum, but so is the definition also, as, in the absence of the datum, the definition will be identical with the enunciation. In fact, what could you say in defining the object of the aforesaid problem except that it is required to find an isosceles triangle of the kind referred to? But that is what the enunciation stated. If then the enunciation does not include, on the one hand, what is given and, on the other, what is sought, there is no setting-out in virtue of there being no datum, and the definition is left out in order to avoid a mere repetition of the enunciation. -

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The constituent parts of an Euclidean proposition will be readily identified by means of the above description. As regards the definition or specification (diorismo/s) it is to be observed that we have here only one of its uses. Here it means a closer definition or description of the object aimed at, by means of the concrete lines or figures set out in the e)/kqesis instead of the general terms used in the enunciation; and its purpose is to rivet the attention better, as Proclus indicates in a later passage (tro/pon tina\ prosexei/as e)sti\n ai)/tios o( diorismo/s)Proclus, p. 208, 21..

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The other technical use of the word to signify the limitations to which the possible solutions of a problem are subject is also described by Proclus, who speaks of diorismoi/ determining whether what is sought is impossible or possible, and how far it is practicable and in how many waysibid. p. 202, 3. -; and the diorismo/s in this sense appears in Euclid as well as in Archimedes and Apollonius. Thus we have in Eucl. 1. 22 the enunciation From three straight lines which are equal to three given straight lines to construct a triangle, - followed immediately by the limiting condition (diorismo/s). Thus two of the straight lines taken together in any manner must be greater than the remaining one. - Similarly in VI. 28 the enunciation To a given straight line to apply a parallelogram equal to a given rectilineal figure and falling short by a parallelogrammic figure similar to a given one - is at once followed by the necessary condition of possibility: Thus the given rectilineal figure must not be greater than that described on half the line and similar to the defect. -

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Tannery supposed that, in giving the other description of the diorismo/s as quoted above, Proclus, or rather his guide, was using the term incorrectly. The diorismo/s in the better known sense of the determination of limits or conditions of possibility was, we are told, invented by Leon. Pappus uses the word in this sense only. The other use of the term might, Tannery thought, be due to a confusion occasioned by the use of the same words (dei= dh/) in introducing the parts of a proposition corresponding to the two meanings of the word diorismo/sLa Géométrie grecque, p. 149 note. Where dei= dh\ introduces the closer description of the problem we may translate, it is then required - or thus it is required - (to constructetc.): when it introduces the condition of possibility we may translate thus it is necessary etc. - Heiberg originally wrote dei= de\ in the latter sense in 1. 22 on the authority of Proclus and Eutocius, and against that of the MSS. Later, on the occasion of XI. 23, he observed that he should have followed the MSS. and written dei= dh\ which he found to be, after all, the right reading in Eutocius (Apollonius, ed. Heiberg, II. p. 178). dei= dh\ is also the expression used by Diophantus for introducing conditions of possibility.. On the other hand it is to be observed that Eutocius distinguishes clearly between the two uses and implies that the difference was well known. See the passage of Eutocius referred to in last note.. The diorismo/s in the sense of condition of possibility follows immediately on the enunciation, is even part of it; the diorismo/s in the other sense of course comes immediately after the setting-out.

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Proclus has a useful observation respecting the conclusion of a propositionProclus, p. 207, 4-25.. The conclusion they are accustomed to make double in a certain way: I mean, by proving it in the given case and then drawing a general inference, passing, that is, from the partial conclusion to the general. For, inasmuch as they do not make use of the individuality of the subjects taken, but only draw an angle or a straight line with a view to placing the datum before our eyes, they consider that this same fact which is established in the case of the particular figure constitutes a conclusion true of every other figure of the same kind. They pass accordingly to the general in order that we may not conceive the conclusion to be partial. And they are justified in so passing, since they use for the demonstration the particular things set out, not quâ particulars, but quâ typical of the rest. For it is not in virtue of such and such a size attaching to the angle which is set out that I effect the bisection of it, but in virtue of its being rectilineal and nothing more. Such and such size is peculiar to the angle set out, but its quality of being rectilineal is common to all rectilineal angles. Suppose, for example, that the given angle is a right angle. If then I had employed in the proof the fact of its being right, I should not have been able to pass to every species of rectilineal angle; but, if I make no use of its being right, and only consider it as rectilineal, the argument will equally apply to rectilineal angles in general. -

- - - - § 6. OTHER TECHNICAL TERMS. -

1. Things said to be given.

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Proclus attaches to his description of the formal divisions of a proposition an explanation of the different senses in which the word given or datum (dedome/non) is used in geometry. Everything that is given is given in one or other of the following ways, in position, in ratio, in magnitude, or in species. The point is given in position only, but a line and the rest may be given in all the sensesProclus, p. 205, 13-15.. -

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The illustrations which Proclus gives of the four senses in which a thing may be given are not altogether happy, and, as regards things which are given in position, in magnitude, and in species, it is best, I think, to follow the definitions given by Euclid himself in his book of Data. Euclid does not mention the fourth class, things given in ratio, nor apparently do any of the great geometers.

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(1) Given in position really needs no definition; and, when Euclid says (Data, Def. 4) that Points, lines and angles are said to be given in position which always occupy the same place, - we are not really the wiser.

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(2) Given in magnitude is defined thus (Data, Def. 1): Areas, lines and angles are called given in magnitude to which we can find equals. - Proclus' illustration is in this case the following: when, he says, two unequal straight lines are given from the greater of which we have to cut off a straight line equal to the lesser, the straight lines are obviously given in magnitude, for greater and less, and finite and infinite are predications peculiar to magnitude. - But he does not explain that part of the implication of the term is that a thing is given in magnitude only, and that, for example, its position is not given and is a matter of indifference

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(3) Given in species. Euclid's definition (Data, Def. 3) is: Rectilineal figures are said to be given in species in which the angles are severally given and the ratios of the sides to one another are given - And this is the recognised use of the term (cf. Pappus, passim) Proclus uses the term in a much wider sense for which I am not aware of any authority. Thus, he says, when we speak of (bisecting) a given rectilineal angle, the angle is given in species by the word rectilineal, which prevents our attempting, by the same method, to bisect a curvilineal angle! On Eucl. 1. 9, to which he here refers, he says that an angle is given in species when e.g. we say that it is right or acute or obtuse or rectilineal or mixed, - but that the actual angle in the proposition is given in species only. As a matter of fact, we should say that the actual angle in the figure of the proposition is given in magnitude and not in species, part of the implication of given in species being that the actual magnitude of the thing given in species is indifferent; an angle cannot be given in species in this sense at all. The confusion in Proclus' mind is shown when, after saying that a right angle is given in species, he describes a third of a right angle as given in magnitude.

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No better example of what is meant by given in species, in its proper sense, as limited to rectilineal figures, can be quoted than the given parallelogram in Eucl. VI. 28, to which the required parallelogram has to be made similar; the former parallelogram is in fact given in species, though its actual size, or scale, is indifferent.

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(4) Given in ratio presumably means something which is given by means of its ratio to some other given thing. This we gather from Proclus' remark (in his note on 1. 9) that an angle may be given in ratio as when we say that it is double and treble of such and such an angle or, generally, greater and less. - The term, however, appears to have no authority and to serve no purpose. Proclus may have derived it from such expressions as in a given ratio - which are common enough.

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- 2. Lemma. -

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The term lemma, - says Proclus - Proclus, pp. 211, 1-212, 4., is often used of any proposition which is assumed for the construction of something else: thus it is a common remark that a proof has been made out of such and such lemmas. But the special meaning of - lemma in geometry is a proposition requiring confirmation. For when, in either construction or demonstration, we assume anything which has not been proved but requires argument, then, because we regard what has been assumed as doubtful in itsėlf and therefore worthy of investigation, we call it a lemma - It would appear, says Tannery (p. 151 n.), that Geminus understood a lemma as being simply lambano/menon, something assumed (cf. the passage of Proclus, p. 73, 4, relating to Menaechmus' view of - elements): hence we cannot consider ourselves authorised in attributing to Geminus the more technical definition of the term here given by Proclus, according to which it is only used of propositions not proved beforehand. This view of a lemma must be considered as relatively modern. It seems to have had its origin in an imperfection of method. - In the course of a demonstration it was necessary to assume a proposition which required proof, but the proof of which would, if inserted in the particular place, break the thread of the demonstration: hence it was necessary either to prove it beforehand as a preliminary proposition or to postpone it to be proved afterwards (w(s e(ch=s deixqh/setai). - When, after the time of Geminus, the progress of original discovery in geometry was arrested, geometers occupied themselves with the study and elucidation of the works of the great mathematicians who had preceded them. This involved the investigation of propositions explicitly quoted or tacitly assumed in the great classical treatises; - and naturally it was found that several such remained to be demonstrated, either because the authors had omitted them as being easy enough to be left to the reader himself to prove, or because books in which they were proved had been lost in the meantime. Hence arose a class of complementary or auxiliary propositions which were called lemmas. - Thus Pappus gives in his Book VII a collection of lemmas in elucidation of the treatises of Euclid and Apollonius included in the so-called Treasury of Analysis - (to/pos a)nalno/menos). When Proclus goes on to distinguish three methods of discovering lemmas, analysis, division, and reductio ad absurdum, he seems to imply that the principal business of contemporary geometers was the investigation of these auxiliary propositions., - differing as it does from the postulate and the axiom in being matter of demonstration, whereas they are immediately taken for granted, without demonstration, for the purpose of confirming other things. Now in the discovery of lemmas the best aid is a mental aptitude for it. For we may see many who are quick at solutions and yet do not work by method; - thus Cratistus in our time was able to obtain the required result from first principles, and those the fewest possible, but it was his natural gift which helped him to the discovery. Nevertheless certain methods have been handed down. The finest is the method which by means of analysis carries the thing sought up to an acknowledged principle, a method which Plato, as they say, communicated to Leodamas - This passage and another from Diogenes Laertius (III. 24, p. 74 ed. Cobet) to the effect that He [Plato] explained (ei)shgh/sato) to Leodamas of Thasos the method of inquiry by analysis - have been commonly understood as ascribing to Plato the invention of the method of analysis; but Tannery points out forcibly (pp. 112, 113) how difficult it is to explain in what Plato's discovery could have consisted if analysis be taken in the sense attributed to it in Pappus, where we can see no more than a series of successive, reductions of a problem until it is finally reduced to a known problem. On the other hand, Proclus' words about carrying up the thing sought to an acknowledged principle - suggest that what he had in mind was the process described at the end of Book VI of the Republic by which the dialectician (unlike the mathematician) uses hypotheses as stepping-stones up to a principle which is not hypothetical, and then is able to descend step by step verifying every one of the hypotheses by which he ascended. This description does not of course refer to mathematical analysis, - but it may have given rise to the idea that analysis was Plato's discovery, since analysis and synthesis following each other are related in the same way as the upward and the downward progression in the dialectician's intellectual method. And it may be that Plato's achievement was to observe the importance, from the point of view of logical rigour, of the confirmatory synthesis following analysis, and to regulariśe in this way and elevate into a completely irrefragable method the partial and uncertain analysis upon which the works of his predecessors depended., - and by which the latter, too, is said to have discovered many things in geometry. The second is the method of division - Here again the successive bipartitions of genera into species such as we find in the Sophist and Republic have very little to say to geometry, and the very fact that they are here mentioned side by side with analysis suggests that Proclus confused the latter with the philosophical method of Rep. VI., - which divides into its parts the genus proposed for consideration and gives a starting-point for the demonstration by means of the elimination of the other elements in the construction of what is proposed, which method also Plato extolled as being of assistance to all sciences. The third is that by means of the reductio ad absurdum, which does not show what is sought directly; but refutes its opposite and discovers the truth incidentally. -

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- 3. Case. -

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The caseTannery rightly remarks (p. 152) that the subdivision of a theorem or problem into several cases is foreign to the really classic form; the ancients preferred, where necessary, to multiply enunciations. As, however, some omissions necessarily occurred, the writers of lemmas naturally added separate cases, which in some instances found their way into the text. A good example is Euclid 1. 7, the second case of which, as it appears in our text-books, was interpolated. On the commentary of Proclus on this proposition Th. Taylor rightly remarks that Euclid everywhere avoids a multitude of cases. - (ptw=sis), - Proclus proceedsProclus, p. 212, 5-11., announces different ways of construction and alteration of positions due to the transposition of points or lines or planes or solids. And, in general, all its varieties are seen in the figure, and this is why it is called case, being a transposition in the construction. -

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- 4. Porism. -

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The term porism is used also of certain problems such as the Porisms written by Euclid. But it is specially used when from what has been demonstrated some other theorem is revealed at the same time without our propounding it, which theorem has on this very account been called a porism (corollary) as being a sort of incidental gain arising from the scientific demonstrationTannery notes however that, so far from distinguishing his corollaries from the conclusions of his propositions, Euclid inserts them before the closing words (being) what it was required to do - or to prove. - In fact the porism-corollary is with Euclid rather a modified form of the regular conclusion than a separate proposition.. - Cf. the note on I. 15.

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5. Objection.

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The objection (e)/nstasis) obstructs the whole course of the argument by appearing as an obstacle (or crying ’halt,’ a)pantw=sa) either to the construction or to the demonstration. There is this difference between the objection and the case, that, whereas he who propounds the case has to prove the proposition to be true of it, he who makes the objection does not need to prove anything: on the contrary it is necessary to destroy the objection and to show that its author is saying what is falseProclus, p. 212, 18-23.. -

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That is, in general the objection endeavours to make it appear that the demonstration is not true in every case; and it is then necessary to prove, in refutation of the objection, either that the supposed case is impossible, or that the demonstration is true even for that case. A good instance is afforded by Eucl. 1. 7. The text-books give a second case which is not in the original text of Euclid. Proclus remarks on the proposition as given by Euclid that the objection may conceivably be raised that what Euclid declares to be impossible may after all be possible in the event of one pair of stiaight lines falling completely within the other pair. Proclus then refutes the objection by proving the impossibility in that case also. His proof then came to be given in the text-books as part of Euclid's proposition.

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The objection is one of the technical terms in Aristotle's logic and its nature is explained in the Prior AnalyticsAnal. prior. II. 26, 69 a 37.. An objection is a proposition contrary to a proposition.... Objections are of two sorts, general or partial.... For when it is maintained that an attribute belongs to every (member of a class), we object either that it belongs to none (of the class) or that there is some one (member of the class) to which it does not belong. -

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6. Reduction.

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This is again an Aristotelian term, explained in the Prior Analyticsibid. II. 25, 69 a 20.. It is well described by Proclus in the following passage:

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Reduction (a)pagwgh/) is a transition from one problem or theorem to another, the solution or proof of which makes that which is propounded manifest also. For example, after the doubling of the cube had been investigated, they transformed the investigation into another upon which it follows, namely the finding of the two means; and from that time forward they inquired how between two given straight lines two mean proportionals could be discovered. And they say that the first to effect the reduction of difficult constructions was Hippocrates of Chios, who also squared a lune and discovered many other things in geometry, being second to none in ingenuity as regards constructionsProclus, pp. 212, 24-213, 11. This passage has frequently been taken as crediting Hippocrates with the discovery of the method of geometrical reduction: cf. Taylor (Translation of Proclus, II. p. 26), Allman (p. 41n., 59), Gow (pp. 169, 170). As Tannery remarks (p. 110), if the particular reduction of the duplication problem to that of the two means is the first noted in history, it is difficult to suppose that it was really the first; for Hippocrates must have found instances of it in the Pythagorean geometry. Bretschneider, I think, comes nearer the truth when he boldly (p. 99) translates: This reduction of the aforesaid construction is said to have been first given by Hippocrates. - The words are prw=ton de/ fasi tw=n a)poroume/nwn diagramma/twn th\n a)pagwgh\n poih/sasqai, which must, literally, be translated as in the text above; but, when Proclus speaks vaguely of difficult constructions, - he probably means to say simply that this first recorded instance of a reduction of a difficult construction is attributed to Hippocrates. -. -

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7. Reductio ad absurdum.

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This is variously called by Aristotle reductio ad absurdum - (h( ei)s to\ a)du/naton a)pagwgh/)Aristotle, Anal. prior. I. 7, 29 b 5; 1. 44, 50 a 30., proof per impossibile - (h( dia\ tou= a)duna/tou dei=cis or a)po/deicis)ibid. I. 21, 39 b 32; I. 29, 45 a 35., proof leading to the impossible - (h( ei)s to\ a)du/naton a)/gousa a)po/deicis)Anal. post. I. 24, 85 a 16 etc.. It is part of proof (starting) from a hypothesisAnal. prior. I. 23, 40 b 25. - (e)c u(poqe/sews). All (syllogisms) which reach the conclusion per impossibile reason out a conclusion which is false, and they prove the original contention (by the method starting) from a hypothesis, when something impossible results from assuming the contradictory of the original contention, as, for example, when it is proved that the diagonal (of a square) is incommensurable because, if it be assumed commensurable, it will follow that odd (numbers) are equal to even (numbers)Anal. prior. I. 23, 41 a 24.. - Or again, proof (leading) to the impossible differs from the direct (deiktikh=s) in that it assumes what it desires to destroy [namely the hypothesis of the falsity of the conclusion] and then reduces it to something admittedly false, whereas the direct proof starts from premisses admittedly trueibid. II. 14, 62 b 29.. -

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Proclus has the following description of the reductio ad absurdum. Proofs by reductio as absurdum in every case reach a conclusion manifestly impossible, a conclusion the contradictory of which is admitted. In some cases the conclusions are found to conflict with the common notions, or the postulates, or the hypotheses (from which we started); in others they contradict propositions previously establishedProclus, p. 254, 22-27. - ...Every reductio ad absurdum assumes what conflicts with the desired result, then, using that as a basis, proceeds until it arrives at an admitted absurdity, and, by thus destroying the hypothesis, establishes the result originally desired. For it is necessary to understand generally that all mathematical arguments either proceed from the first principles or lead back to them, as Porphyry somewhere says. And those which proceed from the first principles are again of two kinds, for they start either from common notions and the clearness of the self-evident alone, or from results previously proved; while those which lead back to the principles are either by way of assuming the principles or by way of destroying them. Those which assume the principles are called analyses, and the opposite of these are syntheses— for it is possible to start from the said principles and to proceed in the regular order to the desired conclusion, and this process is synthesis—while the arguments which would destroy the principles are called reductiones ad absurdum. For it is the function of this method to upset something admitted as clearProclus, p. 255, 8-26.. -

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8. Analysis and Synthesis.

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It will be seen from the note on Eucl. XIII. I that the MSS. of the Elements contain definitions of Analysis and Synthesis followed by alternative proofs of XIII. 1-5 after that method. The definitions and alternative proofs are interpolated, but they have great historical interest because of the possibility that they represent an ancient method of dealing with these propositions, anterior to Euclid. The propositions give properties of a line cut in extreme and mean ratio, - and they are preliminary to the construction and comparison of the five regular solids. Now Pappus, in the section of his Collection dealing with the latter subjectPappus, v. p. 410 sqq., says that he will give the comparisons between the five figures, the pyramid, cube, octahedron, dodecahedron and icosahedron, which have equal surfaces, not by means of the so-called analytical inquiry, by which some of the ancients worked out the proofs, but by the synthetical methodibid. pp. 410, 27-412, 2..... - The conjecture of Bretschneider that the matter interpolated in Eucl. XIII. is a survival of investigations due to Eudoxus has at first sight much to commend itBretschneider, p. 168. See however Heiberg's recent suggestion (Paralipomena zu Euklid in Hermes, XXXVIII., 1903) that the author was Heron. The suggestion is based on a comparison with the remarks on analysis and synthesis quoted from Heron by an-Nairĩzĩ (ed. Curtze, p. 89) at the beginning of his commentary on Eucl. Book II. On the whole, this suggestion commends itself to me more than that of Bretschneider.. In the first place, we are told by Proclus that Eudoxus greatly added to the number of the theorems which Plato originated regarding the section, and employed in them the method of analysisProclus, p. 67, 6.. - It is obvious that the section - was some particular section which by the time of Plato had assumed great importance; and the one section of which this can safely be said is that which was called the golden section, - namely, the division of a straight line in extreme and mean ratio which appears in Eucl. II. 11 and is therefore most probably Pythagorean. Secondly, as Cantor points outCantor, Gesch. d. Math. I_{3}, p. 241., Eudoxus was the founder of the theory of proportions in the form in which we find it in Euclid V., VI., and it was no doubt through meeting, in the course of his investigations, with proportions not expressible by whole numbers that he came to realise the necessity for a new theory of proportions which should be applicable to incommensurable as well as commensurable magnitudes. The golden section - would furnish such a case. And it is even mentioned by Proclus in this connexion. He is explainingProclus, p. 60, 7-9. that it is only in arithmetic that all quantities bear rational - ratios (r(hto\s lo/gos) to one another, while in geometry there are irrational - ones (a)/rrhtos) as well. Theorems about sections like those in Euclid's second Book are common to both [arithmetic and geometry] except that in which the straight line is cut in extreme and mean ratioibid. p. 60, 16-19.. -

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The definitions of Analysis and Synthesis interpolated in Eucl. XIII. are as follows (I adopt the reading of B and V, the only intelligible one, for the second).

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“Analysis is an assumption of that which is sought as if it were admitted <and the passage> through its consequences to something admitted (to be) true.

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Synthesis is an assumption of that which is admitted <and the passage> through its consequences to the finishing or attainment of what is sought. -

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The language is by no means clear and has, at the best, to be filled out.

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Pappus has a fuller accountPappus, VII. pp. 634-6.:

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“The so-called a)naluo/menos (’Treasury of Analysis’) is, to put it shortly, a special body of doctrine provided for the use of those who, after finishing the ordinary Elements, are desirous of acquiring the power of solving problems which may be set them involving (the construction of) lines, and it is useful for this alone. It is the work of three men, Euclid the author of the Elements, Apollonius of Perga, and Aristaeus the elder, and proceeds by way of analysis and synthesis.

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“Analysis then takes that which is sought as if it were admitted and passes from it through its successive consequences to something which is admitted as the result of synthesis: for in analysis we assume that which is sought as if it were (already) done (gegono/s), and we inquire what it is from which this results, and again what is the antecedent cause of the latter, and so on, until by so retracing our steps we come upon something already known or belonging to the class of first principles, and such a method we call analysis as being solution backwards (a)na/palin lu/sin).

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“But in synthesis, reversing the process, we take as already done that which was last arrived at in the analysis and, by arranging in their natural order as consequences what were before antecedents, and successively connecting them one with another, we arrive finally at the construction of what was sought; and this we call synthesis.

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Now analysis is of two kinds, the one directed to searching for the truth and called theoretical, the other directed to finding what we are told to find and called problematical. (1) In the theoretical kind we assume what is sought as if it were existent and true, after which we pass through its successive consequences, as if they too were true and established by virtue of our hypothesis, to something admitted: then (a), if that something admitted is true, that which is sought will also be true and the proof will correspond in the reverse order to the analysis, but (b), if we come upon something admittedly false, that which is sought will also be false. (2) In the problematical kind we assume that which is propounded as if it were known, after which we pass through its successive consequences, taking them as true, up to something admitted: if then (a) what is admitted is possible and obtainable, that is, what mathematicians call given, what was originally proposed will also be possible, and the proof will again correspond in reverse order to the analysis, but if (b) we come upon something admittedly impossible, the problem will also be impossible. -

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The ancient Analysis has been made the subject of careful studies by several writers during the last half-century, the most complete being those of Hankel, Duhamel and Zeuthen; others by Ofterdinger and Cantor should also be mentionedHankel, Zur Geschichte der Mathematik in Alterthum und Mittelalter, 1874, pp. 137-150; Duhamel, Des'méthodes dans les sciences de raisonnement, Part I., 3 ed., Paris, 1885, pp. 39-68; Zeuthen, Geschichte der Mathematik im Altertnm und Mittelalter, 1896, pp. 92-104; Ofterdinger, Beiträge zur Geschichte der griechischen Mathematik, Ulm, 1860; Cantor, Geschichte der Mathematik, I_{3}, pp. 220-2..

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The method is as follows. It is required, let us say, to prove that a certain proposition A is true. We assume as a hypothesis that A is true and, starting from this we find that, if A is true, a certain other proposition B is true; if B is true, then C; and so on until we arrive at a proposition K which is admittedly true. The object of the method is to enable us to infer, in the reverse order, that, since K is true, the proposition A originally assumed is true. Now Aristotle had already made it clear that false hypotheses might lead to a conclusion which is true. There is therefore a possibility of error unless a certain precaution is taken. While, for example, B may be a necessary consequence of A, it may happen that A is not a necessary consequence of B. Thus, in order that the reverse inference from the truth of K that A is true may be logically justified, it is necessary that each step in the chain of inferences should be unconditionally convertible. As a matter of fact, a very large number of theorems in elementary geometry are unconditionally convertible, so that in practice the difficulty in securing that the successive steps shall be convertible is not so great as might be supposed. But care is always necessary. For example, as Hankel saysHankel, p. 139., a proposition may not be unconditionally convertible in the form in which it is generally quoted. Thus the proposition The vertices of all triangles having a common base and constant vertical angle lie on a circle - cannot be converted into the proposition that All triangles with common base and vertices lying on a circle have a constant vertical angle -; for this is only true if the further conditions are satisfied (1) that the circle passes through the extremities of the common base and (2) that only that part of the circle is taken as the locus of the vertices which lies on one side of the base. If these conditions are added, the proposition is unconditionally convertible. Or again, as Zeuthen remarksZeuthen, p. 103., K may be obtained by a series of inferences in which A or some other proposition in the series is only apparently used; this would be the case e.g. when the method of modern algebra is being employed and the expressions on each side of the sign of equality have been inadvertently multiplied by some composite magnitude which is in reality equal to zero.

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Although the above extract from Pappus does not make it clear that each step in the chain of argument must be convertible in the case taken, he almost implies this in the second part of the definition of Analysis where, instead of speaking of the consequences B, C... successively following from A, he suddenly changes the expression and says that we inquire what it is (B) from which A follows (A being thus the consequence of B, instead of the reverse), and then what (viz. C) is the antecedent cause of B; and in practice the Greeks secured what was wanted by always insisting on the analysis being confirmed by subsequent synthesis, that is, they laboriously worked backwards the whole way from K to A, reversing the order of the analysis, which process would undoubtedly bring to light any flaw which had crept into the argument through the accidental neglect of the necessary precautions.

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Reductio ad absurdum a variety of analysis.

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In the process of analysis starting from the hypothesis that a proposition A is true and passing through B, C... as successive consequences we may arrive at a proposition K which, instead of being admittedly true, is either admittedly false or the contradictory of the original hypothesis A or of some one or more of the propositions B, C... intermediate between A and K. Now correct inference from a true proposition cannot lead to a false proposition; and in this case therefore we may at once conclude, without any inquiry whether the various steps in the argument are convertible or not, that the hypothesis A is false, for, if it were true, all the consequences correctly inferred from it would be true and no incompatibility could arise. This method of proving that a given hypothesis is false furnishes an indirect method of proving that a given hypothesis A is true, since we have only to take the contradictory of A and to prove that it is false. This is the method of reductio ad absurdum, which is therefore a variety of analysis. The contradictory of A, or not-A, will generally include more than one case and, in order to prove its falsity, each of the cases must be separately disposed of: e.g., if it is desired to prove that a certain part of a figure is equal to some other part, we take separately the hypotheses (1) that it is greater, (2) that it is less, and prove that each of these hypotheses leads to a conclusion either admittedly false or contradictory to the hypothesis itself or to some one of its consequences.

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- Analysis as applied to problems. -

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It is in relation to problems that the ancient analysis has the greatest significance, because it was the one general method which the Greeks used for solving all the more abstruse problems - (ta\ a)safe/stera tw=n problhma/twn)Proclus, p. 242, 16, 17..

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We have, let us suppose, to construct a figure satisfying a certain set of conditions If we are to proceed at all methodically and not by mere guesswork, it is first necessary to analyse - those conditions. To enable this to be done we must get them clearly in our minds, which is only possible by assuming all the conditions to be actually fulfilled, in other words, by supposing the problem solved. Then we have to transform those conditions, by all the means which practice in such cases has taught us to employ, into other conditions which are necessarily fulfilled if the original conditions are, and to continue this transformation until we at length arrive at conditions which we are in a position to satisfyZeuthen, p. 93.. In other words, we must arrive at some relation which enables us to construct a particular part of the figure which, it is true, has been hypothetically assumed and even drawn, but which nevertheless really requires to be found in order that the problem may be solved. From that moment the particular part of the figure becomes one of the data, and a fresh relation has to be found which enables a fresh part of the figure to be determined by means of the original data and the new one together. When this is done, the second new part of the figure also belongs to the data; and we proceed in this way until all the parts of the required figure are foundHankel, p. 141.. The first part of the analysis down to the point of discovery of a relation which enables us to say that a certain new part of the figure not belonging to the original data is given, Hankel calls the transformation; the second part, in which it is proved that all the remaining parts of the figure are given, - he calls the resolution. Then follows the synthesis, which also consists of two parts, (1) the construction, in the order in which it has to be actually carried out, and in general following the course of the second part of the analysis, the resolution; (2) the demonstration that the figure obtained does satisfy all the given conditions, which follows the steps of the first part of the analysis, the transformation, but in the reverse order. The second part of the analysis, the resolution, would be much facilitated and shortened by the existence of a systematic collection of Data such as Euclid's book bearing that title, consisting of propositions proving that, if in a figure certain parts or relations are given, other parts or relations are also given. As regards the first part of the analysis, the transformation, the usual rule applies that every step in the chain must be unconditionally convertible; and any failure to observe this condition will be brought to light by the subsequent synthesis. The second part, the resolution, can be directly turned into the construction since that only is given which can be constructed by the means provided in the Elements.

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It would be difficult to find a better illustration of the above than the example chosen by Hankel from PappusPappus, VII. pp. 830-2..

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Given a circle ABC and two points D, E external to it, to draw straight lines DB, EB from D, E to a point B on the circle such that, if DB, EB produced meet the circle again in C, A, AC shall be parallel to DE.

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- Analysis. -

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Suppose the problem solved and the tangent at A drawn, meeting ED produced in F.

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(Part I. Transformation.)

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Then, since AC is parallel to DE, the angle at C is equal to the angle CDE.

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But, since FA is a tangent, the angle at C is equal to the angle FAE.

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Therefore the angle FAE is equal to the angle CDE, whence A, B, D, F are concyclic.

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Therefore the rectangle AE, EB is equal to the rectangle FE, ED.

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(Part II. Resolution.)

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But the rectangle AE, EB is given, because it is equal to the square on the tangent from E.

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Therefore the rectangle FE, ED is given; and, since ED is given, FE is given (in length). [Data, 57.]

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But FE is given in position also, so that F is also given. [Data, 27.]

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Now FA is the tangent from a given point F to a circle ABC given in position; therefore FA is given in position and magnitude. [Data, 90.]

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And F is given; therefore A is given.

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But E is also given; therefore the straight line AE is given in position. [Data, 26.]

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And the circle ABC is given in position; therefore the point B is also given. [Data, 25.]

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But the points D, E are also given; therefore the straight lines DB, BE are also given in position.

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- Synthesis. -

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(Part I. Construction.)

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Suppose the circle ABC and the points D, E given.

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Take a rectangle contained by ED and by a certain straight line EF equal to the square on the tangent to the circle from E.

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From F draw FA touching the circle in A; join ABE and then DB, producing DB to meet the circle at C. Join AC.

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I say then that AC is parallel to DE.

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(Part II. Demonstration.)

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Since, by hypothesis, the rectangle FE, ED is equal to the square on the tangent from E, which again is equal to the rectangle AE, EB, the rectangle AE, EB is equal to the rectangle FE, ED.

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Therefore A, B, D, F are concyclic, whence the angle FAE is equal to the angle BDE.

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But the angle FAE is equal to the angle ACB in the alternate segment; therefore the angle ACB is equal to the angle BDE.

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Therefore AC is parallel to DE.

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In cases where a diorismo/s is necessary, i.e. where a solution is only possible under certain conditions, the analysis will enable those conditions to be ascertained. Sometimes the diorismo/s is stated and proved at the end of the analysis, e.g. in Archimedes, On the Sphere and Cylinder, II. 7; sometimes it is stated in that place and the proof postponed till after the end of the synthesis, e.g. in the solution of the problem subsidiary to On the Sphere and Cylinder, II. 4, preserved in Eutocius' commentary on that proposition. The analysis should also enable us to determine the number of solutions of which the problem is susceptible.

- - - - § 7. THE DEFINITIONS. - - General. Real - and Nominal - Definitions. -

It is necessary, says Aristotle, whenever any one treats of any whole subject, to divide the genus into its primary constituents, those which are indivisible in species respectively: e.g. number must be divided into triad and dyad; then an attempt must be made in this way to obtain definitions, e.g. of a straight line, of a circle, and of a right angleAnal. post. II. 13, 96 b 15..

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The word for definition is o(/ros. The original meaning of this word seems to have been boundary, - landmark. - Then we have it in Plato and Aristotle in the sense of standard or determining principle (id quo alicuius rei natura constituitur vel definitur, - Index Aristotelicus)Cf. De anima, I. 2, 404 2 9, where breathing - is spoken of as the o)/ros of life, - and the many passages in the Politics where the word is used to denote that which gives its special character to the several forms of government (virtue being the o(/ros of aristocracy, wealth of oligarchy, liberty of democracy, 1294 a 10); Plato, Republic, VIII. 551 c.; and closely connected with this is the sense of definition. Aristotle uses both o(/ros and o(rismo/s for definition, the former occurring more frequently in the Topics, the latter in the Metaphysics.

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Let us now first be clear as to what a definition does not do. There is nothing in connexion with definitions which Aristotle takes more pains to emphasise than that a definition asserts nothing as to the existence or non-existence of the thing defined. It is an answer to the question what a thing is (ti/ e)sti), and does not say that it is (o(/ti e)/sti). The existence of the various things defined has to be proued, except in the case of a few primary things in each science, the existence of which is indemonstrable and must be assumed among the first principles of each science; e.g. points and lines in geometry must be assumed to exist, but the existence of everything else must be proved. This is stated clearly in the long passage quoted above under First PrinciplesAnal. past. I. 10, 76 a 31 sqq.. It is reasserted in such passages as the following. The (answer to the question) what is a man and the fact that a man exists are different thingsibid. II. 7, 92 b 10.. - It is clear that, even according to the view of definitions now current, those who define things do not prove that they existibid. 92 b 19.. - We say that it is by demonstration that we must show that everything exists, except essence (ei) mh\ ou)si/a ei)/h). But the existence of a thing is never essence; for the existent is not a genus. Therefore there must be demonstration that a thing exists. Thus, what is meant by triangle the geometer assumes, but that it exists he has to proveibid. 92 b 12 sqq.. - Anterior knowledge of two sorts is necessary: for it is necessary to presuppose, with regard to some things, that they exist; in other cases it is necessary to understand what the thing described is, and in other cases it is necessary to do both. Thus, with the fact that one of two contradictories must be true, we must know that it exists (is true); of the triangle we must know that it means such and such a thing; of the unit we must know both what it means and that it existsAnal. post. I. 1, 71 a II sqq.. - What is here so much insisted on is the very fact which Mill pointed out in his discussion of earlier views of Definitions, where he says that the so-called real definitions or definitions of things do not constitute a different kind of definition from nominal definitions, or definitions of names; the former is simply the latter plus something else, namely a covert assertion that the thing defined exists. This covert assertion is not a definition but a postulate. The definition is a mere identical proposition which gives information only about the use of language, and from which no conclusion affecting matters of fact can possibly be drawn. The accompanying postulate, on the other hand, affirms a fact which may lead to consequences of every degree of importance. It affirms the actual or possible existence of Things possessing the combination of attributes set forth in the definition: and this, if true, may be foundation sufficient on which to build a whole fabric of scientific truthMill's System of Logic, Bk. I. ch. Viii.. - This statement really adds nothing to Aristotle's doctrineIt is true that it was in opposition to the ideas of most of the Aristotelian logicians - (rather than of Aristotle himself) that Mill laid such stress on his point of view. Cf. his observation: We have already made, and shall often have to repeat, the remark, that the philosophers who overthrew Realism by no means got rid of the consequences of Realism, but retained long afterwards, in their own philosophy, numerous propositions which could only have a rational meaning as part of a Realistic system. It had been handed down from Aristotle, and probably from earlier times, as an obvious truth, that the science of geometry is deduced from definitions. This, so long as a definition was considered to be a proposition ’unfolding the nature of the thing,’ did well enough. But Hobbes followed and rejected utterly the notion that a definition declares the nature of the thing, or does anything but state the meaning of a name; yet he continued to affirm as broadly as any of his predecessors that the a)rxai/, principia, or original premisses of mathematics, and even of all science, are definitions; producing the singular paradox that systems of scientific truth, nay, all truths whatever at which we arrive by reasoning, are deduced from the arbitrary conventions of mankind concerning the signification of words. - Aristotle was guilty of no such paradox; on the contrary, he exposed it as plainly as did Mill.: it has even the slight disadvantage, due to the use of the word postulate - to describe the covert assertion - in all cases, of not definitely pointing out that there are cases where existence has to be proued as distinct from those where it must be assumed. It is true that the existence of a definiend may have to be taken for granted provisionally until the time comes for proving it; but, so far as regards any case where existence must be proved sooner or later, the provisional assumption would be for Aristotle, not a postulate, but a hypothesis. In modern times, too, Mill's account of the true distinction between real and nominal definitions had been fully anticipated by SaccheriThis has been fully brought out in two papers by G. Vailati, La teoria Aristotelica della definizione (Riuista di Filosofia e scienze affini, 1903), and Di un' opera dimenticata del P. Gerolamo Saccheri (Logica Demonstrativa, - 1697) (in Rivista Filosofica, 1903)., the editor of Euclides ab omni naevo vindicatus (1733), famous in the history of non-Euclidean geometry. In his Logica Demonstrativa (to which he also refers in his Euclid) Saccheri lays down the clear distinction between what he calls definitiones quid nominis or nominales, and definitiones quid rei or reales, namely that the former are only intended to explain the meaning that is to be attached to a given term, whereas the latter, besides declaring the meaning of a word, affirm at the same time the existence of the thing defined or, in geometry, the possibility of constructing it. The definitio quid nominis becomes a definitio quid rei - by means of a postulate, or when we come to the question whether the thing exists and it is answered affirmatively - Definitio quid nominis nata est evadere definitio quid rei per postulatum vel dum venitur ad quaestionem an est et respondetur affirmative. -. - Definitiones quid nominis are in themselves quite arbitrary, and neither require nor are capable of proof; they are merely provisional and are only intended to be turned as quickly as possible into definitiones quid rei, either (1) by means of a postulate in which it is asserted or conceded that what is defined exists or can be constructed, e.g. in the case of straight lines and circles, to which Euclid's first three postulates refer, or (2) by means of a demonstration reducing the construction of the figure defined to the successive carrying-out of a certain number of those elementary constructions, the possibility of which is postulated. Thus definitiones quid rei are in general obtained as the result of a series of demonstrations. Saccheri gives as an instance the construction of a square in Euclid I. 46. Suppose that it is objected that Euclid had no right to define a square, as he does at the beginning of the Book, when it was not certain that such a figure exists in nature; the objection, he says, could only have force if, before proving and making the construction, Euclid had assumed the aforesaid figure as given. That Euclid is not guilty of this error is clear from the fact that he never presupposes the existence of the square as defined until after I. 46.

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Confusion between the nominal and the real definition as thus described, i.e. the use of the former in demonstration before it has been turned into the latter by the necessary proof that the thing defined exists, is according to Saccheri one of the most fruitful sources of illusory demonstration, and the danger is greater in proportion to the complexity - of the definition, i.e. the number and variety of the attributes belonging to the thing defined. For the greater is the possibility that there may be among the attributes some that are incompatible, i.e. the simultaneous presence of which in a given figure can be proved, by means of other postulates etc. forming part of the basis of the science, to be impossible.

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The same thought is expressed by Leibniz also. If, - he says, we give any definition, and it is not clear from it that the idea, which we ascribe to the thing, is possible, we cannot rely upon the demonstrations which we have derived from that definition, because, if that idea by chance involves a contradiction, it is possible that even contradictories may be true of it at one and the same time, and thus our demonstrations will be useless. Whence it is clear that definitions are not arbitrary. And this is a secret which is hardly sufficiently knownOpuscules et fragments inédits de Leibniz, Paris, Alcan, 1903, p. 431. Quoted by Vailati.. - Leibniz' favourite illustration was the regular polyhedron with ten faces, - the impossibility of which is not obvious at first sight.

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It need hardly be added that, speaking generally, Euclid's definitions, and his use of them, agree with the doctrine of Aristotle that the definitions themselves say nothing as to the existence of the things defined, but that the existence of each of them must be proved or (in the case of the principles -) assumed. In geometry, says Aristotle, the existence of points and lines only must be assumed, the existence of the rest being proved. Accordingly Euclid's first three postulates declare the possibility of constructing straight lines and circles (the only lines - except straight lines used in the Elements). Other things are defined and afterwards constructed and proved to exist: e.g. in Book I., Def. 20, it is explained what is meant by an equilateral triangle; then (I. 1) it is proposed to construct it, and, when constructed, it is proved to agree with the definition. When a square is defined (I. Def. 22), the question whether such a thing really exists is left open until, in I. 46, it is proposed to construct it and, when constructed, it is proved to satisfy the definitionTrendelenburg, Elementa Logices Aristoteleae, § 50.. Similarly with the right angle (I. Def. 10, and I. 11) and parallels (I. Def. 23, and I. 27-29). The greatest care is taken to exclude mere presumption and imagination. The transition from the subjective definition of names to the objective definition of things is made, in geometry, by means of constructions (the first principles of which are postulated), as in other sciences it is made by means of experienceTrendelenburg, Erläuterungen zu den Elementen der aristotelischen Logik, 3 ed. p. 107. On construction as proof of existence in ancient geometry cf. H. G. Zeuthen, Die geometrische Construction als <quote>Existenzbeweis</quote> - in der antiken Geometrie (in Mathematische Annalen, 47. Band)..

- - - Aristotle's requirements in a definition. -

We now come to the positive characteristics by which, according to Aristotle, scientific definitions must be marked.

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First, the different attributes in a definition, when taken separately, cover more than the notion defined, but the combination of them does not. Aristotle illustrates this by the triad, - into which enter the several notions of number, odd and prime, and the last in both its two senses (a) of not being measured by any (other) number (w(s mh\ metrei=sqai a)riqmw=|) and (b) of not being obtainable by adding numbers together - (w(s mh\ sugkei=sqai e)c a)riqmw=n), a unit not being a number. Of these attributes some are present in all other odd numbers as well, while the last [primeness in the second sense] belongs also to the dyad, but in nothing but the triad are they all presentAnal. post. II. 13, 96 a 33—b 1..” - The fact can be equally well illustrated from geometry. Thus, e.g. into the definition of a square (Eucl. I., Def. 22) there enter the several notions of figure, four-sided, equilateral, and right-angled, each of which covers more than the notion into which all enter as attributesTrendelenburg, Erläuterungen, p. 108..

-

Secondly, a definition must be expressed in terms of things which are prior to, and better known than, the things definedTopics VI. 4, 141 a 26 sqq.. This is clear, since the object of a definition is to give us knowledge of the thing defined, and it is by means of things prior and better known that we acquire fresh knowledge, as in the course of demonstrations. But the terms prior - and better known - are, as usual susceptible of two meanings; they may mean (1) absolutely or logically prior and better known, or (2) better known relatively to us. In the absolute sense, or from the standpoint of reason, a point is better known than a line, a line than a plane, and a plane than a solid, as also a unit is better known than number (for the unit is prior to, and the first principle of, any number). Similarly, in the absolute sense, a letter is prior to a syllable. But the case is sometimes different relatively to us; for example, a solid is more easily realised by the senses than a plane, a plane than a line, and a line than a point. Hence, while it is more scientific to begin with the absolutely prior, it may, perhaps, be permissible, in case the learner is not capable of following the scientific order, to explain things by means of what is more intelligible to him. Among the definitions framed on this principle are those of the point, the line and the plane; all these explain what is prior by means of what is posterior, for the point is described as the extremity of a line, the line of a plane, the plane of a solid. - But, if it is asserted that such definitions by means of things which are more intelligible relatively only to a particular individual are really definitions, it will follow that there may be many definitions of the same thing, one for each individual for whom a thing is being defined, and even different definitions for one and the same individual at different times, since at first sensible objects are more intelligible, while to a better trained mind they become less so. It follows therefore that a thing should be defined by means of the absolutely prior and not the relatively prior, in order that there may be one sole and immutable definition. This is further enforced by reference to the requirement that a good definition must state the genus and the differentiae, for these are among the things which are, in the absolute sense, better known than, and prior to, the species (tw=n a(plw=s gnwrimwte/rwn kai\ prote/rwn tou= ei)/dous e)sti/n). For to destroy the genus and the differentia is to destroy the species, so that the former are prior to the species; they are also better known, for, when the species is known, the genus and the differentia must necessarily be known also, e.g. he who knows man - must also know animal - and land-animal, - but it does not follow, when the genus and differentia are known, that the species is known too, and hence the species is less known than they areTopics VI. 4, 141 b 25-34.. It may be frankly admitted that the scientific definition will require superior mental powers for its apprehension; and the extent of its use must be a matter of discretion. So far Aristotle; and we have here the best possible explanation why Euclid supplemented his definition of a point by the statement in I. Def. 3 that the extremities of a line are points and his definition of a surface by I. Def. 6 to the effect that the extremities of a surface are lines. The supplementary explanations do in fact enable us to arrive at a better understanding of the formal definitions of a point and a line respectively, as is well explained by Simson in his note on Def. 1. Simson says, namely, that we must consider a solid, that is, a magnitude which has length, breadth and thickness, in order to understand aright the definitions of a point, a line and a surface. Consider, for instance, the boundary common to two solids which are contiguous or the boundary which divides one solid into two contiguous parts; this boundary is a surface. We can prove that it has no thickness by taking away either solid, when it remains the boundary of the other; for, if it had thickness, the thickness must either be a part of one solid or of the other, in which case to take away one or other solid would take away the thickness and therefore the boundary itself: which is impossible. Therefore the boundary or the surface has no thickness. In exactly the same way, regarding a line as the boundary of two contiguous surfaces, we prove that the line has no breadth; and, lastly, regarding a point as the common boundary or extremity of two lines, we prove that a point has no length, breadth or thickness.

- - - Aristotle on unscientific definitions. -

Aristotle distinguishes three kinds of definition which are unscientific because founded on what is not prior (mh\ e)k prote/rwn). The first is a definition of a thing by means of its opposite, e.g. of good - by means of bad -; this is wrong because opposites are naturally evolved together, and the knowledge of opposites is not uncommonly regarded as one and the same, so that one of the two opposites cannot be better known than the other. It is true that, in some cases of opposites, it would appear that no other sort of definition is possible: e.g. it would seem impossible to define double apart from the half and, generally, this would be the case with things which in their very nature (kaq) au(ta/) are relative terms (pro/s ti le/getai), since one cannot be known without the other, so that in the notion of either the other must be comprised as wellTopics VI. 4, 142 a 22-31.. The second kind of definition which is based on what is not prior is that in which there is a complete circle through the unconscious use in the definition itself of the notion to be defined though not of the nameibid. 142 a 34—b 6.. Trendelenburg illustrates this by two current definitions, (1) that of magnitude as that which can be increased or diminished, which is bad because the positive and negative comparatives more - and less - presuppose the notion of the positive great, - (2) the famous Euclidean definition of a straight line as that which lies evenly with the points on itself - (e)c i)/sou toi=s e)f) e(auth=s shmei/ois kei=tai), where lies evenly - can only be understood with the aid of the very notion of a straight line which is to be definedTrendelenburg, Erläuterungen, p. 115.. The third kind of vicious definition from that which is not prior is the definition of one of two coordinate species by means of its coordinate (a)ntidih|rhme/non), e.g. a definition of odd - as that which exceeds the even by a unit (the second alternative in Eucl. VII. Def. 7); for odd - and even - are coordinates, being differentiae of numberTopics VI. 4, 142 b 7-10.. This third kind is similar to the first. Thus, says Trendelenburg, it would be wrong to define a square as a rectangle with equal sides. -

- - - Aristotle's third requirement. -

A third general observation of Aristotle which is specially relevant to geometrical definitions is that to know what a thing is (ti/ e)stin) is the same as knowing why it is (dia\ ti/ e)stin)Anal. post. II. 2, 90 a 31.. - What is an eclipse? A deprivation of light from the moon through the interposition of the earth. Why does an eclipse take place? Or why is the moon eclipsed? Because the light fails through the earth obstructing it. What is harmony? A ratio of numbers in high or low pitch. Why does the high-pitched harmonise with the low-pitched? Because the high and the low have a numerical ratio to one anotherAnal. post. II. 2, 90 a 15-21.. - We seek the cause (to\ dio/ti) when we are already in possession of the fact (to\ o(/ti). Sometimes they both become evident at the same time, but at all events the cause cannot possibly be known [as a cause] before the fact is knownibid. II. 8, 93 a 17.. - It is impossible to know what a thing is if we do not know that it isibid. 93 a 20. - Trendelenburg paraphrases: The definition of the notion does not fulfil its purpose until it is made genetic. It is the producing cause which first reveals the essence of the thing.... . The nominal definitions of geometry have only a provisional significance and are superseded as soon as they are made genetic by means of construction. - E.g. the genetic definition of a parallelogram is evolved from Eucl. I. 31 (giving the construction for parallels) and I. 33 about the lines joining corresponding ends of two straight lines parallel and equal in length. Where existence is proved by construction, the cause and the fact appear togetherTrendelenburg, Erläuterungen, p. 110..

-

Again, it is not enough that the defining statement should set forth the fact, as most definitions do; it should also contain and present the cause; whereas in practice what is stated in the definition is usually no more than a conclusion (sumpe/rasma). For example, what is quadrature? The construction of an equilateral right-angled figure equal to an oblong. But such a definition expresses merely the conclusion. Whereas, if you say that quadrature is the discovery of a mean proportional, then you state the reasonDe anima II. 2, 413 a 13-20,. - This is better understood if we compare the statement elsewhere that the cause is the middle term, and this is what is sought in all casesAnal. post. II. 2, 90 a 6,, - and the illustration of this by the case of the proposition that the angle in a semicircle is a right angle. Here the middle term which it is sought to establish by means of the figure is that the angle in the semi-circle is equal to the half of two right angles. We have then the syllogism: Whatever is half of two right angles is a right angle; the angle in a semi-circle is the half of two right angles; therefore (conclusion) the angle in a semi-circle is a right angleibid. II. 11, 94 a 28.. As with the demonstration, so it should be with the definition. A definition which is to show the genesis of the thing defined should contain the middle term or cause; otherwise it is a mere statement of a conclusion. Consider, for instance, the definition of quadrature - as making a square equal in area to a rectangle with unequal sides. - This gives no hint as to whether a solution of the problem is possible or how it is solved: but, if you add that to find the mean proportional between two given straight lines gives another straight line such that the square on it is equal to the rectangle contained by the first two straight lines, you supply the necessary middle term or causeOther passages in Aristotle may be quoted to the like effect: e.g. Anal. post. I. 2, 71 b 9 We consider that we know a particular thing in the absolute sense, as distinct from the sophistical and incidental sense, when we consider that we know the cause on account of which the thing is, in the sense of knowing that it is the cause of that thing and that it cannot be otherwise, - ibid. I. 13, 79 a 2 For here to know the fact is the function of those who are concerned with sensible things, to know the cause is the function of the mathematician; it is he who possesses the proofs of the causes, and often he does not know the fact. - In view of such passages it is difficult to see how Proclus came to write (p. 202, 11) that Aristotle was the originator (*)aristote/lous kata/rcantos) of the idea of Amphinomus and others that geometry does not investigate the cause and the why (to\ dia\ ti/). To this Geminus replied that the investigation of the cause does, on the contrary, appear in geometry. For how can it be maintained that it is not the business of the geometer to inquire for what reason, on the one hand, an infinite number of equilateral polygons are inscribed in a circle, but, on the other hand, it is not possible to inscribe in a sphere an infinite number of polyhedral figures, equilateral, equiangular, and made up of similar plane figures? Whose business is it to ask this question and find the answer to it if it is not that of the geometer? Now when geometers reason per impossibile they are content to discover the property, but when they argue by direct proof, if such proof be only partial (e)pi\ me/rous), this does not suffice for showing the cause; if however it is general and applies to all like cases, the why (to\ dia\ ti/) is at once and concurrently made evident. -.

- - - Technical terms not defined by Euclid. -

It will be observed that what is here defined, quadrature - or squaring - (tetragwnismo/s), is not a geometrical figure, or an attribute of such a figure or a part of a figure, but a technical term used to describe a certain problem. Euclid does not define such things; but the fact that Aristotle alludes to this particular definition as well as to definitions of deflection (kekla/sqai) and of verging (neu/ein) seems to show that earlier text-books included among definitions explanations of a number of technical terms, and that Euclid deliberately omitted these explanations from his Elements as surplusage. Later the tendency was again in the opposite direction, as we see from the much expanded Definitions of Heron, which, for example, actually include a definition of a deflected line (keklasme/nh grammh/)Heron, Def. 12 (vol. IV. Heib. pp. 22-24).. Euclid uses the passive of kla=n occasionallye.g. in III. 20 and in Data 89., but evidently considered it unnecessary to explain such terms, which had come to bear a recognised meaning.

-

The mention too by Aristotle of a definition of verging (neu/ein) suggests that the problems indicated by this term were not excluded from elementary text-books before Euclid. The type of problem (neu=sis) was that of placing a straight line across two lines, e.g. two straight lines, or a straight line and a circle, so that it shall verge to a given point (i.e. pass through it if produced) and at the same time the intercept on it made by the two given lines shall be of given length. In general, the use of conics is required for the theoretical solution of these problems, or a mechanical contrivance for their practical solutionCf. the chapter on neu/seis in The Works of Archimedes, pp. c—cxxii.. Zeuthen, following Oppermann, gives reasons for supposing, not only that mechanical constructions were practically used by the older Greek geometers for solving these problems, but that they were theoretically recognised as a permissible means of solution when the solution could not be effected by means of the straight line and circle, and that it was only in later times that it was considered necessary to use conics in every case where that was possibleZeuthen, Die Lehre von den Kegelschnitten im Altertum, ch. 12, p. 262.. HeibergHeiberg, Mathematisches zu Aristoteles, p. 16. suggests that the allusion of Aristotle to neu/seis perhaps confirms this supposition, as Aristotle nowhere shows the slightest acquaintance with conics. I doubt whether this is a safe inference, since the problems of this type included in the elementary text-books might easily have been limited to those which could be solved by plane - methods (i.e. by means of the straight line and circle). We know, e.g., from Pappus that Apollonius wrote two Books on plane - neu/seisPappus VII. pp. 670-2.. But one thing is certain, namely that Euclid deliberately excluded this class of problem, doubtless as not being essential in a book of Elements.

- - - Definitions not afterwards used. -

Lastly, Euclid has definitions of some terms which he never afterwards uses, e.g. oblong (e(tero/mhkes), rhombus, rhomboid. The oblong - occurs in Aristotle; and it is certain that all these definitions are survivals from earlier books of Elements.

- - - - + + - - +
+
+ BOOK I. DEFINITIONS. @@ -1741,7 +167,7 @@ 16 - +

And the point is called the centre of the circle.

 @@ -1794,7 +220,7 @@ 5 - +

That, if a straight line falling on two straight lines make the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles.

 @@ -1822,7 +248,7 @@ - + BOOK I. PROPOSITIONS. Proposition 1. @@ -1837,7 +263,7 @@

Again, since the point B is the centre of the circle CAE, BC is equal to BA. [Def. 15]

But CA was also proved equal to AB; therefore each of the straight lines CA, CB is equal to AB.

And things which are equal to the same thing are also equal to one another; [C.N. 1] therefore CA is also equal to CB.

-

Therefore the three straight lines CA, AB, BC are equal to one another.

+

Therefore the three straight lines CA, AB, BC are equal to one another.

Therefore the triangle ABC is equilateral; and it has been constructed on the given finite straight line AB.

@@ -1873,7 +299,7 @@ - + Proposition 2.

To place at a given point (as an extremity) a straight line equal to a given straight line.

@@ -1901,7 +327,7 @@

It will be observed that in this first application of Postulate 2, and again in I. 5, Euclid speaks of the continuation of the straight line as that which is produced in such cases, e)kbeblh/sqwsan and prosekbeblh/sqwsan meaning little more than drawing straight lines in a straight line with - the given straight lines. The first place in which Euclid uses phraseology exactly corresponding to ours when speaking of a straight line being produced is in I. 16: let one side of it, BC, be produced to D + the given straight lines. The first place in which Euclid uses phraseology exactly corresponding to ours when speaking of a straight line being produced is in I. 16: let one side of it, BC, be produced to D (prosekbeblh/sqw au)tou= mi/a pleura\ h( *b*g e)pi\ to\ *d).

 @@ -1917,7 +343,7 @@

Let AB, C be the-two given unequal straight lines, and let AB be the greater of them.

Thus it is required to cut off from AB the greater a straight line equal to C the less.

-

At the point A let AD be placed equal to the straight line C; [I. 2] and with centre A and distance AD let the circle DEF be described. [Post. 3] Now, since the point A is the centre of the circle DEF, AE is equal to AD. [Def. 15] But C is also equal to AD. Therefore each of the straight lines AE, C is equal to AD; so that AE is also equal to C. [C.N. 1]

+

At the point A let AD be placed equal to the straight line C; [I. 2] and with centre A and distance AD let the circle DEF be described. [Post. 3] Now, since the point A is the centre of the circle DEF, AE is equal to AD. [Def. 15] But C is also equal to AD. Therefore each of the straight lines AE, C is equal to AD; so that AE is also equal to C. [C.N. 1]

Therefore, given the two straight lines AB, C, from AB the greater AE has been cut off equal to C the less.

 @@ -1933,7 +359,7 @@

Let ABC, DEF be two triangles having the two sides AB, AC equal to the two sides DE, DF respectively, namely AB to DE and AC to DF, and the angle BAC equal to the angle EDF.

I say that the base BC is also equal to the base EF, the triangle ABC will be equal to the triangle DEF, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend, that is, the angle ABC to the angle DEF, and the angle ACB to the angle DFE.

For, if the triangle ABC be applied to the triangle DEF, and if the point A be placed on the point D and the straight line AB on DE, then the point B will also coincide with E, because AB is equal to DE.

-

+

Again, AB coinciding with DE, the straight line AC will also coincide with DF, because the angle BAC is equal to the angle EDF; hence the point C will also coincide with the point F, because AC is again equal to DF.

But B also coincided with E; hence the base BC will coincide with the base EF.

@@ -1961,7 +387,7 @@

Here we have the word base used for the first time in the Elements. Proclus explains it (p. 236, 12-15) as meaning (1), when no side of a triangle has been mentioned before, the side which is on a level with the sight - (th\n pro\s th=| o)/yei keime/nhn), and (2), when two sides have already been mentioned, the third side. Proclus thus avoids the mistake made by some modern editors who explain the term exclusively with reference to the case where two sides have been mentioned before. That this is an error is proved (1) by the occurrence of the term in the enunciations of I. 37 etc. about triangles on the same base and equal bases, (2) by the application of the same term to the bases of parallelograms in I. 35 etc. The truth is that the use of the term must have been suggested by the practice of drawing the particular side horizontally, as it were, and the rest of the figure above it. The base of a figure was therefore spoken of, primarily, in the same sense as the base of anything else, e.g. of a pedestal or column; but when, as in I. 5, two triangles were compared occupying other than the normal positions which gave rise to the name, and when two sides had been previously mentioned, the base was, as Proclus says, necessarily the third side.

+ (th\n pro\s th=| o)/yei keime/nhn), and (2), when two sides have already been mentioned, the third side. Proclus thus avoids the mistake made by some modern editors who explain the term exclusively with reference to the case where two sides have been mentioned before. That this is an error is proved (1) by the occurrence of the term in the enunciations of I. 37 etc. about triangles on the same base and equal bases, (2) by the application of the same term to the bases of parallelograms in I. 35 etc. The truth is that the use of the term must have been suggested by the practice of drawing the particular side horizontally, as it were, and the rest of the figure above it. The base of a figure was therefore spoken of, primarily, in the same sense as the base of anything else, e.g. of a pedestal or column; but when, as in I. 5, two triangles were compared occupying other than the normal positions which gave rise to the name, and when two sides had been previously mentioned, the base was, as Proclus says, necessarily the third side.

u(potei/nein u(po/, to stretch under, @@ -1987,7 +413,7 @@ - + Proposition 5.

In isosceles triangles the angles at the base are equal to one another, and, if the equal straight lines be produced further, the angles under the base will be equal to one another.

@@ -1999,7 +425,7 @@

Let a point F be taken at random on BD;

from AE the greater let AG be cut off equal to AF the less; [I. 3] and let the straight lines FC, GB be joined. [Post. 1]

Then, since AF is equal to AG and AB to AC, the two sides FA, AC are equal to the two sides GA, AB, respectively; and they contain a common angle, the angle FAG. Therefore the base FC is equal to the base GB, and the triangle AFC is equal to the triangle AGB, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend, that is, the angle ACF to the angle ABG, and the angle AFC to the angle AGB. [I. 4]

And, since the whole AF is equal to the whole AG, and in these AB is equal to AC, the remainder BF is equal to the remainder CG.

-

But FC was also proved equal to GB; therefore the two sides BF, FC are equal to the two sides CG, GB respectively; and the angle BFC is equal to the angle CGB, while the base BC is common to them; therefore the triangle BFC is also equal to the triangle CGB, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend; therefore the angle FBC is equal to the angle GCB, and the angle BCF to the angle CBG. +

But FC was also proved equal to GB; therefore the two sides BF, FC are equal to the two sides CG, GB respectively; and the angle BFC is equal to the angle CGB, while the base BC is common to them; therefore the triangle BFC is also equal to the triangle CGB, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend; therefore the angle FBC is equal to the angle GCB, and the angle BCF to the angle CBG.

Accordingly, since the whole angle ABG was proved equal to the angle ACF, and in these the angle CBG is equal to the angle BCF, the remaining angle ABC is equal to the remaining angle ACB; and they are at the base of the triangle ABC. But the angle FBC was also proved equal to the angle GCB; and they are under the base.

Therefore etc.

@@ -2047,7 +473,7 @@

Let AB be greater; and from AB the greater let DB be cut off equal to AC the less;

let DC be joined.

Then, since DB is equal to AC, and BC is common, the two sides DB, BC are equal to the two sides AC, CB respectively; - and the angle DBC is equal to the angle ACB; therefore the base DC is equal to the base AB, and the triangle DBC will be equal to the triangle ACB, the less to the greater: which is absurd. Therefore AB is not unequal to AC; it is therefore equal to it.

+ and the angle DBC is equal to the angle ACB; therefore the base DC is equal to the base AB, and the triangle DBC will be equal to the triangle ACB, the less to the greater: which is absurd. Therefore AB is not unequal to AC; it is therefore equal to it.

Therefore etc.

@@ -2061,7 +487,7 @@

For, if possible, given two straight lines AC, CB constructed on the straight line AB and meeting at the point C, let two other straight lines

- AD, DB be constructed on the same straight line AB, on the same side of it, meeting in another point D and equal to the former two respectively, namely each to that which has the same extremity with it, so that CA is equal to DA which has the same extremity A with it, and CB to DB which has the same extremity B with it; and let CD be joined.

+ AD, DB be constructed on the same straight line AB, on the same side of it, meeting in another point D and equal to the former two respectively, namely each to that which has the same extremity with it, so that CA is equal to DA which has the same extremity A with it, and CB to DB which has the same extremity B with it; and let CD be joined.

Then, since AC is equal to AD, the angle ACD is also equal to the angle ADC; [I. 5] therefore the angle ADC is greater than the angle DCB; therefore the angle CDB is much greater than the angle DCB.

Again, since CB is equal to DB, the angle CDB is also equal to the angle DCB. But it was also proved much greater than it: which is impossible.

@@ -2081,7 +507,7 @@ (ai) ou)=n a)po\ tw=n *h*k a)nago/menai grammai\ e)n tou/tw| tw=| lo/gw| ou) sustaqh/soutai tou= e)f) w)=| *a h(mikukli/ou pro\s a)/llo kai\ a)/llo shmei=on).

If a paraphrase is allowed instead of a translation adhering as closely as possible to the original, Simson's is the best that could be found, since the fact that the straight lines form triangles on the same base is really conveyed in the Greek. Simson's enunciation is, Upon the same base, and on the same side of it, there cannot be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise those which are terminated at the other extremity. Th. Taylor (the translator of Proclus) attacks Simson's alteration as indiscreet and as detracting from the beauty and accuracy of Euclid's enunciation which are enlarged upon by Proclus in his commentary. Yet, when Taylor says Whatever difficulty learners may find in conceiving this proposition abstractedly is easily removed by its exposition in the figure, - he really gives his case away. The fact is that Taylor, always enthusiastic over his author, was nettled by Simson's slighting remarks on Proclus' comments on the proposition. Simson had said, with reference to Proclus' explanation of the bearing of the second part of I. 5 on I. 7, that it was not worth while to relate his trifles at full length, + he really gives his case away. The fact is that Taylor, always enthusiastic over his author, was nettled by Simson's slighting remarks on Proclus' comments on the proposition. Simson had said, with reference to Proclus' explanation of the bearing of the second part of I. 5 on I. 7, that it was not worth while to relate his trifles at full length, to which Taylor retorts But Mr Simson was no philosopher; and therefore the greatest part of these Commentaries must be considered by him as trifles, from the want of a philosophic genius to comprehend their meaning, and a taste superior to that of a mere mathematician, to discover their beauty and elegance.

@@ -2104,7 +530,7 @@

Let ABC, DEF be two triangles having the two sides AB, AC equal to the two sides DE, DF respectively, namely AB to DE, and AC to DF; and let them have the base BC equal to the base EF;

I say that the angle BAC is also equal to the angle EDF.

For, if the triangle ABC be applied to the triangle DEF, and if the point B be placed on the point E and the straight line BC on EF, the point C will also coincide with F, because BC is equal to EF. -

+

Then, BC coinciding with EF, BA, AC will also coincide with ED, DF; for, if the base BC coincides with the base EF, and the sides BA, AC do not coincide with ED, DF but fall beside them as EG, GF, then, given two straight lines constructed on a straight line (from its extremities) and meeting in a point, there will have been constructed on the same straight line (from its extremities), and on the same side of it, two other straight lines meeting in another point and equal to the former two respectively, namely each to that which has the same extremity with it. But they cannot be so constructed. [I. 7]

Therefore it is not possible that, if the base BC be applied to the base EF, the sides BA, AC should not coincide with ED, DF; they will therefore coincide, @@ -2162,7 +588,7 @@ - + Proposition 11.

To draw a straight line at right angles to a given straight line from a given point on it.

@@ -2190,7 +616,7 @@

To a given infinite straight line, from a given point which is not on it, to draw a perpendicular straight line.

-

Let AB be the given infinite straight line, and C the given point which is not on it; thus it is required to draw to the given infinite straight line AB, from the given point C which is not on it, a perpendicular straight line.

+

Let AB be the given infinite straight line, and C the given point which is not on it; thus it is required to draw to the given infinite straight line AB, from the given point C which is not on it, a perpendicular straight line.

For let a point D be taken at random on the other side of the straight line AB, and with centre C and distance CD let the circle EFG be described; [Post. 3]

let the straight line EG be bisected at H, [I. 10] and let the straight lines CG, CH, CE be joined. [Post. 1]

@@ -2228,7 +654,7 @@

Again, since the angle DBA is equal to the two angles DBE, EBA, let the angle ABC be added to each; therefore the angles DBA. ABC are equal to the three angles DBE, EBA, ABC. [C. N. 2]

But the angles CBE, EBD were also proved equal to the same three angles; and things which are equal to the same thing are also equal to one another; [C. N. 1] therefore the angles CBE, EBD are also equal to the angles DBA, ABC. - But the angles CBE, EBD are two right angles; therefore the angles DBA, ABC are also equal to two right angles.

+ But the angles CBE, EBD are two right angles; therefore the angles DBA, ABC are also equal to two right angles.

Therefore etc.

@@ -2256,7 +682,7 @@

Then, since the straight line AB stands on the straight line CBE,

the angles ABC, ABE are equal to two right angles. [I. 13] But the angles ABC, ABD are also equal to two right angles; therefore the angles CBA, ABE are equal to the angles CBA, ABD. [Post. 4 and C.N. 1]

Let the angle CBA be subtracted from each; therefore the remaining angle ABE is equal to the remaining angle ABD, [C.N. 3] the less to the greater: which is impossible. Therefore BE is not in a straight line with CB.

-

Similarly we can prove that neither is any other straight line except BD. Therefore CB is in a straight line with BD.

+

Similarly we can prove that neither is any other straight line except BD. Therefore CB is in a straight line with BD.

Therefore etc.

@@ -2286,7 +712,7 @@

I say that the angle AEC is equal to the angle DEB, and the angle CEB to the angle AED.

For, since the straight line AE stands on the straight line CD, making the angles CEA, AED, the angles CEA, AED are equal to two right angles [I. 13] -

+

Again, since the straight line DE stands on the straight line AB, making the angles AED, DEB, the angles AED, DEB are equal to two right angles. [I. 13]

But the angles CEA, AED were also proved equal to two right angles; therefore the angles CEA, AED are equal to the angles AED @@ -2317,7 +743,7 @@

Let AC be bisected at E [I. 10], and let BE be joined and produced in a straight line to F;

let EF be made equal to BE[I. 3], let FC be joined [Post. 1], and let AC be drawn through to G [Post. 2].

Then, since AE is equal to EC, and BE to EF, the two sides AE, EB are equal to the two sides CE, EF respectively; and the angle AEB is equal to the angle FEC, for they are vertical angles. [I. 15] Therefore the base AB is equal to the base FC, and the triangle ABE is equal to the triangle CFE, and the remaining angles are equal to the remaining angles respectively, namely those which the equal sides subtend; [I. 4] therefore the angle BAE is equal to the angle ECF. -

+

But the angle ECD is greater than the angle ECF; [C. N. 5] therefore the angle ACD is greater than the angle BAE.

Similarly also, if BC be bisected, the angle BCG, that is, the angle ACD [I. 15], can be proved greater than the angle ABC as well.

Therefore etc.

@@ -2350,7 +776,7 @@

Let ABC be a triangle; I say that two angles of the triangle ABC taken together in any manner are less than two right angles.

For let BC be produced to D. [Post. 2]

Then, since the angle ACD is an exterior angle of the triangle ABC,

-

it is greater than the interior and opposite angle ABC. [I. 16] Let the angle ACB be added to each; therefore the angles ACD, ACB are greater than the angles ABC, BCA.

But the angles ACD, ACB are equal to two right angles. [I. 13]

+

it is greater than the interior and opposite angle ABC. [I. 16] Let the angle ACB be added to each; therefore the angles ACD, ACB are greater than the angles ABC, BCA.

But the angles ACD, ACB are equal to two right angles. [I. 13]

Therefore the angles ABC, BCA are less than two right angles.

Similarly we can prove that the angles BAC, ACB are also less than two right angles, and so are the angles CAB, ABC as well.

Therefore etc.

@@ -2363,7 +789,7 @@ - + Proposition 18.

In any triangle the greater side subtends the greater angle.

@@ -2415,7 +841,7 @@

Then, since DA is equal to AC, the angle ADC is also equal to the angle ACD; [I. 5]

therefore the angle BCD is greater than the angle ADC. [C.N. 5]

And, since DCB is a triangle having the angle BCD greater than the angle BDC, and the greater angle is subtended by the greater side, [I. 19] therefore DB is greater than BC. -

+

But DA is equal to AC; therefore BA, AC are greater than BC.

Similarly we can prove that AB, BC are also greater than CA, and BC, CA than AB.

Therefore etc.

@@ -2430,7 +856,7 @@ - + Proposition 21.

If on one of the sides of a triangle, from its extremities, there be constructed two straight lines meeting within the triangle, the straight lines so constructed will be less than the remaining two sides of the triangle, but will contain a greater angle.

@@ -2470,7 +896,7 @@

Let the three given straight lines be A, B, C, and of these let two taken together in any manner be greater than the remaining one, namely A, B greater than C, A, C greater than B, and B, C greater than A; thus it is required to construct a triangle out of straight lines equal to A, B, C.

Let there be set out a straight line DE, terminated at D but of infinite length in the direction of E, and let DF be made equal to A, FG equal to B, and GH equal to C. [I. 3]

With centre F and distance FD let the circle DKL be described; again, with centre G and distance GH let the circle KLH be described; and let KF, KG be joined;

-

I say that the triangle KFG has been constructed out of three straight lines equal to A, B, C.

+

I say that the triangle KFG has been constructed out of three straight lines equal to A, B, C.

For, since the point F is the centre of the circle DKL, FD is equal to FK.

But FD is equal to A; therefore KF is also equal to A.

Again, since the point G is the centre of the circle LKH, GH is equal to GK.

@@ -2504,7 +930,7 @@

Let AB be the given straight line, A the point on it, and the angle DCE the given rectilineal angle;

thus it is required to construct on the given straight line AB, and at the point A on it, a rectilineal angle equal to the given rectilineal angle DCE.

-

On the straight lines CD, CE respectively let the points D, E be taken at random; let DE be joined, and out of three straight lines which are equal to the three straight lines CD, DE, CE let the triangle AFG be constructed in such a way that CD is equal to AF, CE to AG, and further DE to FG.

+

On the straight lines CD, CE respectively let the points D, E be taken at random; let DE be joined, and out of three straight lines which are equal to the three straight lines CD, DE, CE let the triangle AFG be constructed in such a way that CD is equal to AF, CE to AG, and further DE to FG.

Then, since the two sides DC, CE are equal to the two sides FA, AG respectively, and the base DE is equal to the base FG, the angle DCE is equal to the angle FAG. [I. 8]

Therefore on the given straight line AB, and at the point A on it, the rectilineal angle FAG has been constructed equal to the given rectilineal angle DCE.

@@ -2522,7 +948,7 @@

I say that the base BC is also greater than the base EF.

For, since the angle BAC is greater than the angle EDF, let there be constructed, on the straight line DE, and at the point D on it, the angle EDG equal to the angle BAC; [I. 23] let DG be made equal to either of the two straight lines AC, DF, and let EG, FG be joined.

-

+

Then, since AB is equal to DE, and AC to DG, the two sides BA, AC are equal to the two sides ED, DG, respectively; and the angle BAC is equal to the angle EDG; therefore the base BC is equal to the base EG. [I. 4]

Again, since DF is equal to DG, the angle DGF is also equal to the angle DFG; [I. 5] therefore the angle DFG is greater than the angle EGF.

Therefore the angle EFG is much greater than the angle EGF.

@@ -2561,7 +987,7 @@ Proposition 26. -

If two triangles have the two angles equal to two angles respectively, and one side equal to one side, namely, either the side adjoining the equal angles, or that subtending one of the equal angles, they will also have the remaining sides equal to the remaining sides and the remaining angle to the remaining angle.

+

If two triangles have the two angles equal to two angles respectively, and one side equal to one side, namely, either the side adjoining the equal angles, or that subtending one of the equal angles, they will also have the remaining sides equal to the remaining sides and the remaining angle to the remaining angle.

Let ABC, DEF be two triangles having the two angles ABC, BCA equal to the two angles DEF, EFD respectively, namely the angle ABC to the angle DEF, and the angle BCA to the angle EFD; and let them also have one side equal to one side, first that adjoining the equal angles, namely BC to EF;

@@ -2570,7 +996,7 @@

Let AB be greater, and let BG be made equal to DE; and let GC be joined.

Then, since BG is equal to DE, and BC to EF, the two sides GB, BC are equal to the two sides DE, EF respectively; and the angle GBC is equal to the angle DEF; therefore the base GC is equal to the base DF, and the triangle GBC is equal to the triangle DEF, and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend; [I. 4] therefore the angle GCB is equal to the angle DFE. But the angle DFE is by hypothesis equal to the angle BCA; therefore the angle BCG is equal to the angle BCA, the less to the greater: which is impossible. Therefore AB is not unequal to DE, and is therefore equal to it.

But BC is also equal to EF; therefore the two sides AB, BC are equal to the two sides DE, EF respectively, and the angle ABC is equal to the angle DEF; therefore the base AC is equal to the base DF, and the remaining angle BAC is equal to the remaining angle EDF. [I. 4] -

+

Again, let sides subtending equal angles be equal, as AB to DE;

I say again that the remaining sides will be equal to the remaining sides, namely AC to DF and BC to EF, and further the remaining angle BAC is equal to the remaining angle EDF.

For, if BC is unequal to EF, one of them is greater.

@@ -2604,7 +1030,7 @@

Let them be produced and meet, in the direction of B, D, at G.

Then, in the triangle GEF, the exterior angle AEF is equal to the interior and opposite angle EFG: which is impossible. [I. 16]

Therefore AB, CD when produced will not meet in the direction of B, D.

-

Similarly it can be proved that neither will they meet towards A, C.

+

Similarly it can be proved that neither will they meet towards A, C.

But straight lines which do not meet in either direction are parallel; [Def. 23] therefore AB is parallel to CD.

Therefore etc.

 @@ -2628,7 +1054,7 @@ Proposition 28. -

If a straight line falling on two straight lines make the exterior angle equal to the interior and opposite angle on the same side, or the interior angles on the same side equal to two right angles, the straight lines will be parallel to one another.

+

If a straight line falling on two straight lines make the exterior angle equal to the interior and opposite angle on the same side, or the interior angles on the same side equal to two right angles, the straight lines will be parallel to one another.

For let the straight line EF falling on the two straight lines AB, CD make the exterior angle EGB equal to the interior and opposite angle GHD, or the interior angles on the same side, namely BGH, GHD, equal to two right angles;

@@ -2652,7 +1078,7 @@

I say that it makes the alternate angles AGH, GHD equal, the exterior angle EGB equal to the interior and opposite angle GHD, and the interior angles on the same side, namely BGH, GHD, equal to two right angles.

For, if the angle AGH is unequal to the angle GHD, one of them is greater.

Let the angle AGH be greater.

-

Let the angle BGH be added to each; therefore the angles AGH, BGH are greater than the angles BGH, GHD.

+

Let the angle BGH be added to each; therefore the angles AGH, BGH are greater than the angles BGH, GHD.

But the angles AGH, BGH are equal to two right angles; [I. 13] therefore the angles BGH, GHD are less than two right angles.

But straight lines produced indefinitely from angles less than two right angles meet; [Post. 5] therefore AB, CD, if produced indefinitely, will meet; but they do not meet, because they are by hypothesis parallel.

Therefore the angle AGH is not unequal to the angle GHD, and is therefore equal to it. @@ -2702,7 +1128,7 @@

Through a given point to draw a straight line parallel to a given straight line.

-

Let A be the given point, and BC the given straight line; thus it is required to draw through the point A a straight line parallel to the straight line BC.

+

Let A be the given point, and BC the given straight line; thus it is required to draw through the point A a straight line parallel to the straight line BC.

Let a point D be taken at random on BC, and let AD be joined; on the straight line DA, and at the point A on it, let the angle DAE be constructed equal to the angle ADC [I. 23]; and let the straight line AF be produced in a straight line with EA.

Then, since the straight line AD falling on the two straight lines BC, EF has made the alternate angles EAD, ADC equal to one another, therefore EAF is parallel to BC. [I. 27]

Therefore through the given point A the straight line EAF has been drawn parallel to the given straight line BC.

@@ -2718,7 +1144,7 @@

Let ABC be a triangle, and let one side of it BC be produced to D;

-

I say that the exterior angle ACD is equal to the two interior and opposite angles CAB, ABC, and the three interior angles of the triangle ABC, BCA, CAB are equal to two right angles.

+

I say that the exterior angle ACD is equal to the two interior and opposite angles CAB, ABC, and the three interior angles of the triangle ABC, BCA, CAB are equal to two right angles.

For let CE be drawn through the point C parallel to the straight line AB. [I. 31]

Then, since AB is parallel to CE, and AC has fallen upon them, the alternate angles BAC, ACE are equal to one another. [I. 29]

@@ -2738,7 +1164,7 @@

The straight lines joining equal and parallel straight lines (at the extremities which are) in the same directions (respectively) are themselves also equal and parallel.

-

Let AB, CD be equal and parallel, and let the straight lines AC, BD join them (at the extremities which are) in the same directions (respectively); I say that AC, BD are also equal and parallel.

+

Let AB, CD be equal and parallel, and let the straight lines AC, BD join them (at the extremities which are) in the same directions (respectively); I say that AC, BD are also equal and parallel.

Let BC be joined.

Then, since AB is parallel to CD, and BC has fallen upon them, the alternate angles ABC, BCD are equal to one another. [I. 29]

And, since AB is equal to CD, and BC is common, the two sides AB, BC are equal to the two sides DC, CB; and the angle ABC is equal to the angle BCD; therefore the base AC is equal to the base BD, and the griangle ABC is equal to the triangle DCB, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend; [I. 4] therefore the angle ACB is equal to the angle CBD.

@@ -2768,7 +1194,7 @@

In parallelogrammic areas the opposite sides and angles are equal to one another, and the diameter bisects the areas.

-

Let ACDB be a parallelogrammic area, and BC its diameter; I say that the opposite sides and angles of the parallelogram ACDB are equal to one another, and the diameter BC bisects it.

+

Let ACDB be a parallelogrammic area, and BC its diameter; I say that the opposite sides and angles of the parallelogram ACDB are equal to one another, and the diameter BC bisects it.

For, since AB is parallel to CD, and the straight line BC has fallen upon them, the alternate angles ABC, BCD are equal to one another. [I. 29]

Again, since AC is parallel to BD, and BChas fallen upon them, the alternate angles ACB, CBD are equal to one another. [I. 29]

Therefore ABC, DCB are two triangles having the two angles ABC, BCA equal to the two angles DCB, CBD respectively, and one side equal to one side, namely that adjoining the equal angles and common to both of them, BC; therefore they will also have the remaining sides equal to the remaining sides respectively, and the remaining angle to the remaining angle; [I. 26] therefore the side AB is equal to CD, and AC to BD, and further the angle BAC is equal to the angle CDB.

@@ -2783,7 +1209,7 @@

Q. E. D.

- +

It is to be observed that, when parallelograms have to be mentioned for the first time, Euclid calls them parallelogrammic areas or, more exactly, parallelogram areas (parallhlo/gramma xwri/a). The meaning is simply areas bounded by parallel straight lines with the further limitation placed upon the term by Euclid that only four-sided figures are so called, although of course there are certain regular polygons which have opposite sides parallel, and which therefore might be said to be areas bounded by parallel straight lines. We gather from Proclus (p. 393) that the word parallelogram @@ -2803,7 +1229,7 @@

Let ABCD, EBCF be parallelograms on the same base BC and in the same parallels AF, BC; I say that ABCD is equal to the parallelogram EBCF.

For, since ABCD is a parallelogram, AD is equal to BC. [I. 34] -

+

For the same reason also EF is equal to BC, so that AD is also equal to EF; [C.N. 1] and DE is common; therefore the whole AE is equal to the whole DF. [C.N. 2]

But AB is also equal to DC; [I. 34] therefore the two sides EA, AB are equal to the two sides FD, DC respectively,

and the angle FDC is equal to the angle EAB, the exterior to the interior; [I. 29] therefore the base EB is equal to the base FC, and the triangle EAB will be equal to the triangle FDC. [I. 4]

@@ -2840,7 +1266,7 @@ - + Proposition 37.

Triangles which are on the same base and in the same parallels are equal to one another.

@@ -2872,7 +1298,7 @@

Let ABC, DEF be triangles on equal bases BC, EF and in the same parallels BF, AD; I say that the triangle ABC is equal to the triangle DEF.

For let AD be produced in both directions to G, H; through B let BG be drawn parallel to CA, [I. 31] and through F let FH be drawn parallel to DE.

-

Then each of the figures GBCA, DEFH is a parallelogram; and GBCA is equal to DEFH;

+

Then each of the figures GBCA, DEFH is a parallelogram; and GBCA is equal to DEFH;

for they are on equal bases BC, EF and in the same parallels BF, GH. [I. 36]

Moreover the triangle ABC is half of the parallelogram GBCA; for the diameter AB bisects it. [I. 34]

And the triangle FED is half of the parallelogram DEFH; for the diameter DF bisects it. [I. 34]

@@ -2903,7 +1329,7 @@

Q. E. D.

- +

[I say that they are also in the same parallels.] Heiberg has proved (Hermes, XXXVIII., 1903, p. 50) from a recently discovered papyrus-fragment (Fayūm towns and their papyri, p. 96, No. IX.) that these words are an interpolation by some one who did not observe that the words And let AD be joined are part of the setting-out (e)/kqesis), but took them as belonging to the construction (kataskeuh/) and consequently thought that a diorismo/s or definition (of the thing to be proved) should precede. The interpolator then altered And @@ -2920,7 +1346,7 @@

For, if not, let AF be drawn through A parallel to BE [I. 31], and let FE be joined.

Therefore the triangle ABC is equal to the triangle FCE; for they are on equal bases BC, CE and in the same parallels BE, AF. [I. 38]

But the triangle ABC is equal to the triangle DCE; therefore the triangle DCE is also equal to the triangle FCE, [C.N. 1] the greater to the less: which is impossible. Therefore AF is not parallel to BE. -

+

Similarly we can prove that neither is any other straight line except AD; therefore AD is parallel to BE.

Therefore etc. Q. E. D.]

@@ -2951,7 +1377,7 @@

Let BC be bisected at E, and let AE be joined; on the straight line EC, and at the point E on it, let the angle CEF be constructed equal to the angle D; [I. 23] through A let AG be drawn parallel to EC, and [I. 31] through C let CG be drawn parallel to EF.

Then FECG is a parallelogram.

And, since BE is equal to EC, the triangle ABE is also equal to the triangle AEC, for they are on equal bases BE, EC and in the same parallels BC, AG; [I. 38] therefore the triangle ABC is double of the triangle AEC. -

+

But the parallelogram FECG is also double of the triangle AEC, for it has the same base with it and is in the same parallels with it; [I. 41] therefore the parallelogram FECG is equal to the triangle ABC.

And it has the angle CEF equal to the given angle D.

Therefore the parallelogram FECG has been constructed equal to the given triangle ABC, in the angle CEF which is equal to D. Q. E. F.

@@ -2979,7 +1405,7 @@

Q. E. D.

- +

complements, paraplhrw/mata, the figures put in to fill up (interstices).

 @@ -2995,7 +1421,7 @@

Let AB be the given straight line, C the given triangle and D the given rectilineal angle; thus it is required to apply to the given straight line AB, in an angle equal to the angle D, a parallelogram equal to the given triangle C.

-

Let the parallelogram BEFG be constructed equal to the triangle C, in the angle EBG which is equal to D [I. 42]; let it be placed so that BE is in a straight line with AB; letFG be drawn through to H, and let AH be drawn through A parallel to either BG or EF. [I. 31]

+

Let the parallelogram BEFG be constructed equal to the triangle C, in the angle EBG which is equal to D [I. 42]; let it be placed so that BE is in a straight line with AB; letFG be drawn through to H, and let AH be drawn through A parallel to either BG or EF. [I. 31]

Let HB be joined.

Then, since the straight line HF falls upon the parallels AH, EF, the angles AHF, HFE are equal to two right angles. [I. 29] Therefore the angles BHG, GFE are less than two right angles; and straight lines produced indefinitely from angles less than two right angles meet; [Post. 5] therefore HB, FE, when produced, will meet.

Let them be produced and meet at K; through the point K let KL be drawn parallel to either EA or FH, [I. 31] and let HA, GB be produced to the points L, M.

@@ -3020,7 +1446,7 @@

Let ABCD be the given rectilineal figure and E the given rectilineal angle; thus it is required to construct, in the given angle E, a parallelogram equal to the rectilineal figure ABCD.

Let DB be joined, and let the parallelogram FH be constructed equal to the triangle ABD, in the angle HKF which is equal to E; [I. 42] let the parallelogram GM equal to the triangle DBC be applied to the straight line GH, in the angle GHM which is equal to E. [I. 44]

Then, since the angle E is equal to each of the angles HKF, GHM, the angle HKF is also equal to the angle GHM. [C.N. 1] -

+

Let the angle KHG be added to each; therefore the angles FKH, KHG are equal to the angles KHG, GHM.

But the angles FKH, KHG are equal to two right angles; [I. 29] therefore the angles KHG, GHM are also equal to two right angles.

Thus, with a straight line GH, and at the point H on it, two straight lines KH, HM not lying on the same side make the adjacent angles equal to two right angles; therefore KH is in a straight line with HM. [I. 14]

@@ -3050,7 +1476,7 @@

Let AB be the given straight line; thus it is required to describe a square on the straight line AB.

-

Let AC be drawn at right angles to the straight line AB from the point A on it [I. 11], and let AD be made equal to AB; through the point D let DE be drawn parallel to AB, and through the point B let BE be drawn parallel to AD. [I. 31]

+

Let AC be drawn at right angles to the straight line AB from the point A on it [I. 11], and let AD be made equal to AB; through the point D let DE be drawn parallel to AB, and through the point B let BE be drawn parallel to AD. [I. 31]

Therefore ADEB is a parallelogram; therefore AB is equal to DE, and AD to BE. [I. 34]

But AB is equal to AD; therefore the four straight lines BA, AD, DE, EB are equal to one another; therefore the parallelogram ADEB is equilateral.

I say next that it is also right-angled.

@@ -3082,7 +1508,7 @@

For the same reason BA is also in a straight line with AH.

And, since the angle DBC is equal to the angle FBA: for each is right: let the angle ABC be added to each; therefore the whole angle DBA is equal to the whole angle FBC. [C.N. 2] -

+

And, since DB is equal to BC, and FB to BA, the two sides AB, BD are equal to the two sides FB, BC respectively, and the angle ABD is equal to the angle FBC; therefore the base AD is equal to the base FC, and the triangle ABD is equal to the triangle FBC. [I. 4]

Now the parallelogram BL is double of the triangle ABD, for they have the same base BD and are in the same parallels BD, AL. [I. 41]

And the square GB is double of the triangle FBC, for they again have the same base FB and are in the same parallels FB, GC. [I. 41]

@@ -3121,7 +1547,7 @@

I say that the angle BAC is right.

For let AD be drawn from the point A at right angles to the straight line AC, let AD be made equal to BA, and let DC be joined.

Since DA is equal to AB, the square on DA is also equal to the square on AB.

-

Let the square on AC be added to each; therefore the squares on DA, AC are equal to the squares on BA, AC.

+

Let the square on AC be added to each; therefore the squares on DA, AC are equal to the squares on BA, AC.

But the square on DC is equal to the squares on DA, AC, for the angle DAC is right; [I. 47] and the square on BC is equal to the squares on BA, AC, for this is the hypothesis; therefore the square on DC is equal to the square on BC, so that the side DC is also equal to BC.

And, since DA is equal to AB, and AC is common, the two sides DA, AC are equal to the two sides BA, AC; and the base DC is equal to the base BC; therefore the angle DAC is equal to the angle BAC. [I. 8] But the angle DAC is right; therefore the angle BAC is also right.

Therefore etc.

@@ -3131,9 +1557,9 @@ - - - +
+
+ BOOK II. DEFINITIONS. @@ -3147,7 +1573,7 @@ - + BOOK II. PROPOSITIONS. Proposition 1. @@ -3161,7 +1587,7 @@

Therefore the rectangle A, BC is equal to the rectangle A, BD, the rectangle A, DE and the rectangle A, EC.

Therefore etc. Q. E. D. - +

the rectangle A, BC. From this point onward I shall translate thus in cases where Euclid leaves out the word contained (periexo/menon). Though the word rectangle is also omitted in the Greek (the neuter article being sufficient to show that the rectangle is meant), it cannot be dispensed with in English. De Morgan advises the use of the expression the rectangle under two lines. This does not seem to me a very good expression, and, if used in a translation from the Greek, it might suggest that u(po/ in to\ u(po/ meant under, which it does not.

@@ -3195,7 +1621,7 @@

For let the square ADEB be described on AB, [I. 46] let BD be joined; through C let CF be drawn parallel to either AD or EB, and through G let HK be drawn parallel to either AB or DE. [I. 31]

Then, since CF is parallel to AD, and BD has fallen on them, the exterior angle CGB is equal to the interior and opposite angle ADB. [I. 29]

-

But the angle ADB is equal to the angle ABD, since the side BA is also equal to AD; [I. 5] therefore the angle CGB is also equal to the angle GBC, so that the side BC is also equal to the side CG. [I. 6]

+

But the angle ADB is equal to the angle ABD, since the side BA is also equal to AD; [I. 5] therefore the angle CGB is also equal to the angle GBC, so that the side BC is also equal to the side CG. [I. 6]

But CB is equal to GK, and CG to KB; [I. 34] therefore GK is also equal to KB; therefore CGKB is equilateral.

I say next that it is also right-angled.

For, since CG is parallel to BK, the angles KBC, GCB are equal to two right angles. [I. 29]

@@ -3233,7 +1659,7 @@

For let the square CEFB be described on CB, [I. 46] and let BE be joined; through D let DG be drawn parallel to either CE or BF, through H again let KM be drawn parallel to either AB or EF, and again through A let AK be drawn parallel to either CL or BM. [I. 31]

Then, since the complement CH is equal to the complement HF, [I. 43] let DM be added to each; therefore the whole CM is equal to the whole DF.

But CM is equal to AL, since AC is also equal to CB; [I. 36] therefore AL is also equal to DF. Let CH be added to each; therefore the whole AH is equal to the gnomon NOP. -

+

But AH is the rectangle AD, DB, for DH is equal to DB, therefore the gnomon NOP is also equal to the rectangle AD, DB.

Let LG, which is equal to the square on CD, be added to each; therefore the gnomon NOP and LG are equal to the rectangle contained by AD, DB and the square on CD.

But the gnomon NOP and LG are the whole square CEFB, which is described on CB; therefore the rectangle contained by AD, DB together with the square on CD is equal to the square on CB.

@@ -3249,7 +1675,7 @@

If a straight line be bisected and a straight line be added to it in a straight line, the rectangle contained by the whole with the added straight line and the added straight line together with the square on the half is equal to the square on the straight line made up of the half and the added straight line.

For let a straight line AB be bisected at the point C, and let a straight line BD be added to it in a straight line;

I say that the rectangle contained by AD, DB together with the square on CB is equal to the square on CD.

-

For let the square CEFD be described on CD, [I. 46] and let DE be joined; through the point B let BG be drawn parallel to either EC or DF, through the point H let KM be drawn parallel to either AB or EF, and further through A let AK be drawn parallel to either CL or DM. [I. 31]

+

For let the square CEFD be described on CD, [I. 46] and let DE be joined; through the point B let BG be drawn parallel to either EC or DF, through the point H let KM be drawn parallel to either AB or EF, and further through A let AK be drawn parallel to either CL or DM. [I. 31]

Then, since AC is equal to CB, AL is also equal to CH. [I. 36] But CH is equal to HF. [I. 43] Therefore AL is also equal to HF.

Let CM be added to each; therefore the whole AM is equal to the gnomon NOP.

But AM is the rectangle AD, DB, for DM is equal to DB; therefore the gnomon NOP is also equal to the rectangle AD, DB.

@@ -3265,7 +1691,7 @@

For let the square ADEB be described on AB, [I. 46] and let the figure be drawn.

Then, since AG is equal to GE, [I. 43] let CF be added to each; therefore the whole AF is equal to the whole CE.

Therefore AF, CE are double of AF.

-

But AF, CE are the gnomon KLM and the square CF; therefore the gnomon KLM and the square CF are double of AF.

+

But AF, CE are the gnomon KLM and the square CF; therefore the gnomon KLM and the square CF are double of AF.

But twice the rectangle AB, BC is also double of AF; for BF is equal to BC; therefore the gnomon KLM and the square CF are equal to twice the rectangle AB, BC.

Let DG, which is the square on AC, be added to each; therefore the gnomon KLM and the squares BG, GD are equal to twice the rectangle contained by AB, BC and the square on AC.

But the gnomon KLM and the squares BG, GD are the whole ADEB and CF, which are squares described on AB, BC; therefore the squares on AB, BC are equal to twice the rectangle contained by AB, BC together with the square on AC.

@@ -3275,7 +1701,7 @@ Proposition 8.

If a straight line be cut at random, four times the rectangle contained by the whole and one of the segments together with the square on the remaining segment is equal to the square described on the whole and the aforesaid segment as on one straight line.

For let a straight line AB be cut at random at the point C;

-

I say that four times the rectangle contained by AB, BC together with the square on AC is equal to the square described on AB, BC as on one straight line.

+

I say that four times the rectangle contained by AB, BC together with the square on AC is equal to the square described on AB, BC as on one straight line.

For let [the straight line] BD be produced in a straight line [with AB], and let BD be made equal to CB; let the square AEFD be described on AD, and let the figure be drawn double.

Then, since CB is equal to BD, while CB is equal to GK, and BD to KN, therefore GK is also equal to KN.

For the same reason QR is also equal to RP.

@@ -3284,7 +1710,7 @@

Again, since CB is equal to BD, while BD is equal to BK, that is CG, and CB is equal to GK, that is GQ, therefore CG is also equal to GQ.

And, since CG is equal to GQ, and QR to RP, AG is also equal to MQ, and QL to RF. [I. 36]

But MQ is equal to QL, for they are complements of the parallelogram ML; [I. 43] therefore AG is also equal to RF; therefore the four areas AG, MQ, QL, RF are equal to one another. Therefore the four are quadruple of AG. But the four areas CK, KD, GR, RN were proved to be quadruple of CK; therefore the eight areas, which contain the gnomon STU, are quadruple of AK.

-

Now, since AK is the rectangle AB, BD, for BK is equal to BD, therefore four times the rectangle AB, BD is quadruple of AK.

+

Now, since AK is the rectangle AB, BD, for BK is equal to BD, therefore four times the rectangle AB, BD is quadruple of AK.

But the gnomon STU was also proved to be quadruple of AK; therefore four times the rectangle AB, BD is equal to the gnomon STU.

Let OH, which is equal to the square on AC, be added to each; therefore four times the rectangle AB, BD together with the square on AC is equal to the gnomon STU and OH.

But the gnomon STU and OH are the whole square AEFD, which is described on AD. therefore four times the rectangle AB, BD together with the square on AC is equal to the square on AD

@@ -3301,7 +1727,7 @@

And, since the angle at C is right, the remaining angles EAC, AEC are equal to one right angle. [I. 32]

And they are equal; therefore each of the angles CEA, CAE is half a right angle.

For the same reason each of the angles CEB, EBC is also half a right angle; therefore the whole angle AEB is right.

-

And, since the angle GEF is half a right angle, and the angle EGF is right, for it is equal to the interior and opposite angle ECB, [I. 29] the remaining angle EFG is half a right angle; [I. 32] therefore the angle GEF is equal to the angle EFG, so that the side EG is also equal to GF. [I. 6]

+

And, since the angle GEF is half a right angle, and the angle EGF is right, for it is equal to the interior and opposite angle ECB, [I. 29] the remaining angle EFG is half a right angle; [I. 32] therefore the angle GEF is equal to the angle EFG, so that the side EG is also equal to GF. [I. 6]

Again, since the angle at B is half a right angle, and the angle FDB is right, for it is again equal to the interior and opposite angle ECB, [I. 29] the remaining angle BFD is half a right angle; [I. 32] therefore the angle at B is equal to the angle DFB, so that the side FD is also equal to the side DB. [I. 6]

Now, since AC is equal to CE, the square on AC is also equal to the square on CE; therefore the squares on AC, CE are double of the square on AC.

But the square on EA is equal to the squares on AC, CE, for the angle ACE is right; [I. 47] therefore the square on EA is double of the square on AC.

@@ -3310,7 +1736,7 @@

But GF is equal to CD; [I. 34] therefore the square on EF is double of the square on CD.

But the square on EA is also double of the square on AC; therefore the squares on AE, EF are double of the squares on AC, CD.

And the square on AF is equal to the squares on AE, EF, for the angle AEF is right; [I. 47] therefore the square on AF is double of the squares on AC, CD.

-

But the squares on AD, DF are equal to the square on AF, for the angle at D is right; [I. 47] therefore the squares on AD, DF are double of the squares on AC, CD.

+

But the squares on AD, DF are equal to the square on AF, for the angle at D is right; [I. 47] therefore the squares on AD, DF are double of the squares on AC, CD.

And DF is equal to DB; therefore the squares on AD, DB are double of the squares on AC, CD.

Therefore etc. Q. E. D.

@@ -3320,7 +1746,7 @@

For let a straight line AB be bisected at C, and let a straight line BD be added to it in a straight line;

I say that the squares on AD, DB are double of the squares on AC, CD.

For let CE be drawn from the point C at right angles to AB [I. 11], and let it be made equal to either AC or CB [I. 3]; let EA, EB be joined; through E let EF be drawn parallel to AD, and through D let FD be drawn parallel to CE. [I. 31]

-

Then, since a straight line EF falls on the parallel straight lines EC, FD, the angles CEF, EFD are equal to two right angles; [I. 29] therefore the angles FEB, EFD are less than two right angles.

+

Then, since a straight line EF falls on the parallel straight lines EC, FD, the angles CEF, EFD are equal to two right angles; [I. 29] therefore the angles FEB, EFD are less than two right angles.

But straight lines produced from angles less than two right angles meet; [I. Post. 5] therefore EB, FD, if produced in the direction B, D, will meet.

Let them be produced and meet at G, and let AG be joined.

Then, since AC is equal to CE, the angle EAC is also equal to the angle AEC; [I. 5] and the angle at C is right; therefore each of the angles EAC, AEC is half a right angle. [I. 32]

@@ -3328,7 +1754,7 @@

And, since the angle EBC is half a right angle, the angle DBG is also half a right angle. [I. 15] But the angle BDG is also right, for it is equal to the angle DCE, they being alternate; [I. 29] therefore the remaining angle DGB is half a right angle; [I. 32] therefore the angle DGB is equal to the angle DBG, so that the side BD is also equal to the side GD. [I. 6]

Again, since the angle EGF is half a right angle, and the angle at F is right, for it is equal to the opposite angle, the angle at C, [I. 34] the remaining angle FEG is half a right angle; [I. 32] therefore the angle EGF is equal to the angle FEG, so that the side GF is also equal to the side EF. [I. 6]

Now, since the square on EC is equal to the square on CA, the squares on EC, CA are double of the square on CA.

-

But the square on EA is equal to the squares on EC, CA; [I. 47] therefore the square on EA is double of the square on AC. [C. N. 1]

+

But the square on EA is equal to the squares on EC, CA; [I. 47] therefore the square on EA is double of the square on AC. [C. N. 1]

Again, since FG is equal to EF, the square on FG is also equal to the square on FE; therefore the squares on GF, FE are double of the square on EF.

But the square on EG is equal to the squares on GF, FE; [I. 47] therefore the square on EG is double of the square on EF.

And EF is equal to CD; [I. 34] therefore the square on EG is double of the square on CD. But the square on EA was also proved double of the square on AC; therefore the squares on AE, EG are double of the squares on AC, CD.

@@ -3346,7 +1772,7 @@

But EF is equal to EB; therefore the rectangle CF, FA together with the square on AE is equal to the square on EB.

But the squares on BA, AE are equal to the square on EB, for the angle at A is right; [I. 47] therefore the rectangle CF, FA together with the square on AE is equal to the squares on BA, AE.

Let the square on AE be subtracted from each; therefore the rectangle CF, FA which remains is equal to the square on AB. -

+

Now the rectangle CF, FA is FK, for AF is equal to FG; and the square on AB is AD; therefore FK is equal to AD.

Let AK be subtracted from each; therefore FH which remains is equal to HD.

And HD is the rectangle AB, BH, for AB is equal to BD; and FH is the square on AH; therefore the rectangle contained by AB, BH is equal to the square on HA. therefore the given straight line AB has been cut at H so as to make the rectangle contained by AB, BH equal to the square on HA. Q. E. F.

@@ -3354,7 +1780,7 @@ Proposition 12.

In obtuse-angled triangles the square on the side subtending the obtuse angle is greater than the squares on the sides containing the obtuse angle by twice the rectangle contained by one of the sides about the obtuse angle, namely that on which the - perpendicular falls, and the straight line cut off outside by the perpendicular towards the obtuse angle.

+ perpendicular falls, and the straight line cut off outside by the perpendicular towards the obtuse angle.

Let ABC be an obtuse-angled triangle having the angle BAC obtuse, and let BD be drawn from the point B perpendicular to CA produced;

I say that the square on BC is greater than the squares on BA, AC by twice the rectangle contained by CA, AD.

For, since the straight line CD has been cut at random at the point A, the square on DC is equal to the squares on CA, AD and twice the rectangle contained by CA, AD. [II. 4]

@@ -3388,7 +1814,7 @@

But GF is equal to GH; therefore the rectangle BE, EF together with the square on GE is equal to the square on GH.

But the squares on HE, EG are equal to the square on GH; [I. 47] therefore the rectangle BE, EF together with the square on GE is equal to the squares on HE, EG.

-

Let the square on GE be subtracted from each; therefore the rectangle contained by BE, EF which remains is equal to the square on EH. +

Let the square on GE be subtracted from each; therefore the rectangle contained by BE, EF which remains is equal to the square on EH.

But the rectangle BE, EF is BD, for EF is equal to ED; therefore the parallelogram BD is equal to the square on HE.

And BD is equal to the rectilineal figure A.

@@ -3403,10 +1829,10 @@ - +
- - +
+ BOOK III. DEFINITIONS. @@ -3448,7 +1874,7 @@ 10 - +

A sector of a circle is the figure which, when an angle is constructed at the centre of the circle, is contained by the straight lines containing the angle and the circumference cut off by them.

 @@ -3457,7 +1883,7 @@ - + BOOK III. PROPOSITIONS. PROPOSITION 1. @@ -3469,7 +1895,7 @@

Then, since AD is equal to DB, and DG is common, the two sides AD, DG are equal to the two sides BD, DG respectively; and the base GA is equal to the base GB, for they are radii; therefore the angle ADG is equal to the angle GDB. [I. 8]

But, when a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right; [I. Def. 10] therefore the angle GDB is right. -

+

But the angle FDB is also right; therefore the angle FDB is equal to the angle GDB, the greater to the less: which is impossible. Therefore G is not the centre of the circle ABC.

Similarly we can prove that neither is any other point except F. Therefore the point F is the centre of the circle ABC. @@ -3491,7 +1917,7 @@

If on the circumference of a circle two points be taken at random, the straight line joining the points will fall within the circle.

Let ABC be a circle, and let two points A, B be taken at random on its circumference; I say that the straight line joined from A to B will fall within the circle.

-

For suppose it does not, but, if possible, let it fall outside, as AEB; let the centre of the circle ABC be taken [III. 1], and let it be D; let DA, DB be joined, and let DFE be drawn through.

+

For suppose it does not, but, if possible, let it fall outside, as AEB; let the centre of the circle ABC be taken [III. 1], and let it be D; let DA, DB be joined, and let DFE be drawn through.

Then, since DA is equal to DB, the angle DAE is also equal to the angle DBE. [I. 5] And, since one side AEB of the triangle DAE is produced, the angle DEB is greater than the angle DAE. [I. 16]

But the angle DAE is equal to the angle DBE; therefore the angle DEB is greater than the angle DBE. And the greater angle is subtended by the greater side; [I. 19] therefore DB is greater than DE. But DB is equal to DF; @@ -3516,7 +1942,7 @@

But, when a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right; [I. Def. 10] therefore each of the angles AFE, BFE is right.

Therefore CD, which is through the centre, and bisects AB which is not through the centre, also cuts it at right angles.

-

Again, let CD cut AB at right angles; I say that it also bisects it. that is, that AF is equal to FB.

+

Again, let CD cut AB at right angles; I say that it also bisects it. that is, that AF is equal to FB.

For, with the same construction, since EA is equal to EB, the angle EAF is also equal to the angle EBF. [I. 5]

But the right angle AFE is equal to the right angle BFE, therefore EAF, EBF are two triangles having two angles equal to two angles and one side equal to one side, namely EF, which is common to them, and subtends one of the equal angles; therefore they will also have the remaining sides equal to the remaining sides; [I. 26] therefore AF is equal to FB.

@@ -3533,7 +1959,7 @@

Then, since a straight line FE through the centre bisects a straight line AC not through the centre, it also cuts it at right angles; [III. 3] therefore the angle FEA is right.

Again, since a straight line FE bisects a straight line BD, it also cuts it at right angles; [III. 3] therefore the angle FEB is right. -

+

But the angle FEA was also proved right; therefore the angle FEA is equal to the angle FEB, the less to the greater: which is impossible.

Therefore AC, BD do not bisect one another.

@@ -3555,7 +1981,7 @@

Therefore etc. Q. E. D.

- + PROPOSITION 6.

If two circles touch one another, they will not have the same centre.

For let the two circles ABC, CDE touch one another at the point C; I say that they will not have the same centre.

@@ -3571,7 +1997,7 @@

Therefore etc. Q. E. D.

- + PROPOSITION 7.

If on the diameter of a circle a point be taken which is not the centre of the circle, and from the point straight lines fall upon the circle, that will be greatest on which the centre is, the remainder of the same diameter will be least, and of the rest the nearer to the straight line through the centre is always greater than the more remote, and only two equal straight lines will fall from the point on the circle, one on each side of the least straight line.

@@ -3590,7 +2016,7 @@

Let EF be subtracted from each; therefore the remainder GF is greater than the remainder FD.

-

Therefore FA is greatest, FD is least, and FB is greater than FC, and FC than FG.

+

Therefore FA is greatest, FD is least, and FB is greater than FC, and FC than FG.

I say also that from the point F only two equal straight lines will fall on the circle ABCD, one on each side of the least FD.

For on the straight line EF, and at the point E on it, let the angle FEH be constructed equal to the angle GEF [I. 23], and let FH be joined.

Then, since GE is equal to EH, and EF is common, the two sides GE, EF are equal to the two sides HE, EF; and the angle GEF is equal to the angle HEF; @@ -3617,7 +2043,7 @@ PROPOSITION 8.

If a point be taken outside a circle and from the point straight lines be drawn through to the circle, one of which is through the centre and the others are drawn at random, then, of the straight lines which fall on the concave circumference, that through the centre is greatest, while of the rest - the nearer to that through the centre is always greater than the more remote, but, of the straight lines falling on the convex circumference, that between the point and the diameter is least, while of the rest the nearer to the least is always less than the more remote, and only two equal straight lines will fall on the circle from the point, one on each side of the least.

+ the nearer to that through the centre is always greater than the more remote, but, of the straight lines falling on the convex circumference, that between the point and the diameter is least, while of the rest the nearer to the least is always less than the more remote, and only two equal straight lines will fall on the circle from the point, one on each side of the least.

Let ABC be a circle, and let a point D be taken outside ABC; let there be drawn through from it straight lines DA, DE, DF, DC, and let DA be through the centre; I say that, of the straight lines falling on the concave circumference AEFC, the straight line DA through the centre is greatest,

while DE is greater than DF and DF than DC.; but, of the straight lines falling on the convex circumference HLKG, the straight line DG between the point and the diameter AG is least; and the nearer to the least DG is always less than the more remote, namely DK than DL, and DL than DH.

For let the centre of the circle ABC be taken [III. 1], and let it be M; let ME, MF, MC, MK, ML, MH be joined.

Then, since AM is equal to EM, let MD be added to each; therefore AD is equal to EM, MD. @@ -3627,7 +2053,7 @@

Again, since ME is equal to MF, and MD is common, therefore EM, MD are equal to FM, MD; and the angle EMD is greater than the angle FMD; therefore the base ED is greater than the base FD. [I. 24]

-

Similarly we can prove that FD is greater than CD; therefore DA is greatest, while DE is greater than DF, and DF than DC.

+

Similarly we can prove that FD is greater than CD; therefore DA is greatest, while DE is greater than DF, and DF than DC.

Next, since MK, KD are greater than MD, [I. 20] and MG is equal to MK, therefore the remainder KD is greater than the remainder GD, so that GD is less than KD.

And, since on MD, one of the sides of the triangle MLD, two straight lines MK, KD were constructed meeting within the triangle, therefore MK, KD are less than ML, LD; [I. 21] and MK is equal to ML; therefore the remainder DK is less than the remainder DL. @@ -3655,7 +2081,7 @@

Therefore each of the angles AED, BED is right; [I. Def. 10] therefore GK cuts AB into two equal parts and at right angles.

And since, if in a circle a straight line cut a straight line into two equal parts and at right angles, the centre of the circle is on the cutting straight line, [III. 1, Por.] the centre of the circle is on GK. -

+

For the same reason the centre of the circle ABC is also on HL.

And the straight lines GK, HL have no other point common but the point D; therefore the point D is the centre of the circle ABC. @@ -3663,7 +2089,7 @@

Therefore etc. Q. E. D.

- + PROPOSITION 10.

A circle does not cut a circle at more points than two.

For, if possible, let the circle ABC cut the circle DEF at more points than two, namely B, C, F, H;

@@ -3689,7 +2115,7 @@

For suppose it does not, but, if possible, let it fall as FGH, and let AF, AG be joined.

Then, since AG, GF are greater than FA, that is, than FH,

-

let FG be subtracted from each; therefore the remainder AG is greater than the remainder GH.

+

let FG be subtracted from each; therefore the remainder AG is greater than the remainder GH.

But AG is equal to GD; therefore GD is also greater than GH, the less than the greater: which is impossible.

Therefore the straight line joined from F to G will not fall outside; therefore it will fall at A on the point of contact. @@ -3706,7 +2132,7 @@ PROPOSITION 12.

If two circles touch one another externally, the straight line joining their centres will pass through the point of contact.

-

For let the two circles ABC, ADE touch one another externally at the point A, and let the centre F of ABC, and the centre G of ADE, be taken; I say that the straight line joined from F to G will pass through the point of contact at A.

+

For let the two circles ABC, ADE touch one another externally at the point A, and let the centre F of ABC, and the centre G of ADE, be taken; I say that the straight line joined from F to G will pass through the point of contact at A.

For suppose it does not, but, if possible, let it pass as FCDG, and let AF, AG be joined.

Then, since the point F is the centre of the circle ABC, FA is equal to FC. @@ -3731,7 +2157,7 @@ joined from G to H will fall on B, D. [III. 11]

Let it so fall, as BGHD.

Then, since the point G is the centre of the circle ABCD, BG is equal to GD; - therefore BG is greater than HD; therefore BH is much greater than HD. + therefore BG is greater than HD; therefore BH is much greater than HD.

Again, since the point H is the centre of the circle EBFD, BH is equal to HD; but it was also proved much greater than it: which is impossible. @@ -3755,7 +2181,7 @@ PROPOSITION 14.

In a circle equal straight lines are equally distant from the centre, and those which are equally distant from the centre are equal to one another.

Let ABDC be a circle, and let AB, CD be equal straight lines in it; I say that AB, CD are equally distant from the centre.

-

For let the centre of the circle ABDC be taken [III. 1], and let it be E; from E let EF, EG be drawn perpendicular to AB, CD, and let AE, EC be joined.

+

For let the centre of the circle ABDC be taken [III. 1], and let it be E; from E let EF, EG be drawn perpendicular to AB, CD, and let AE, EC be joined.

Then, since a straight line EF through the centre cuts a straight line AB not through the centre at right angles, it also bisects it. [III. 3]

Therefore AF is equal to FB; therefore AB is double of AF.

@@ -3768,7 +2194,7 @@

Next, let the straight lines AB, CD be equally distant from the centre; that is, let EF be equal to EG.

I say that AB is also equal to CD.

For, with the same construction, we can prove, similarly, that AB is double of AF, and CD of CG.

-

And, since AE is equal to CE, the square on AE is equal to the square on CE. But the squares on EF, FA are equal to the square on AE, and the squares on EG, GC equal to the square on CE. [I. 47]

+

And, since AE is equal to CE, the square on AE is equal to the square on CE. But the squares on EF, FA are equal to the square on AE, and the squares on EG, GC equal to the square on CE. [I. 47]

Therefore the squares on EF, FA are equal to the squares on EG, GC, of which the square on EF is equal to the square on EG, for EF is equal to EG; therefore the square on AF which remains is equal to the square on CG; therefore AF is equal to CG. And AB is double of AF, and CD double of CG; therefore AB is equal to CD.

Therefore etc. Q. E. D.

@@ -3784,7 +2210,7 @@

Then, since EH is equal to EL, BC is also equal to MN. [III. 14]

Again, since AE is equal to EM, and ED to EN, AD is equal to ME, EN. -

+

But ME, EN are greater than MN, [I. 20] and MN is equal to BC; therefore AD is greater than BC.

And, since the two sides ME, EN are equal to the two sides FE, EG, and the angle MEN greater than the angle FEG, therefore the base MN is greater than the base FG. [I. 24]

@@ -3802,7 +2228,7 @@ PROPOSITION 16.

The straight line drawn at right angles to the diameter of a circle from its extremity will fall outside the circle, and into the space between the straight line and the circumference another straight line cannot be interposed; further the angle of the semicircle is greater, and the remaining angle less, than any acute rectilineal angle.

-

Let ABC be a circle about D as centre and AB as diameter; I say that the straight line drawn from A at right angles to AB from its extremity will fall outside the circle.

+

Let ABC be a circle about D as centre and AB as diameter; I say that the straight line drawn from A at right angles to AB from its extremity will fall outside the circle.

For suppose it does not, but, if possible, let it fall within as CA, and let DC be joined.

Since DA is equal to DC,

the angle DAC is also equal to the angle ACD. [I. 5] @@ -3819,7 +2245,7 @@

Therefore another straight line cannot be interposed into the space between the straight line and the circumference.

I say further that the angle of the semicircle contained by the straight line BA and the circumference CHA is greater than any acute rectilineal angle, and the remaining angle contained by the circumference CHA and the straight line AE is less than any acute rectilineal angle.

-

For, if there is any rectilineal angle greater than the angle contained by the straight line BA and the circumference CHA, and any rectilineal angle less than the angle contained by the circumference CHA and the straight line AE, then into the space between the circumference and the straight line AE a straight line will be interposed such as will make an angle contained by straight lines which is greater than the angle contained by the straight line BA and the circumference CHA, and another angle contained by straight lines which is less than the angle contained by the circumference CHA and the straight line AE.

+

For, if there is any rectilineal angle greater than the angle contained by the straight line BA and the circumference CHA, and any rectilineal angle less than the angle contained by the circumference CHA and the straight line AE, then into the space between the circumference and the straight line AE a straight line will be interposed such as will make an angle contained by straight lines which is greater than the angle contained by the straight line BA and the circumference CHA, and another angle contained by straight lines which is less than the angle contained by the circumference CHA and the straight line AE.

But such a straight line cannot be interposed;

therefore there will not be any acute angle contained by straight lines which is greater than the angle contained by the straight line BA and the circumference CHA, nor yet any acute angle contained by straight lines which is less than the angle contained by the circumference CHA and the straight line AE.—

@@ -3858,7 +2284,7 @@

Similarly we can prove that neither is any other straight line except FC; therefore FC is perpendicular to DE.

Therefore etc. Q. E. D. - +

the tangent, h( e)faptome/nh.

@@ -3888,7 +2314,7 @@

For the same reason the angle FEC is also double of the angle EAC.

Therefore the whole angle BEC is double of the whole angle BAC.

-

Again let another straight line be inflected, and let there be another angle BDC; let DE be joined and produced to G.

+

Again let another straight line be inflected, and let there be another angle BDC; let DE be joined and produced to G.

Similarly then we can prove that the angle GEC is double of the angle EDC, of which the angle GEB is double of the angle EDB; therefore the angle BEC which remains is double of the angle BDC.

Therefore etc. Q. E. D. @@ -3913,7 +2339,7 @@

The opposite angles of quadrilaterals in circles are equal to two right angles.

Let ABCD be a circle, and let ABCD be a quadrilateral in it; I say that the opposite angles are equal to two right angles.

Let AC, BD be joined.

-

Then, since in any triangle the three angles are equal to two right angles, [I. 32] the three angles CAB, ABC, BCA of the triangle ABC are equal to two right angles. +

Then, since in any triangle the three angles are equal to two right angles, [I. 32] the three angles CAB, ABC, BCA of the triangle ABC are equal to two right angles.

But the angle CAB is equal to the angle BDC, for they are in the same segment BADC; [III. 21] and the angle ACB is equal to the angle

ADB, for they are in the same segment ADCB; therefore the whole angle ADC is equal to the angles BAC, ACB.

@@ -3925,7 +2351,7 @@ PROPOSITION 23.

On the same straight line there cannot be constructed two similar and unequal segments of circles on the same side.

For, if possible, on the same straight line AB let two similar and unequal segments of circles ACB, ADB be constructed on the same side;

let ACD be drawn through, and let CB, DB be joined.

-

Then, since the segment ACB is similar to the segment ADB, and similar segments of circles are those which admit equal angles, [III. Def. 11] the angle ACB is equal to the angle ADB, the exterior to the interior: which is impossible. [I. 16] +

Then, since the segment ACB is similar to the segment ADB, and similar segments of circles are those which admit equal angles, [III. Def. 11] the angle ACB is equal to the angle ADB, the exterior to the interior: which is impossible. [I. 16]

Therefore etc. Q. E. D. @@ -3936,7 +2362,7 @@ PROPOSITION 24.

Similar segments of circles on equal straight lines are equal to one another.

For let AEB, CFD be similar segments of circles on equal straight lines AB, CD; I say that the segment AEB is equal to the segment CFD.

-

For, if the segment AEB be applied to CFD, and if the point A be placed on C and the straight line AB on CD, the point B will also coincide with the point D, because AB is equal to CD; +

For, if the segment AEB be applied to CFD, and if the point A be placed on C and the straight line AB on CD, the point B will also coincide with the point D, because AB is equal to CD; and, AB coinciding with CD, the segment AEB will also coincide with CFD.

@@ -3952,7 +2378,7 @@ PROPOSITION 25.

Given a segment of a circle, to describe the complete circle of which it is a segment.

Let ABC be the given segment of a circle; thus it is required to describe the complete circle belonging to the segment ABC, that is, of which it is a segment.

-

For let AC be bisected at D, let DB be drawn from the point D at right angles to AC, and let AB. be joined; the angle ABD is then greater than, equal to, or less than the angle BAD. +

For let AC be bisected at D, let DB be drawn from the point D at right angles to AC, and let AB. be joined; the angle ABD is then greater than, equal to, or less than the angle BAD.

First let it be greater; and on the straight line BA, and at the point A on it, let the angle BAE be constructed equal to the angle ABD; let DB be drawn through to E, and let EC be joined.

Then, since the angle ABE is equal to the angle BAE,

@@ -3966,7 +2392,7 @@

And it is manifest that the segment ABC is less than a semicircle, because the centre E happens to be outside it.

Similarly, even if the angle ABD be equal to the angle BAD, AD being equal to each of the two BD, DC, the three straight lines DA, DB, DC will be equal to one another,

D will be the centre of the completed circle, and ABC will clearly be a semicircle. -

+

But, if the angle ABD be less than the angle BAD, and if we construct, on the straight line BA and at the point A on it, an angle equal to the angle ABD, the centre will fall on DB within the segment ABC, and the segment

ABC will clearly be greater than a semicircle.

Therefore, given a segment of a circle, the complete circle has been described. Q. E. F. @@ -3976,7 +2402,7 @@ PROPOSITION 26. -

In equal circles equal angles stand on equal circumferences, whether they stand at the centres or at the circumferences.

+

In equal circles equal angles stand on equal circumferences, whether they stand at the centres or at the circumferences.

Let ABC, DEF be equal circles, and in them let there be equal angles, namely at the centres the angles BGC, EHF, and at the circumferences the angles BAC, EDF; I say that the circumference BKC is equal to the circumference ELF.

For let BC, EF be joined.

@@ -3999,7 +2425,7 @@

But EF is equal to BC; therefore BK is also equal to BC, the less to the greater : which is impossible.

Therefore the angle BGC is not unequal to the angle EHF; therefore it is equal to it. -

+

And the angle at A is half of the angle BGC, and the angle at D half of the angle EHF; [III. 20] therefore the angle at A is also equal to the angle at D.

Therefore etc. Q. E. D.

@@ -4012,7 +2438,7 @@

Now, since the circles are equal, the radii are also equal; therefore the two sides AK, KB are equal to the two sides DL, LE; and the base AB is equal to the base DE; therefore the angle AKB is equal to the angle DLE. [I. 8]

But equal angles stand on equal circumferences, when they are at the centres; [III. 26] therefore the circumference AGB is equal to DHE. -

+

And the whole circle ABC is also equal to the whole circle DEF; therefore the circumference ACB which remains is also equal to the circumference DFE which remains.

Therefore etc. Q. E. D.

@@ -4041,7 +2467,7 @@ PROPOSITION 31. -

In a circle the angle in the semicircle is right, that in a greater segment less than a right angle, and that in a less segment greater than a right angle; and further the angle of the greater segment is greater than a right angle, and the angle of the less segment less than a right angle.

+

In a circle the angle in the semicircle is right, that in a greater segment less than a right angle, and that in a less segment greater than a right angle; and further the angle of the greater segment is greater than a right angle, and the angle of the less segment less than a right angle.

Let ABCD be a circle, let BC be its diameter, and E its centre, and let BA, AC, AD, DC be joined; I say that the angle BAC in the semicircle BAC is right, the angle ABC in the segment ABC greater than the semicircle is less

than a right angle, and the angle ADC in the segment ADC less than the semicircle is greater than a right angle.

Let AE be joined, and let BA be carried through to F.

Then, since BE is equal to EA, the angle ABE is also equal to the angle BAE. [I. 5] @@ -4052,7 +2478,7 @@

But the angle FAC exterior to the triangle ABC is also equal to the two angles ABC, ACB; [I. 32] therefore the angle BAC is also equal to the angle FAC; therefore each is right; [I. Def. 10] therefore the angle BAC in the semicircle BAC is right.

Next, since in the triangle ABC the two angles ABC, BAC are less than two right angles, [I. 17] and the angle BAC is a right angle, the angle ABC is less than a right angle; and it is the angle in the segment ABC greater than the semicircle.

-

Next, since ABCD is a quadrilateral in a circle, and the opposite angles of quadrilaterals in circles are equal to two right angles, [III. 22] while the angle ABC is less than a right angle, therefore the angle ADC which remains is greater than a right angle; and it is the angle in the segment ADC less than the semicircle.

+

Next, since ABCD is a quadrilateral in a circle, and the opposite angles of quadrilaterals in circles are equal to two right angles, [III. 22] while the angle ABC is less than a right angle, therefore the angle ADC which remains is greater than a right angle; and it is the angle in the segment ADC less than the semicircle.

I say further that the angle of the greater segment, namely that contained by the circumference ABC and the straight line AC, is greater than a right angle; and the angle of the less segment, namely that contained by the circumference ADC and the straight line AC, is less than a right angle.

This is at once manifest. For, since the angle contained by the straight lines BA, AC is right, the angle contained by the circumference ABC and the straight line AC is greater than a right angle.

@@ -4063,7 +2489,7 @@ PROPOSITION 32.

If a straight line touch a circle, and from the point of contact there be drawn across, in the circle, a straight line cutting the circle, the angles which it makes with the tangent will be equal to the angles in the alternate segments of the circle.

-

For let a straight line EF touch the circle ABCD at the point B, and from the point B let there be drawn across, in the circle ABCD, a straight line BD cutting it; I say that the angles which BD makes with the tangent EF will be equal to the angles in the alternate segments of the circle, that is, that the angle FBD is equal to the angle constructed in the segment BAD, and the angle EBD is equal to the angle constructed in the segment DCB.

+

For let a straight line EF touch the circle ABCD at the point B, and from the point B let there be drawn across, in the circle ABCD, a straight line BD cutting it; I say that the angles which BD makes with the tangent EF will be equal to the angles in the alternate segments of the circle, that is, that the angle FBD is equal to the angle constructed in the segment BAD, and the angle EBD is equal to the angle constructed in the segment DCB.

For let BA be drawn from B at right angles to EF, let a point C be taken at random on the circumference BD,

and let AD, DC, CB be joined.

Then, since a straight line EF touches the circle ABCD at B, and BA has been drawn from the point of contact at right angles to the tangent, the centre of the circle ABCD is on BA. [III. 19]

Therefore BA is a diameter of the circle ABCD; therefore the angle ADB, being an angle in a semicircle, is right. [III. 31] @@ -4084,7 +2510,7 @@

The angle at C is then acute, or right, or obtuse.

First let it be acute, and, as in the first figure, on the straight line AB, and at the point A, let the angle BAD be constructed equal to the angle at C; therefore the angle BAD is also acute.

-

Let AE be drawn at right angles to DA, let AB be bisected at F, let FG be drawn from the point F at right angles to AB, and let GB be joined.

+

Let AE be drawn at right angles to DA, let AB be bisected at F, let FG be drawn from the point F at right angles to AB, and let GB be joined.

Then, since AF is equal to FB, and FG is common, the two sides AF, FG are equal to the two sides BF, FG; and the angle AFG is equal to the angle BFG; therefore the base AG is equal to the base BG. [I. 4]

Therefore the circle described with centre G and distance GA will pass through B also.

@@ -4096,7 +2522,7 @@

But the angle DAB is equal to the angle at C; therefore the angle at C is also equal to the angle AEB.

Therefore on the given straight line AB the segment AEB of a circle has been described admitting the angle AEB equal to the given angle, the angle at C.

Next let the angle at C be right;

and let it be again required to describe on AB a segment of a circle admitting an angle equal to the right angle at C.

-

Let the angle BAD be constructed equal to the right angle at C, as is the case in the second figure; let AB be bisected at F, and with centre F and distance either FA or FB let the circle AEB be described.

+

Let the angle BAD be constructed equal to the right angle at C, as is the case in the second figure; let AB be bisected at F, and with centre F and distance either FA or FB let the circle AEB be described.

Therefore the straight line AD touches the circle ABE, because the angle at A is right. [III. 16, Por.]

And the angle BAD is equal to the angle in the segment AEB, for the latter too is itself a right angle, being an angle in a semicircle. [III. 31]

But the angle BAD is also equal to the angle at C.

@@ -4109,7 +2535,7 @@

Now, since AD is drawn at right angles to the diameter AE from its extremity, AD touches the circle AEB. [III. 16, Por.]

And AB has been drawn across from the point of contact at A; therefore the angle BAD is equal to the angle constructed in the alternate segment AHB of the circle. [III. 32] -

+

But the angle BAD is equal to the angle at C.

Therefore the angle in the segment AHB is also equal to the angle at C.

Therefore on the given straight line AB the segment AHB of a circle has been described admitting an angle equal to the angle at C. Q. E. F.

@@ -4119,7 +2545,7 @@

From a given circle to cut off a segment admitting an angle equal to a given rectilineal angle.

Let ABC be the given circle, and the angle at D the given rectilineal angle; thus it is required to cut off from the circle ABC a segment admitting an angle equal to the given rectilineal angle, the angle at D.

Let EF be drawn touching ABC at the point B, and on the straight line FB, and at the point B on it, let the angle FBC be constructed equal to the angle at D. [I. 23]

-

Then, since a straight line EF touches the circle ABC, and BC has been drawn across from the point of contact at B, the angle FBC is equal to the angle constructed in the alternate segment BAC. [III. 32]

+

Then, since a straight line EF touches the circle ABC, and BC has been drawn across from the point of contact at B, the angle FBC is equal to the angle constructed in the alternate segment BAC. [III. 32]

But the angle FBC is equal to the angle at D; therefore the angle in the segment BAC is equal to the angle at D.

@@ -4130,7 +2556,7 @@

If in a circle two straight lines cut one another, the rectangle contained by the segments of the one is equal to the rectangle contained by the segments of the other.

For in the circle ABCD let the two straight lines AC, BD cut one another at the point E; I say that the rectangle contained by AE, EC is equal to the rectangle contained by DE, EB.

-

If now AC, BD are through the centre, so that E is the centre of the circle ABCD, it is manifest that, AE, EC, DE, EB being equal, the rectangle contained by AE, EC is also equal to the rectangle contained by DE, EB.

+

If now AC, BD are through the centre, so that E is the centre of the circle ABCD, it is manifest that, AE, EC, DE, EB being equal, the rectangle contained by AE, EC is also equal to the rectangle contained by DE, EB.

Next let AC, DB not be through the centre; let the centre of ABCD be taken, and let it be F; from F let FG, FH be drawn perpendicular to the straight lines AC, DB, and let FB, FC, FE be joined.

Then, since a straight line GF through the centre cuts a straight line AC not through the centre at right angles, it also bisects it; [III. 3] therefore AG is equal to GC. @@ -4147,7 +2573,7 @@ PROPOSITION 36. -

If a point be taken outside a circle and from it there fall on the circle two straight lines, and if one of them cut the circle and the other touch it, the rectangle contained by the whole of the straight line which cuts the circle and the straight line intercepted on it outside between the point and the convex circumference will be equal to the square on the tangent.

+

If a point be taken outside a circle and from it there fall on the circle two straight lines, and if one of them cut the circle and the other touch it, the rectangle contained by the whole of the straight line which cuts the circle and the straight line intercepted on it outside between the point and the convex circumference will be equal to the square on the tangent.

For let a point D be taken outside the circle ABC, and from D let the two straight lines DCA, DB fall on the circle ABC; let DCA cut the circle ABC and let BD touch it; I say that the rectangle contained by AD, DC is equal to the square on DB.

Then DCA is either through the centre or not through the centre.

@@ -4161,7 +2587,7 @@

Then the angle EBD is right. [III. 18]

And, since a straight line EF through the centre cuts a straight line AC not through the centre at right angles, it also bisects it; [III. 3] therefore AF is equal to FC. -

+

Now, since the straight line AC has been bisected at the point F, and CD is added to it, the rectangle contained by AD, DC together with the square on FC is equal to the square on FD. [II. 6]

Let the square on FE be added to each; therefore the rectangle AD, DC together with the squares on CF, FE is equal to the squares on FD, FE.

But the square on EC is equal to the squares on CF, FE, for the angle EFC is right; [I. 47] and the square on ED is equal to the squares on DF, FE; therefore the rectangle AD, DC together with the square on EC is equal to the square on ED.

@@ -4172,7 +2598,7 @@ PROPOSITION 37. -

If a point be taken outside a circle and from the point there fall on the circle two straight lines, if one of them cut the circle, and the other fall on it, and if further the rectangle contained by the whole of the straight line which cuts the circle and the straight line intercepted on it outside between the point and the convex circumference be equal to the square on the straight line which falls on the circle, the straight line which falls on it will touch the circle.

+

If a point be taken outside a circle and from the point there fall on the circle two straight lines, if one of them cut the circle, and the other fall on it, and if further the rectangle contained by the whole of the straight line which cuts the circle and the straight line intercepted on it outside between the point and the convex circumference be equal to the square on the straight line which falls on the circle, the straight line which falls on it will touch the circle.

For let a point D be taken outside the circle ABC; from D let the two straight lines DCA, DB fall on the circle ACB; let DCA cut the circle and DB fall on it; and let the rectangle AD, DC be equal to the square on DB.

I say that DB touches the circle ABC.

@@ -4191,9 +2617,9 @@

Therefore etc. Q. E. D.

- - - +
+
+ BOOK IV. DEFINITIONS. @@ -4227,7 +2653,7 @@ - + BOOK IV. PROPOSITIONS PROPOSITION 1. @@ -4247,7 +2673,7 @@ PROPOSITION 2.

In a given circle to inscribe a triangle equiangular with a given triangle.

Let ABC be the given circle, and DEF the given triangle; thus it is required to inscribe in the circle ABC a triangle equiangular with the triangle DEF.

-

Let GH be drawn touching the circle ABC at A [III. 16, Por.]; on the straight line AH, and at the point A on it, let the angle HAC be constructed equal to the angle DEF, and on the straight line AG, and at the point A on it, let the angle GAB be constructed equal to the angle DFE; [I. 23] let BC be joined.

+

Let GH be drawn touching the circle ABC at A [III. 16, Por.]; on the straight line AH, and at the point A on it, let the angle HAC be constructed equal to the angle DEF, and on the straight line AG, and at the point A on it, let the angle GAB be constructed equal to the angle DFE; [I. 23] let BC be joined.

Then, since a straight line AH touches the circle ABC, and from the point of contact at A the straight line AC is drawn across in the circle, therefore the angle HAC is equal to the angle ABC in the alternate segment of the circle. [III. 32]

But the angle HAC is equal to the angle DEF; therefore the angle ABC is also equal to the angle DEF.

@@ -4260,7 +2686,7 @@

Let ABC be the given circle, and DEF the given triangle; thus it is required to circumscribe about the circle ABC a triangle equiangular with the triangle DEF.

Let EF be produced in both directions to the points G, H, let the centre K of the circle ABC be taken [III. 1], and let the straight line KB be drawn across at random; on the straight line KB, and at the point K on it, let the angle BKA be constructed equal to the angle DEG, and the angle BKC equal to the angle DFH; [I. 23] and through the points A, B, C let LAM, MBN, NCL be drawn touching the circle ABC. [III. 16, Por.]

-

Now, since LM, MN, NL touch the circle ABC at the points A, B, C, and KA, KB, KC have been joined from the centre K to the points A, B, C, therefore the angles at the points A, B, C are right. [III. 18]

+

Now, since LM, MN, NL touch the circle ABC at the points A, B, C, and KA, KB, KC have been joined from the centre K to the points A, B, C, therefore the angles at the points A, B, C are right. [III. 18]

And, since the four angles of the quadrilateral AMBK are equal to four right angles, inasmuch as AMBK is in fact divisible into two triangles, and the angles KAM, KBM are right, therefore the remaining angles AKB, AMB are equal to two right angles.

But the angles DEG, DEF are also equal to two right angles; [I. 13] therefore the angles AKB, AMB are equal to the angles DEG, DEF, of which the angle AKB is equal to the angle DEG; therefore the angle AMB which remains is equal to the angle DEF which remains. @@ -4293,7 +2719,7 @@

Therefore the three straight lines DE, DF, DG are equal to one another; therefore the circle described with centre D and distance one of the straight lines DE, DF, DG will pass also through the remaining points, and will touch the straight lines AB, BC, CA, because the angles at the points E, F, G are right.

-

For, if it cuts them, the straight line drawn at right angles to the diameter of the circle from its extremity will be found to fall within the circle : which was proved absurd; [III. 16] therefore the circle described with centre D and distance one of the straight lines DE, DF, DG will not cut the straight lines AB, BC, CA; +

For, if it cuts them, the straight line drawn at right angles to the diameter of the circle from its extremity will be found to fall within the circle : which was proved absurd; [III. 16] therefore the circle described with centre D and distance one of the straight lines DE, DF, DG will not cut the straight lines AB, BC, CA; therefore it will touch them, and will be the circle inscribed in the triangle ABC. [IV. Def. 5]

Let it be inscribed, as FGE.

@@ -4310,7 +2736,7 @@

About a given triangle to circumscribe a circle.

Let ABC be the given triangle; thus it is required to circumscribe a circle about the given triangle ABC.

-

Let the straight lines AB, AC be bisected at the points D, E [I. 10], and from the points D, E let DF, EF be drawn at right angles to AB, AC; they will then meet within the triangle ABC, or on the straight line BC, or outside BC.

+

Let the straight lines AB, AC be bisected at the points D, E [I. 10], and from the points D, E let DF, EF be drawn at right angles to AB, AC; they will then meet within the triangle ABC, or on the straight line BC, or outside BC.

First let them meet within at F, and let FB, FC, FA be joined.

Then, since AD is equal to DB, and DF is common and at right angles, therefore the base AF is equal to the base FB. [I. 4]

Similarly we can prove that CF is also equal to AF; so that FB is also equal to FC; therefore the three straight lines FA, FB, FC are equal to one another. @@ -4323,10 +2749,10 @@

Then again, since AD is equal to DB, and DF is common and at right angles, therefore the base AF is equal to the base BF. [I. 4]

Similarly we can prove that CF is also equal to AF; so that BF is also equal to FC; therefore the circle described with centre F and distance one of the straight lines FA, FB, FC will pass also through the remaining points, and will have been circumscribed about the triangle ABC.

Therefore about the given triangle a circle has been circumscribed. Q. E. F.

-

And it is manifest that, when the centre of the circle falls within the triangle, the angle BAC, being in a segment greater than the semicircle, is less than a right angle; when the centre falls on the straight line BC, the angle BAC, being in a semicircle, is right; and when the centre of the circle falls outside the triangle, the angle BAC, being in a segment less than the semicircle, is greater than a right angle. [III. 31]

+

And it is manifest that, when the centre of the circle falls within the triangle, the angle BAC, being in a segment greater than the semicircle, is less than a right angle; when the centre falls on the straight line BC, the angle BAC, being in a semicircle, is right; and when the centre of the circle falls outside the triangle, the angle BAC, being in a segment less than the semicircle, is greater than a right angle. [III. 31]

- + PROPOSITION 6.

In a given circle to inscribe a square.

Let ABCD be the given circle; thus it is required to inscribe a square in the circle ABCD.

@@ -4354,7 +2780,7 @@

For the same reason the angles at the points B, C, D are also right.

Now, since the angle AEB is right, and the angle EBG is also right, therefore GH is parailel to AC. [I. 28] -

+

For the same reason AC is also parallel to FK, so that GH is also parallel to FK. [I. 30]

Similarly we can prove that each of the straight lines GF, HK is parallel to BED. @@ -4371,7 +2797,7 @@

Therefore about the given circle a square has been circumscribed. Q. E. F.

 - + PROPOSITION 8.

In a given square to inscribe a circle.

Let ABCD be the given square; thus it is required to inscribe a circle in the given square ABCD.

@@ -4406,7 +2832,7 @@

Therefore about the given square a circle has been circumscribed. Q. E. F.

- + PROPOSITION 10.

To construct an isosceles triangle having each of the angles at the base double of the remaining one.

Let any straight line AB be set out, and let it be cut at the point C so that the rectangle contained by AB, BC is equal to the square on CA; [II. 11] with centre A and distance AB let the circle BDE be described, and let there be fitted in the circle BDE the straight line BD equal to the straight line AC which is not greater than the diameter of the circle BDE. [IV. 1]

@@ -4416,7 +2842,7 @@

And, since a point B has been taken outside the circle ACD, and from B the two straight lines BA, BD have fallen on the circle ACD, and one of them cuts it, while the other falls on it, and the rectangle AB, BC is equal to the square on BD, therefore BD touches the circle ACD. [III. 37]

Since, then, BD touches it, and DC is drawn across from the point of contact at D, therefore the angle BDC is equal to the angle DAC in the alternate segment of the circle. [III. 32]

-

Since, then, the angle BDC is equal to the angle DAC, let the angle CDA be added to each; therefore the whole angle BDA is equal to the two angles CDA, DAC.

+

Since, then, the angle BDC is equal to the angle DAC, let the angle CDA be added to each; therefore the whole angle BDA is equal to the two angles CDA, DAC.

But the exterior angle BCD is equal to the angles CDA, DAC; [I. 32] therefore the angle BDA is also equal to the angle BCD.

But the angle BDA is equal to the angle CBD, since the side AD is also equal to AB; [I. 5] so that the angle DBA is also equal to the angle BCD.

Therefore the three angles BDA, DBA, BCD are equal to one another.

@@ -4433,7 +2859,7 @@

Let ABCDE be the given circle; thus it is required to inscribe in the circle ABCDE an equilateral and equiangular pentagon.

Let the isosceles triangle FGH be set out having each of the angles at G, H double of the angle at F; [IV. 10]

let there be inscribed in the circle ABCDE the triangle ACD equiangular with the triangle FGH, so that the angle CAD is equal to the angle at F and the angles at G, H respectively equal to the angles ACD, CDA; [IV. 2] therefore each of the angles ACD, CDA is also double of the angle CAD.

Now let the angles ACD, CDA be bisected respectively by the straight lines CE, DB [I. 9], and let AB, BC, DE, EA be joined.

-

Then, since each of the angles ACD, CDA is double of the angle CAD, and they have been bisected by the straight lines CE, DB, therefore the five angles DAC, ACE, ECD, CDB, BDA are equal to one another.

+

Then, since each of the angles ACD, CDA is double of the angle CAD, and they have been bisected by the straight lines CE, DB, therefore the five angles DAC, ACE, ECD, CDB, BDA are equal to one another.

But equal angles stand on equal circumferences; [III. 26] therefore the five circumferences AB, BC, CD, DE, EA are equal to one another.

But equal circumferences are subtended by equal straight lines; [III. 29] therefore the five straight lines AB, BC, CD, DE, EA are equal to one another; therefore the pentagon ABCDE is equilateral.

@@ -4454,7 +2880,7 @@

Then, since the straight line KL touches the circle ABCDE at C, and FC has been joined from the centre F to the point of contact at C, therefore FC is perpendicular to KL; [III. 18] therefore each of the angles at C is right.

For the same reason the angles at the points B, D are also right. -

+

And, since the angle FCK is right, therefore the square on FK is equal to the squares on FC, CK.

For the same reason [I. 47] the square on FK is also equal to the squares on FB, BK; so that the squares on FC, CK are equal to the squares on FB, BK, of which the square on FC is equal to the square on FB; therefore the square on CK which remains is equal to the square on BK.

Therefore BK is equal to CK.

@@ -4467,7 +2893,7 @@

But the angle FCK is also equal to the angle FCL; therefore FKC, FLC are two triangles having two angles equal to two angles and one side equal to one side, namely FC which is common to them; therefore they will also have the remaining sides equal to the remaining sides, and the remaining angle to the remaining angle; [I. 26] therefore the straight line KC is equal to CL, and the angle FKC to the angle FLC.

-

And, since KC is equal to CL, therefore KL is double of KC.

+

And, since KC is equal to CL, therefore KL is double of KC.

For the same reason it can be proved that HK is also double of BK.

And BK is equal to KC; therefore HK is also equal to KL. @@ -4485,14 +2911,14 @@ PROPOSITION 13.

In a given pentagon, which is equilateral and equiangular, to inscribe a circle.

Let ABCDE be the given equilateral and equiangular pentagon; thus it is required to inscribe a circle in the pentagon ABCDE.

-

For let the angles BCD, CDE be bisected by the straight lines CF, DF respectively; and from the point F, at which the straight lines CF, DF meet one another, let the straight lines FB, FA, FE be joined.

+

For let the angles BCD, CDE be bisected by the straight lines CF, DF respectively; and from the point F, at which the straight lines CF, DF meet one another, let the straight lines FB, FA, FE be joined.

Then, since BC is equal to CD, and CF common, the two sides BC, CF are equal to the two sides DC, CF;

and the angle BCF is equal to the angle DCF; therefore the base BF is equal to the base DF, and the triangle BCF is equal to the triangle DCF, and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend. [I. 4]

Therefore the angle CBF is equal to the angle CDF.

And, since the angle CDE is double of the angle CDF, and the angle CDE is equal to the angle ABC, while the angle CDF is equal to the angle CBF; therefore the angle CBA is also double of the angle CBF; therefore the angle ABF is equal to the angle FBC; therefore the angle ABC has been bisected by the straight line BF.

Similarly it can be proved that the angles BAE, AED have also been bisected by the straight lines FA, FE respectively.

Now let FG, FH, FK, FL, FM be drawn from the point F perpendicular to the straight lines AB, BC, CD, DE, EA.

Then, since the angle HCF is equal to the angle KCF, and the right angle FHC is also equal to the angle FKC, FHC, FKC are two triangles having two angles equal to two angles and one side equal to one side, namely FC which is common to them and subtends one of the equal angles; therefore they will also have the remaining sides equal to the remaining sides; [I. 26] therefore the perpendicular FH is equal to the perpendicular FK.

-

Similarly it can be proved that each of the straight lines FL, FM, FG is also equal to each of the straight lines FH, FK; therefore the five straight lines FG, FH, FK, FL, FM are equal to one another.

+

Similarly it can be proved that each of the straight lines FL, FM, FG is also equal to each of the straight lines FH, FK; therefore the five straight lines FG, FH, FK, FL, FM are equal to one another.

Therefore the circle described with centre F and distance one of the straight lines FG, FH, FK, FL, FM will pass also through the remaining points; and it will touch the straight lines AB, BC, CD, DE, EA, because the angles at the points G, H, K, L, M are right.

For, if it does not touch them. but cuts them, it will result that the straight line drawn at right angles to the diameter of the circle from its extremity falls within the circle: which was proved absurd. [III. 16]

Therefore the circle described with centre F and distance one of the straight lines FG, FH, FK, FL, FM will not cut the straight lines AB, BC, CD, DE, EA; therefore it will touch them. @@ -4506,7 +2932,7 @@

Let ABCDE be the given pentagon, which is equilateral and equiangular; thus it is required to circumscribe a circle about the pentagon ABCDE.

Let the angles BCD, CDE be bisected by the straight lines CF, DF respectively, and from the point F, at which the straight lines meet, let the straight lines FB, FA, FE be joined to the points B, A, E.

-

Then in manner similar to the preceding it can be proved that the angles CBA, BAE, AED have also been bisected by the straight lines FB, FA, FE respectively.

+

Then in manner similar to the preceding it can be proved that the angles CBA, BAE, AED have also been bisected by the straight lines FB, FA, FE respectively.

Now, since the angle BCD is equal to the angle CDE, and the angle FCD is half of the angle BCD, and the angle CDF half of the angle CDE, therefore the angle FCD is also equal to the angle CDF, so that the side FC is also equal to the side FD. [I. 6]

Similarly it can be proved that each of the straight lines FB, FA, FE is also equal to each of the straight lines FC, FD; therefore the five straight lines FA, FB, FC, FD, FE are equal to one another.

@@ -4521,7 +2947,7 @@

Let the diameter AD of the circle ABCDEF be drawn; let the centre G of the circle be taken, and with centre D and distance DG let the circle EGCH be described;

let EG, CG be joined and carried through to the points B, F, and let AB, BC, CD, DE, EF, FA be joined.

I say that the hexagon ABCDEF is equilateral and equiangular.

For, since the point G is the centre of the circle ABCDEF, GE is equal to GD. -

+

Again, since the point D is the centre of the circle GCH, DE is equal to DG.

But GE was proved equal to GD; therefore GE is also equal to ED; therefore the triangle EGD is equilateral; and therefore its three angles EGD, GDE, DEG are equal to one another, inasmuch as, in isosceles triangles, the angles at the base are equal to one another. [I. 5]

@@ -4536,7 +2962,7 @@

I say next that it is also equiangular.

For, since the circumference FA is equal to the circumference ED, let the circumference ABCD be added to each; therefore the whole FABCD is equal to the whole EDCBA; - and the angle FED stands on the circumference FABCD, and the angle AFE on the circumference EDCBA; therefore the angle AFE is equal to the angle DEF. [III. 27] + and the angle FED stands on the circumference FABCD, and the angle AFE on the circumference EDCBA; therefore the angle AFE is equal to the angle DEF. [III. 27]

Similarly it can be proved that the remaining angles of the hexagon ABCDEF are also severally equal to each of the angles AFE, FED; therefore the hexagon ABCDEF is equiangular.

@@ -4550,7 +2976,7 @@ - + PROPOSITION 16.

In a given circle to inscribe a fifteen-angled figure which shall be both equilateral and equiangular.

Let ABCD be the given circle; thus it is required to inscribe in the circle ABCD a fifteenangled figure which shall be both equilateral and equiangular.

@@ -4563,9 +2989,9 @@

And further, by proofs similar to those in the case of the pentagon, we can both inscribe a circle in the given fifteenangled figure and circumscribe one about it. Q. E. F.

- - - +
+
+ BOOK V. DEFINITIONS. @@ -4579,7 +3005,7 @@ 3 - +

A ratio is a sort of relation in respect of size between two magnitudes of the same kind.

 @@ -4624,7 +3050,7 @@ 14 - +

Composition of a ratio means taking the antecedent together with the consequent as one in relation to the consequent by itself.

 @@ -4646,7 +3072,7 @@ - + BOOK V. PROPOSITIONS. PROPOSITION I. @@ -4657,7 +3083,7 @@

Let AB be divided into the magnitudes AG, GB equal to E, and CD into CH, HD equal to F; then the multitude of the magnitudes AG, GB will be equal to the multitude of the magnitudes CH, HD.

Now, since AG is equal to E, and CH to F, therefore AG is equal to E, and AG, CH to E, F.

For the same reason

-

GB is equal to E, and GB, HD to E, F; therefore, as many magnitudes as there are in AB equal to E, so many also are there in AB, CD equal to E, F; therefore, whatever multiple AB is of E, that multiple will AB, CD also be of E, F.

+

GB is equal to E, and GB, HD to E, F; therefore, as many magnitudes as there are in AB equal to E, so many also are there in AB, CD equal to E, F; therefore, whatever multiple AB is of E, that multiple will AB, CD also be of E, F.

Therefore etc. Q. E. D.

 @@ -4665,7 +3091,7 @@

If a first magnitude be the same multiple of a second that a third is of a fourth, and a fifth also be the same multiple of the second that a sixth is of the fourth, the sum of the first and fifth will also be the same multiple of the second that the sum of the third and sixth is of the fourth.

Let a first magnitude, AB, be the same multiple of a second, C, that a third, DE, is of a fourth, F, and let a fifth, BG, also be the same multiple of the second, C, that a sixth, EH, is of the fourth F;

I say that the sum of the first and fifth, AG, will be the same multiple of the second, C, that the sum of the third and sixth, DH, is of the fourth, F.

For, since AB is the same multiple of C that DE is of F, therefore, as many magnitudes as there are in AB equal to C, so many also are there in DE equal to F.

-

For the same reason also, as many as there are in BG equal to C, so many are there also in EH equal to F; therefore, as many as there are in the whole AG equal to C, so many also are there in the whole DH equal to F.

+

For the same reason also, as many as there are in BG equal to C, so many are there also in EH equal to F; therefore, as many as there are in the whole AG equal to C, so many also are there in the whole DH equal to F.

Therefore, whatever multiple AG is of C, that multiple also is DH of F.

Therefore the sum of the first and fifth, AG, is the same multiple of the second, C, that the sum of the third and sixth, DH, is of the fourth, F.

Therefore etc. Q.E.D.

@@ -4674,7 +3100,7 @@ PROPOSITION 3.

If a first magnitude be the same multiple of a second that a third is of a fourth, and if equimultiples be taken of the first and third, then also ex aequali the magnitudes taken will be equimultiples respectively, the one of the second and the other of the fourth.

Let a first magnitude A be the same multiple of a second B that a third C is of a fourth D, and let equimultiples EF, GH be taken of A, C; I say that EF is the same multiple of B that GH is of D.

-

For, since EF is the same multiple of A that GH is of C, therefore, as many magnitudes as there are in EF equal to A, so many also are there in GH equal to C.

+

For, since EF is the same multiple of A that GH is of C, therefore, as many magnitudes as there are in EF equal to A, so many also are there in GH equal to C.

Let EF be divided into the magnitudes EK, KF equal to A, and GH into the magnitudes GL, LH equal to C; then the multitude of the magnitudes EK, KF will be equal to the multitude of the magnitudes GL, LH.

And, since A is the same multiple of B that C is of D, while EK is equal to A, and GL to C, therefore EK is the same multiple of B that GL is of D.

@@ -4691,7 +3117,7 @@

For let equimultiples K, L be taken of E, F, and other, chance, equimultiples M, N of G, H.

Since E is the same multiple of A that F is of C, and equimultiples K, L of E, F have been taken, therefore K is the same multiple of A that L is of C. [V. 3]

For the same reason M is the same multiple of B that N is of D. -

+

And, since, as A is to B, so is C to D, and of A, C equimultiples K, L have been taken, and of B, D other, chance, equimultiples M, N, therefore, if K is in excess of M, L also is in excess of N, if it is equal, equal, and if less, less. [V. Def. 5]

And K, L are equimultiples of E, F, and M, N other, chance, equimultiples of G, H; therefore, as E is to G, so is F to H. [V. Def. 5]

Therefore etc. Q. E. D.

@@ -4706,7 +3132,7 @@

But, by the assumption, AE is the same multiple of CF that AB is of CD.

Therefore AB is the same multiple of each of the magnitudes GF, CD; therefore GF is equal to CD.

-

Let CF be subtracted from each; therefore the remainder GC is equal to the remainder FD.

+

Let CF be subtracted from each; therefore the remainder GC is equal to the remainder FD.

And, since AE is the same multiple of CF that EB is of GC, and GC is equal to DF, therefore AE is the same multiple of CF that EB is of FD.

But, by hypothesis,

AE is the same multiple of CF that AB is of CD; therefore EB is the same multiple of FD that AB is of CD.

@@ -4725,7 +3151,7 @@

Since AG is the same multiple of E that CH is of F, while GB is equal to E and KC to F, therefore AB is the same multiple of E that KH is of F. [V. 2]

But, by hypothesis, AB is the same multiple of E that CD is of F; therefore KH is the same multiple of F that CD is of F.

Since then each of the magnitudes KH, CD is the same multiple of F, therefore KH is equal to CD. -

+

Let CH be subtracted from each; therefore the remainder KC is equal to the remainder HD.

But F is equal to KC; therefore HD is also equal to F.

Hence, if GB is equal to E, HD is also equal to F.

@@ -4740,7 +3166,7 @@

For let equimultiples D, E of A, B be taken, and of C another, chance, multiple F.

Then, since D is the same multiple of A that E is of B, while A is equal to B, therefore D is equal to E.

-

But F is another, chance, magnitude.

+

But F is another, chance, magnitude.

If therefore D is in excess of F, E is also in excess of F, if equal to it, equal; and, if less, less.

And D, E are equimultiples of A, B, while F is another, chance, multiple of C; therefore, as A is to C, so is B to C. [V. Def. 5]

@@ -4757,7 +3183,7 @@ PROPOSITION 8. -

Of unequal magnitudes, the greater has to the same a greater ratio than the less has; and the same has to the less a greater ratio than it has to the greater.

+

Of unequal magnitudes, the greater has to the same a greater ratio than the less has; and the same has to the less a greater ratio than it has to the greater.

Let AB, C be unequal magnitudes, and let AB be greater; let D be another, chance, magnitude; I say that AB has to D a greater ratio than C has to D, and D has to C a greater ratio than it has to AB.

For, since AB is greater than C, let BE be made equal to C; then the less of the magnitudes AE, EB, if multiplied, will sometime be greater than D. [V. Def. 4]

@@ -4768,7 +3194,7 @@

But FG is the same multiple of AE that K is of C; therefore FH is the same multiple of AB that K is of C; therefore FH, K are equimultiples of AB, C.

Again, since GH is the same multiple of EB that K is of C, and EB is equal to C, therefore GH is equal to K. -

+

But K is not less than M; therefore neither is GH less than M.

And FG is greater than D; therefore the whole FH is greater than D, M together.

@@ -4785,7 +3211,7 @@

Then the less, EB, if multiplied, will sometime be greater than D. [V. Def. 4]

Let it be multiplied, and let GH be a multiple of EB and greater than D; and, whatever multiple GH is of EB, let FG be made the same multiple of AE, and K

of C.

-

Then we can prove similarly that FH, K are equimultiples of AB, C; and, similarly, let N be taken a multiple of D but the first that is greater than FG, so that FG is again not less than M.

+

Then we can prove similarly that FH, K are equimultiples of AB, C; and, similarly, let N be taken a multiple of D but the first that is greater than FG, so that FG is again not less than M.

But GH is greater than D; therefore the whole FH is in excess of D, M, that is, of N.

Now K is not in excess of N, inasmuch as FG also, which is greater than GH, that is, than K, is not in excess of N.

And in the same manner, by following the above argument, we complete the demonstration.

@@ -4793,7 +3219,7 @@ PROPOSITION 9. -

Magnitudes which have the same ratio to the same are equal to one another; and magnitudes to which the same has the same ratio are equal.

+

Magnitudes which have the same ratio to the same are equal to one another; and magnitudes to which the same has the same ratio are equal.

For let each of the magnitudes A, B have the same ratio to C; I say that A is equal to B.

For, otherwise, each of the magnitudes A, B would not have had the same ratio to C; [V. 8] but it has; therefore A is equal to B. @@ -4818,7 +3244,7 @@

Again, let C have to B a greater ratio than C has to A; I say that B is less than A.

For, if not, it is either equal or greater.

Now B is not equal to A; for in that case C would have had the same ratio to each of the magnitudes A, B; [V. 7] but it has not; therefore A is not equal to B. -

+

Nor again is B greater than A; for in that case C would have had to B a less ratio than it has to A; [V. 8] but it has not; therefore B is not greater than A.

But it was proved that it is not equal either; therefore B is less than A. @@ -4826,7 +3252,7 @@

Therefore etc. Q. E. D.

- + PROPOSITION 11.

Ratios which are the same with the same ratio are also the same with one another.

For, as A is to B, so let C be to D, and, as C is to D, so let E be to F; I say that, as A is to B, so is E to F.

@@ -4845,7 +3271,7 @@

Let any number of magnitudes A, B, C, D, E, F be proportional, so that, as A is to B, so is C to D and E to F; I say that, as A is to B, so are A, C, E to B, D, F.

For of A, C, E let equimultiples G, H, K be taken, and of B, D, F other, chance, equimultiples L, M, N.

-

Then since, as A is to B, so is C to D, and E to F, and of A, C, E equimultiples G, H, K have been taken, and of B, D, F other, chance, equimultiples L, M, N, therefore, if G is in excess of L, H is also in excess of M, and K of N, if equal, equal, and if less, less; so that, in addition, if G is in excess of L, then G, H, K are in excess of L, M, N, if equal, equal, and if less, less.

+

Then since, as A is to B, so is C to D, and E to F, and of A, C, E equimultiples G, H, K have been taken, and of B, D, F other, chance, equimultiples L, M, N, therefore, if G is in excess of L, H is also in excess of M, and K of N, if equal, equal, and if less, less; so that, in addition, if G is in excess of L, then G, H, K are in excess of L, M, N, if equal, equal, and if less, less.

Now G and G, H, K are equimultiples of A and A, C, E, since, if any number of magnitudes whatever are respectively equimultiples of any magnitudes equal in multitude, whatever multiple one of the magnitudes is of one, that multiple also will all be of all. [V. 1]

For the same reason L and L, M, N are also equimultiples of B and B, D, F; therefore, as A is to B, so are A, C, E to B, D, F. [V. Def. 5]

@@ -4856,7 +3282,7 @@

If a first magnitude have to a second the same ratio as a third to a fourth, and the third have to the fourth a greater ratio than a fifth has to a sixth, the first will also have to the second a greater ratio than the fifth to the sixth.

For let a first magnitude A have to a second B the same ratio as a third C has to a fourth D, and let the third C have to the fourth D a greater ratio than a fifth E has to a sixth F; I say that the first A will also have to the second B a greater ratio than the fifth E to the sixth F.

-

For, since there are some equimultiples of C, E, and of D, F other, chance, equimultiples, such that the multiple of C is in excess of the multiple of D, while the multiple of E is not in excess of the multiple of F, [V. Def. 7] let them be taken, and let G, H be equimultiples of C, E, and K, L other, chance, equimultiples of D, F, so that G is in excess of K, but H is not in excess of L; and, whatever multiple G is of C, let M be also that multiple of A, and, whatever multiple K is of D, let N be also that multiple of B.

+

For, since there are some equimultiples of C, E, and of D, F other, chance, equimultiples, such that the multiple of C is in excess of the multiple of D, while the multiple of E is not in excess of the multiple of F, [V. Def. 7] let them be taken, and let G, H be equimultiples of C, E, and K, L other, chance, equimultiples of D, F, so that G is in excess of K, but H is not in excess of L; and, whatever multiple G is of C, let M be also that multiple of A, and, whatever multiple K is of D, let N be also that multiple of B.

Now, since, as A is to B, so is C to D, and of A, C equimultiples M, G have been taken, and of B, D other, chance, equimultiples N, K, therefore, if M is in excess of N, G is also in excess of K, if equal, equal, and if less, less. [V. Def. 5]

But G is in excess of K; therefore M is also in excess of N.

But H is not in excess of L; and M, H are equimultiples of A, E, and N, L other, chance, equimultiples of B, F; therefore A has to B a greater ratio than E has to F. [V. Def. 7] @@ -4885,7 +3311,7 @@

Let AB be divided into the magnitudes AG, GH, HB equal to C, and DE into the magnitudes DK, KL, LE equal to F; then the multitude of the magnitudes AG, GH, HB will be equal to the multitude of the magnitudes DK, KL, LE.

And, since AG. GH, HB are equal to one another, and DK, KL, LE are also equal to one another, therefore, as AG is to DK, so is GH to KL, and HB to LE. [V. 7]

Therefore, as one of the antecedents is to one of the consequents, so will all the antecedents be to all the consequents; [V. 12] therefore, as AG is to DK, so is AB to DE. -

+

But AG is equal to C and DK to F; therefore, as C is to F, so is AB to DE.

Therefore etc. Q. E. D.

@@ -4901,7 +3327,7 @@

Again, since G, H are equimultiples of C, D, therefore, as C is to D, so is G to H. [V. 15]

But, as C is to D, so is E to F; therefore also, as E is to F, so is G to H. [V. 11]

But, if four magnitudes be proportional, and the first be greater than the third, the second will also be greater than the fourth; if equal, equal; and if less, less. [V. 14]

-

Therefore, if E is in excess of G, F is also in excess of H, if equal, equal, and if less, less.

+

Therefore, if E is in excess of G, F is also in excess of H, if equal, equal, and if less, less.

Now E, F are equimultiples of A, B, and G, H other, chance, equimultiples of C, D; therefore, as A is to C, so is B to D. [V. Def. 5]

Therefore etc. Q. E. D. @@ -4919,7 +3345,7 @@ componendo, they will also be proportional separando.

Let AB, BE, CD, DF be magnitudes proportional componendo, so that, as AB is to BE, so is CD to DF; I say that they will also be proportional separando, that is, as AE is to EB, so is CF to DF.

-

For of AE, EB, CF, FD let equimultiples GH, HK, LM, MN be taken, and of EB, FD other, chance, equimultiples, KO, NP.

+

For of AE, EB, CF, FD let equimultiples GH, HK, LM, MN be taken, and of EB, FD other, chance, equimultiples, KO, NP.

Then, since GH is the same multiple of AE that HK is of EB, therefore GH is the same multiple of AE that GK is of AB. [V. 1]

But GH is the same multiple of AE that LM is of CF; therefore GK is the same multiple of AB that LM is of CF.

@@ -4932,7 +3358,7 @@

Let GK be in excess of HO; then, if HK be subtracted from each, GH is also in excess of KO.

But we saw that, if GK was in excess of HO, LN was also in excess of MP; therefore LN is also in excess of MP, - and, if MN be subtracted from each, LM is also in excess of NP; so that, if GH is in excess of KO, LM is also in excess of NP.

+ and, if MN be subtracted from each, LM is also in excess of NP; so that, if GH is in excess of KO, LM is also in excess of NP.

Similarly we can prove that, if GH be equal to KO, LM will also be equal to NP, and if less, less.

And GH, LM are equimultiples of AE, CF, while KO, NP are other, chance, equimultiples of EB, FD; therefore, as AE is to EB, so is CF to FD.

@@ -4955,7 +3381,7 @@

But the first CG is greater than the third CF; therefore the second GD is also greater than the fourth FD. [V. 14]

But it is also less: which is impossible.

-

Therefore, as AB is to BE, so is not CD to a less magnitude than FD.

+

Therefore, as AB is to BE, so is not CD to a less magnitude than FD.

Similarly we can prove that neither is it in that ratio to a greater; it is therefore in that ratio to FD itself.

Therefore etc. Q. E. D.

@@ -4982,7 +3408,7 @@ PROPOSITION 20.

If there be three magnitudes, and others equal to them in multitude, which taken two and two are in the same ratio, and if ex aequali - the first be greater than the third, the fourth will also be greater than the sixth; if equal, equal; and, if less, less.

+ the first be greater than the third, the fourth will also be greater than the sixth; if equal, equal; and, if less, less.

Let there be three magnitudes A, B, C, and others D, E, F equal to them in multitude, which taken two and two are in the same ratio, so that, as A is to B, so is D to E, and as B is to C, so is E to F; and let A be greater than C ex aequali; I say that D will also be greater than F; if A is equal to C, equal; and, if less, less.

For, since A is greater than C, and B is some other magnitude, and the greater has to the same a greater ratio than the less has, [V. 8] therefore A has to B a greater ratio than C has to B.

@@ -5002,7 +3428,7 @@

For, since A is greater than C, and B is some other magnitude, therefore A has to B a greater ratio than C has to B. [V. 8]

But, as A is to B, so is E to F, and, as C is to B, inversely, so is E to D. Therefore also E has to F a greater ratio than E has to D. [V. 13]

But that to which the same has a greater ratio is less; [V. 10] therefore F is less than D; therefore D is greater than F. -

+

Similarly we can prove that, if A be equal to C, D will also be equal to F; and if less, less.

Therefore etc. Q. E. D.

@@ -5010,7 +3436,7 @@ PROPOSITION 22.

If there be any number of magnitudes whatever, and others equal to them in multitude, which taken two and two together are in the same ratio, they will also be in the same ratio ex aequali.

Let there be any number of magnitudes A, B, C, and others D, E, F equal to them in multitude, which taken two and two together are in the same ratio, so that, as A is to B, so is D to E, and, as B is to C, so is E to F; I say that they will also be in the same ratio ex aequali, <that is, as A is to C, so is D to F>. -

+

For of A, D let equimultiples G, H be taken, and of B, E other, chance, equimultiples K, L; and, further, of C, F other, chance, equimultiples M, N.

Then, since, as A is to B, so is D to E, and of A, D equimultiples G, H have been taken, and of B, E other, chance, equimultiples K, L, therefore, as G is to K, so is H to L. [V. 4] @@ -5036,7 +3462,7 @@

Next, since, as B is to C, so is D to E, alternately, also, as B is to D, so is C to E. [V. 16]

And, since H, K are equimultiples of B, D, and parts have the same ratio as their equimultiples, therefore, as B is to D, so is H to K. [V. 15] -

+

But, as B is to D, so is C to E; therefore also, as H is to K, so is C to E. [V. 11]

Again, since L, M are equimultiples of C, E, therefore, as C is to E, so is L to M. [V. 15] @@ -5052,7 +3478,7 @@ PROPOSITION 24.

If a first magnitude have to a second the same ratio as a third has to a fourth, and also a fifth have to the second the same ratio as a sixth to the fourth, the first and fifth added together will have to the second the same ratio as the third and sixth have to the fourth.

-

Let a first magnitude AB have to a second C the same ratio as a third DE has to a fourth F; and let also a fifth BG have to

the second C the same ratio as a sixth EH has to the fourth F; I say that the first and fifth added together, AG, will have to the second C the same ratio as the third and sixth, DH, has to the fourth F.

+

Let a first magnitude AB have to a second C the same ratio as a third DE has to a fourth F; and let also a fifth BG have to

the second C the same ratio as a sixth EH has to the fourth F; I say that the first and fifth added together, AG, will have to the second C the same ratio as the third and sixth, DH, has to the fourth F.

For since, as BG is to C, so is EH to F, inversely, as C is to BG, so is F to EH.

Since, then, as AB is to C, so is DE to F, and, as C is to BG, so is F to EH, therefore, ex aequali, as AB is to BG, so is DE to EH. [V. 22]

And, since the magnitudes are proportional separando, they will also be proportional componendo; [V. 18] therefore, as AG is to GB, so is DH to HE. @@ -5061,7 +3487,7 @@

Therefore etc. Q. E. D.

- + PROPOSITION 25.

If four magnitudes be proportional, the greatest and the least are greater than the remaining two.

Let the four magnitudes AB, CD, E, F be proportional so that, as AB is to CD, so is E to F, and let AB be the greatest of them and F the least; I say that AB, F are greater than CD, E.

@@ -5079,12 +3505,12 @@

Therefore etc. Q. E. D.

- - - +
+
+ BOOK VI. - + DEFINITIONS. 1 @@ -5108,7 +3534,7 @@ - + BOOK VI. PROPOSITIONS. PROPOSITION 1. @@ -5119,7 +3545,7 @@

Then, since CB, BG, GH are equal to one another, the triangles ABC, AGB, AHG are also equal to one another. [I. 38]

Therefore, whatever multiple the base HC is of the base BC, that multiple also is the triangle AHC of the triangle ABC.

-

For the same reason, whatever multiple the base LC is of the base CD, that multiple also is the triangle ALC of the triangle ACD; and, if the base HC is equal to the base CL, the triangle AHC is also equal to the triangle ACL, [I. 38] if the base HC is in excess of the base CL, the triangle AHC +

For the same reason, whatever multiple the base LC is of the base CD, that multiple also is the triangle ALC of the triangle ACD; and, if the base HC is equal to the base CL, the triangle AHC is also equal to the triangle ACL, [I. 38] if the base HC is in excess of the base CL, the triangle AHC is also in excess of the triangle ACL, and, if less, less.

Thus, there being four magnitudes, two bases BC, CD and two triangles ABC, ACD, equimultiples have been taken of the base BC and the triangle ABC, namely the base HC and the triangle AHC, and of the base CD and the triangle ADC other, chance, equimultiples, namely the base LC and the triangle ALC; and it has been proved that, if the base HC is in excess of the base CL, the triangle AHC is also in excess of the triangle ALC; if equal, equal; and, if less, less.

@@ -5153,7 +3579,7 @@

But equals have the same ratio to the same; [V. 7] therefore, as the triangle BDE is to the triangle ADE, so is the triangle CDE to the triangle ADE.

But, as the triangle BDE is to ADE, so is BD to DA; for, being under the same height, the perpendicular drawn from E to AB, they are to one another as their bases. [VI. 1]

For the same reason also, as the triangle CDE is to ADE, so is CE to EA.

-

Therefore also, as BD is to DA, so is CE to EA. [V. 11]

+

Therefore also, as BD is to DA, so is CE to EA. [V. 11]

Again, let the sides AB, AC of the triangle ABC be cut proportionally, so that, as BD is to DA, so is CE to EA; and let DE be joined.

I say that DE is parallel to BC.

For, with the same construction, since, as BD is to DA, so is CE to EA, but, as BD is to DA, so is the triangle BDE to the triangle ADE, and, as CE is to EA, so is the triangle CDE to the triangle ADE, [VI. 1] therefore also, as the triangle BDE is to the triangle ADE, so is the triangle CDE to the triangle ADE. [V. 11] @@ -5166,7 +3592,7 @@ PROPOSITION 3. -

If an angle of a triangle be bisected and the straight line cutting the angle cut the base also, the segments of the base will have the same ratio as the remaining sides of the triangle; and, if the segments of the base have the same ratio as the remaining sides of the triangle, the straight line joined from the vertex to the point of section will bisect the angle of the triangle.

+

If an angle of a triangle be bisected and the straight line cutting the angle cut the base also, the segments of the base will have the same ratio as the remaining sides of the triangle; and, if the segments of the base have the same ratio as the remaining sides of the triangle, the straight line joined from the vertex to the point of section will bisect the angle of the triangle.

Let ABC be a triangle, and let the angle BAC be bisected by the straight line AD; I say that, as BD is to CD, so is BA to AC.

For let CE be drawn through C parallel to DA, and let BA

be carried through and meet it at E.

@@ -5180,7 +3606,7 @@

And, since AD has been drawn parallel to EC, one of the sides of the triangle BCE, therefore, proportionally, as BD is to DC, so is BA to AE.

But AE is equal to AC; [VI. 2] therefore, as BD is to DC, so is BA to AC.

Again, let BA be to AC as BD to DC, and let AD be joined; I say that the angle BAC has been bisected by the straight line A.D.

-

For, with the same construction, since, as BD is to DC, so is BA to AC, and also, as BD is to DC, so is BA to AE : for AD has been drawn parallel to EC, one of the sides of the triangle BCE: [VI. 2] therefore also, as BA is to AC, so is BA to AE. [V. 11]

+

For, with the same construction, since, as BD is to DC, so is BA to AC, and also, as BD is to DC, so is BA to AE : for AD has been drawn parallel to EC, one of the sides of the triangle BCE: [VI. 2] therefore also, as BA is to AC, so is BA to AE. [V. 11]

Therefore AC is equal to AE, [V. 9] so that the angle AEC is also equal to the angle ACE. [I. 5]

But the angle AEC is equal to the exterior angle BAD, [I. 29] and the angle ACE is equal to the alternate angle CAD; [id.] therefore the angle BAD is also equal to the angle CAD.

@@ -5189,7 +3615,7 @@ PROPOSITION 4. -

In equiangular triangles the sides about the equal angles are proportional, and those are corresponding sides which subtend the equal angles.

+

In equiangular triangles the sides about the equal angles are proportional, and those are corresponding sides which subtend the equal angles.

Let ABC, DCE be equiangular triangles having the angle ABC equal to the angle DCE, the angle BAC to the angle CDE, and further the angle ACB to the angle CED; I say that in the triangles ABC, DCE the sides about the equal angles are proportional, and those are corresponding sides which subtend the equal angles.

For let BC be placed in a straight line with CE.

@@ -5203,7 +3629,7 @@

And, since AC has been drawn parallel to FE, one side of the triangle FBE, therefore, as BA is to AF, so is BC to CE. [VI. 2]

But AF is equal to CD; therefore, as BA is to CD, so is BC to CE, and alternately, as AB is to BC, so is DC to CE. [V. 16]

Again, since CD is parallel to BF, therefore, as BC is to CE, so is FD to DE. [VI. 2]

-

But FD is equal to AC; therefore, as BC is to CE, so is AC to DE, and alternately, as BC is to CA, so is CE to ED. [V. 16]

+

But FD is equal to AC; therefore, as BC is to CE, so is AC to DE, and alternately, as BC is to CA, so is CE to ED. [V. 16]

Since then it was proved that, as AB is to BC, so is DC to CE, and, as BC is to CA, so is CE to ED; therefore, ex aequali, as BA is to AC, so is CD to DE. [V. 22]

Therefore etc. Q. E. D.

@@ -5211,7 +3637,7 @@ PROPOSITION 5.

If two triangles have their sides proportional, the triangles will be equiangular and will have those angles equal which the corresponding sides subtend.

Let ABC, DEF be two triangles having their sides proportional, so that, as AB is to BC, so is DE to EF, as BC is to CA, so is EF to FD, and further, as BA is to AC, so is ED to DF; I say that the triangle ABC is equiangular with the triangle DEF, and they will have those angles equal which the corresponding sides subtend, namely the angle ABC to the angle DEF, the angle BCA to the angle EFD, and further the angle BAC to the angle EDF.

-

For on the straight line EF, and at the points E, F on it, let there be constructed the angle FEG equal to the angle ABC, and the angle EFG equal to the angle ACB; [I. 23] therefore the remaining angle at A is equal to the remaining angle at G. [I. 32] +

For on the straight line EF, and at the points E, F on it, let there be constructed the angle FEG equal to the angle ABC, and the angle EFG equal to the angle ACB; [I. 23] therefore the remaining angle at A is equal to the remaining angle at G. [I. 32]

Therefore the triangle ABC is equiangular with the triangle GEF.

@@ -5226,7 +3652,7 @@

Since then DE is equal to EG, and EF is common, the two sides DE, EF are equal to the two sides GE, EF; and the base DF is equal to the base FG; therefore the angle DEF is equal to the angle GEF, [I. 8] and the triangle DEF is equal to the triangle GEF, and the remaining angles are equal to the remaining angles, namely those which the equal sides subtend. [I. 4]

Therefore the angle DFE is also equal to the angle GFE, and the angle EDF to the angle EGF.

-

And, since the angle FED is equal to the angle GEF, while the angle GEF is equal to the angle ABC, therefore the angle ABC is also equal to the angle DEF.

+

And, since the angle FED is equal to the angle GEF, while the angle GEF is equal to the angle ABC, therefore the angle ABC is also equal to the angle DEF.

For the same reason the angle ACB is also equal to the angle DFE, and further, the angle at A to the angle at D; therefore the triangle ABC is equiangular with the triangle DEF.

Therefore etc. Q. E. D.

@@ -5236,7 +3662,7 @@

If two triangles have one angle equal to one angle and the sides about the equal angles proportional, the triangles will be equiangular and will have those angles equal which the corresponding sides subtend.

Let ABC, DEF be two triangles having one angle BAC equal to one angle EDF and the sides about the equal angles proportional, so that, as BA is to AC, so is ED to DF; I say that the triangle ABC is equiangular with the triangle DEF, and will have the angle ABC equal to the angle DEF, and the angle ACB to the angle DFE.

For on the straight line DF, and at the points D, F on it, let there be constructed the angle FDG equal to either of the angles BAC, EDF, and the angle DFG equal to the angle ACB; [I. 23] therefore the remaining angle at B is equal to the remaining angle at G. [I. 32] -

+

Therefore the triangle ABC is equiangular with the triangle DGF.

Therefore, proportionally, as BA is to AC, so is GD to DF. [VI. 4]

But, by hypothesis, as BA is to AC, so also is ED to DF; therefore also, as ED is to DF, so is GD to DF. [V. 11]

@@ -5250,7 +3676,7 @@

Therefore etc. Q. E. D.

- + PROPOSITION 7.

If two triangles have one angle equal to one angle, the sides about other angles proportional, and the remaining angles either both less or both not less than a right angle, the triangles will be equiangular and will have those angles equal, the sides about which are proportional.

Let ABC, DEF be two triangles having one angle equal to one angle, the angle BAC to the angle EDF, the sides about other angles ABC, DEF proportional, so that, as AB is to BC, so is DE to EF, and, first, each of the remaining angles at C, F less than a right angle;

I say that the triangle ABC is equiangular with the triangle DEF, the angle ABC will be equal to the angle DEF, and the remaining angle, namely the angle at C, equal to the remaining angle, the angle at F.

@@ -5259,7 +3685,7 @@

Then, since the angle A is equal to D, and the angle ABG to the angle DEF, therefore the remaining angle AGB is equal to the remaining angle DFE. [I. 32]

Therefore the triangle ABG is equiangular with the triangle DEF.

Therefore, as AB is to BG, so is DE to EF [VI. 4]

-

But, as DE is to EF, so by hypothesis is AB to BC; therefore AB has the same ratio to each of the straight lines BC, BG; [V. 11] therefore BC is equal to BG, [V. 9] so that the angle at C is also equal to the angle BGC. [I. 5]

+

But, as DE is to EF, so by hypothesis is AB to BC; therefore AB has the same ratio to each of the straight lines BC, BG; [V. 11] therefore BC is equal to BG, [V. 9] so that the angle at C is also equal to the angle BGC. [I. 5]

But, by hypothesis, the angle at C is less than a right angle; therefore the angle BGC is also less than a right angle; so that the angle AGB adjacent to it is greater than a right angle. [I. 13]

And it was proved equal to the angle at F; therefore the angle at F is also greater than a right angle.

But it is by hypothesis less than a right angle : which is absurd.

@@ -5272,14 +3698,14 @@

But the angle at C is not less than a right angle; therefore neither is the angle BGC less than a right angle.

Thus in the triangle BGC the two angles are not less than two right angles: which is impossible. [I. 17]

Therefore, once more, the angle ABC is not unequal to the angle DEF; therefore it is equal to it.

-

But the angle at A is also equal to the angle at D; therefore the remaining angle at C is equal to the remaining angle at F. [I. 32]

+

But the angle at A is also equal to the angle at D; therefore the remaining angle at C is equal to the remaining angle at F. [I. 32]

Therefore the triangle ABC is equiangular with the triangle DEF.

Therefore etc. Q. E. D.

PROPOSITION 8.

If in a right-angled triangle a perpendicular be drawn from the right angle to the base, the triangles adjoining the perpendicular are similar both to the whole and to one another.

-

Let ABC be a right-angled triangle having the angle BAC right, and let AD be drawn from A perpendicular to BC; I say that each of the triangles ABD, ADC is similar to the whole ABC and, further, they are similar to one another.

+

Let ABC be a right-angled triangle having the angle BAC right, and let AD be drawn from A perpendicular to BC; I say that each of the triangles ABD, ADC is similar to the whole ABC and, further, they are similar to one another.

For, since the angle BAC is equal to the angle ADB, for each is right, and the angle at B is common to the two triangles ABC and ABD, therefore the remaining angle ACB is equal to the remaining angle BAD; [I. 32]

therefore the triangle ABC is equiangular with the triangle ABD.

Therefore, as BC which subtends the right angle in the triangle ABC is to BA which subtends the right angle in the triangle ABD, so is AB itself which subtends the angle at C in the triangle ABC to BD which subtends the equal angle BAD in the triangle ABD, and so also is AC to AD which subtends the angle at B common to the two triangles. [VI. 4]

Therefore the triangle ABC is both equiangular to the triangle ABD and has the sides about the equal angles proportional.

@@ -5287,7 +3713,7 @@

Similarly we can prove that the triangle ABC is also similar to the triangle ADC; therefore each of the triangles ABD, ADC is similar to the whole ABC.

I say next that the triangles ABD, ADC are also similar to one another.

For, since the right angle BDA is equal to the right angle ADC, and moreover the angle BAD was also proved equal to the angle at C, therefore the remaining angle at B is also equal to the remaining angle DAC; [I. 32] therefore the triangle ABD is equiangular with the triangle ADC.

-

Therefore, as BD which subtends the angle BAD in the triangle ABD is to DA which subtends the angle at C in the triangle ADC equal to the angle BAD, so is AD itself which subtends the angle at B in the triangle ABD to DC which subtends the angle DAC in the triangle ADC equal to the angle at B, and so also is BA to AC, these sides subtending the right angles; [VI. 4] therefore the triangle ABD is similar to the triangle ADC. [VI. Def. 1]

+

Therefore, as BD which subtends the angle BAD in the triangle ABD is to DA which subtends the angle at C in the triangle ADC equal to the angle BAD, so is AD itself which subtends the angle at B in the triangle ABD to DC which subtends the angle DAC in the triangle ADC equal to the angle at B, and so also is BA to AC, these sides subtending the right angles; [VI. 4] therefore the triangle ABD is similar to the triangle ADC. [VI. Def. 1]

Therefore etc.

PORISM. @@ -5297,7 +3723,7 @@ PROPOSITION 9.

From a given straight line to cut off a prescribed part.

-

Let AB be the given straight line; thus it is required to cut off from AB a prescribed part.

+

Let AB be the given straight line; thus it is required to cut off from AB a prescribed part.

Let the third part be that prescribed.

Let a straight line AC be drawn through from A containing with AB any angle; let a point D be taken at random on AC, and let DE, EC be made equal to AD. [I. 3]

@@ -5318,7 +3744,7 @@

Let AB be the given uncut straight line, and AC the straight line cut at the points D, E; and let them be so placed as to contain any angle; let CB be joined, and through D, E let DF, EG be drawn parallel to BC, and through D let DHK be drawn parallel to AB. [I. 31]

Therefore each of the figures FH, HB is a parallelogram; therefore DH is equal to FG and HK to GB. [I. 34]

-

Now, since the straight line HE has been drawn parallel to KC, one of the sides of the triangle DKC, therefore, proportionally, as CE is to ED, so is KH to HD. [VI. 2]

+

Now, since the straight line HE has been drawn parallel to KC, one of the sides of the triangle DKC, therefore, proportionally, as CE is to ED, so is KH to HD. [VI. 2]

But KH is equal to BG, and HD to GF; therefore, as CE is to ED, so is BG to GF.

Again, since FD has been drawn parallel to GE, one of the sides of the triangle AGE, therefore, proportionally, as ED is to DA, so is GF to FA. [VI. 2]

But it was also proved that, as CE is to ED, so is BG to GF; therefore, as CE is to ED, so is BG to GF, and, as ED is to DA, so is GF to FA. @@ -5340,7 +3766,7 @@

- + PROPOSITION 12.

To three given straight lines to find a fourth proportional.

Let A, B, C be the three given straight lines; thus it is required to find a fourth proportional to A, B, C.

@@ -5361,7 +3787,7 @@ PROPOSITION 14. -

In equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional; and equiangular parallelograms in which the sides about the equal angles are reciprocally proportional are equal.

+

In equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional; and equiangular parallelograms in which the sides about the equal angles are reciprocally proportional are equal.

Let AB, BC be equal and equiangular parallelograms having the angles at B equal, and let DB, BE be placed in a straight line; therefore FB, BG are also in a straight line. [I. 14]

I say that, in AB, BC, the sides about the equal angles are reciprocally proportional, that is to say, that, as DB is to BE, so is GB to BF.

@@ -5378,7 +3804,7 @@ PROPOSITION 15.

In equal triangles which have one angle equal to one angle the sides about the equal angles are reciprocally proportional; and those triangles which have one angle equal to one angle, and in which the sides about the equal angles are reciprocally proportional, are equal.

Let ABC, ADE be equal triangles having one angle equal to one angle, namely the angle BAC to the angle DAE; I say that in the triangles ABC, ADE the sides about the equal angles are reciprocally proportional, that is to say, that, as CA is to AD, so is EA to AB. -

+

For let them be placed so that CA is in a straight line with AD; therefore EA is also in a straight line with AB. [I. 14]

Let BD be joined.

Since then the triangle ABC is equal to the triangle ADE, and BAD is another area, therefore, as the triangle CAB is to the triangle BAD, so is the triangle EAD to the triangle BAD. [V. 7]

@@ -5400,7 +3826,7 @@

Let AG, CH be drawn from the points A, C at right angles to the straight lines AB, CD, and let AG be made equal to F, and CH equal to E.

Let the parallelograms BG, DH be completed.

Then since, as AB is to CD, so is E to F, while E is equal to CH, and F to AG, therefore, as AB is to CD, so is CH to AG.

-

Therefore in the parallelograms BG, DH the sides about the equal angles are reciprocally proportional.

+

Therefore in the parallelograms BG, DH the sides about the equal angles are reciprocally proportional.

But those equiangular parallelograms in which the sides about the equal angles are reciprocally proportional are equal; [VI. 14] therefore the parallelogram BG is equal to the parallelogram DH.

And BG is the rectangle AB, F, for AG is equal to F; and DH is the rectangle CD, E, for E is equal to CH; therefore the rectangle contained by AB, F is equal to the rectangle contained by CD, E.

Next, let the rectangle contained by AB, F be equal to the rectangle contained by CD, E; I say that the four straight lines will be proportional, so that, as AB is to CD, so is E to F.

@@ -5422,7 +3848,7 @@

But, if four straight lines be proportional, the rectangle contained by the extremes is equal to the rectangle contained by the means. [VI. 16]

Therefore the rectangle A, C is equal to the rectangle B, D.

But the rectangle B, D is the square on B, for B is equal to D; therefore the rectangle contained by A, C is equal to the square on B.

-

Next, let the rectangle A, C be equal to the square on B; I say that, as A is to B, so is B to C.

+

Next, let the rectangle A, C be equal to the square on B; I say that, as A is to B, so is B to C.

For, with the same construction, since the rectangle A, C is equal to the square on B, while the square on B is the rectangle B, D, for B is equal to D, therefore the rectangle A, C is equal to the rectangle B, D.

But, if the rectangle contained by the extremes be equal to that contained by the means, the four straight lines are proportional. [VI. 16]

Therefore, as A is to B, so is D to C.

@@ -5438,7 +3864,7 @@

Let DF be joined, and on the straight line AB, and at the points A, B on it, let the angle GAB be constructed equal to the angle at C, and the angle ABG equal to the angle CDF. [I. 23]

Therefore the remaining angle CFD is equal to the angle AGB; [I. 32] therefore the triangle FCD is equiangular with the triangle GAB.

-

Therefore, proportionally, as FD is to GB, so is FC to GA, and CD to AB.

+

Therefore, proportionally, as FD is to GB, so is FC to GA, and CD to AB.

Again, on the straight line BG, and at the points B, G on it, let the angle BGH be constructed equal to the angle DFE, and the angle GBH equal to the angle FDE. [I. 23]

Therefore the remaining angle at E is equal to the remaining angle at H; [I. 32] therefore the triangle FDE is equiangular with the triangle GBH; therefore, proportionally, as FD is to GB, so is FE to GH, and ED to HB. [VI. 4]

@@ -5459,7 +3885,7 @@

Let ABC, DEF be similar triangles having the angle at B equal to the angle at E, and such that, as AB is to BC, so is DE to EF, so that BC corresponds to EF; [V. Def. 11] I say that the triangle ABC has to the triangle DEF a ratio duplicate of that which BC has to EF.

For let a third proportional BG be taken to BC, EF, so that, as BC is to EF, so is EF to BG; [VI. 11] and let AG be joined.

-

Since then, as AB is to BC, so is DE to EF, therefore, alternately, as AB is to DE, so is BC to EF. [V. 16]

+

Since then, as AB is to BC, so is DE to EF, therefore, alternately, as AB is to DE, so is BC to EF. [V. 16]

But, as BC is to EF, so is EF to BG; therefore also, as AB is to DE, so is EF to BG. [V. 11]

Therefore in the triangles ABG, DEF the sides about the equal angles are reciprocally proportional.

But those triangles which have one angle equal to one angle, and in which the sides about the equal angles are reciprocally proportional, are equal; [VI. 15] therefore the triangle ABG is equal to the triangle DEF.

@@ -5486,7 +3912,7 @@

Now, since the polygon ABCDE is similar to the polygon FGHKL, the angle BAE is equal to the angle GFL; and, as BA is to AE, so is GF to FL. [VI. Def. 1]

-

Since then ABE, FGL are two triangles having one angle equal to one angle and the sides about the equal angles proportional, therefore the triangle ABE is equiangular with the triangle FGL; [VI. 6] so that it is also similar; [VI. 4 and Def. 1] therefore the angle ABE is equal to the angle FGL.

+

Since then ABE, FGL are two triangles having one angle equal to one angle and the sides about the equal angles proportional, therefore the triangle ABE is equiangular with the triangle FGL; [VI. 6] so that it is also similar; [VI. 4 and Def. 1] therefore the angle ABE is equal to the angle FGL.

But the whole angle ABC is also equal to the whole angle FGH because of the similarity of the polygons; therefore the remaining angle EBC is equal to the angle LGH.

And, since, because of the similarity of the triangles ABE, FGL, as EB is to BA, so is LG to GF, and moreover also, because of the similarity of the polygons, as AB is to BC, so is FG to GH, therefore, ex aequali, as EB is to BC, so is LG to GH; [V. 22] that is, the sides about the equal angles EBC, LGH are proportional; therefore the triangle EBC is equiangular with the triangle LGH, [VI. 6] so that the triangle EBC is also similar to the triangle LGH. [VI. 4 and Def. 1]

@@ -5497,7 +3923,7 @@

For let AC, FH be joined.

Then since, because of the similarity of the polygons, the angle ABC is equal to the angle FGH, and, as AB is to BC, so is FG to GH, the triangle ABC is equiangular with the triangle FGH; [VI. 6] therefore the angle BAC is equal to the angle GFH, and the angle BCA to the angle GHF.

-

And, since the angle BAM is equal to the angle GFN, and the angle ABM is also equal to the angle FGN, therefore the remaining angle AMB is also equal to the remaining angle FNG; [I. 32] therefore the triangle ABM is equiangular with the triangle FGN.

+

And, since the angle BAM is equal to the angle GFN, and the angle ABM is also equal to the angle FGN, therefore the remaining angle AMB is also equal to the remaining angle FNG; [I. 32] therefore the triangle ABM is equiangular with the triangle FGN.

Similarly we can prove that the triangle BMC is also equiangular with the triangle GNH.

Therefore, proportionally, as AM is to MB, so is FN to NG, and, as BM is to MC, so is GN to NH; so that, in addition, ex aequali, as AM is to MC, so is FN to NH.

@@ -5507,7 +3933,7 @@

For the same reason also, as FN is to NH, so is the triangle FGL to the triangle GLH.

And, as AM is to MC, so is FN to NH; therefore also, as the triangle ABE is to the triangle BEC, so is the triangle FGL to the triangle GLH; and, alternately, as the triangle ABE is to the triangle FGL, so is the triangle BEC to the triangle GLH.

Similarly we can prove, if BD, GK be joined, that, as the triangle BEC is to the triangle LGH, so also is the triangle ECD to the triangle LHK.

-

And since, as the triangle ABE is to the triangle FGL, so is EBC to LGH, and further ECD to LHK, therefore also, as one of the antecedents is to one of the consequents so are all the antecedents to all the consequents; [V. 12 therefore, as the triangle ABE is to the triangle FGL, so is the polygon ABCDE to the polygon FGHKL.

+

And since, as the triangle ABE is to the triangle FGL, so is EBC to LGH, and further ECD to LHK, therefore also, as one of the antecedents is to one of the consequents so are all the antecedents to all the consequents; [V. 12 therefore, as the triangle ABE is to the triangle FGL, so is the polygon ABCDE to the polygon FGHKL.

But the triangle ABE has to the triangle FGL a ratio duplicate of that which the corresponding side AB has to the corresponding side FG; for similar triangles are in the duplicate ratio of the corresponding sides. [VI. 19]

Therefore the polygon ABCDE also has to the polygon FGHKL a ratio duplicate of that which the corresponding side AB has to the corresponding side FG.

Therefore etc.

@@ -5531,7 +3957,7 @@ PROPOSITION 21.

Figures which are similar to the same rectilineal figure are also similar to one another.

-

For let each of the rectilineal figures A, B be similar to C; I say that A is also similar to B.

+

For let each of the rectilineal figures A, B be similar to C; I say that A is also similar to B.

For, since A is similar to C, it is equiangular with it and has the sides about the equal angles proportional. [VI. Def. 1]

Again, since B is similar to C, it is equiangular with it and has the sides about the equal angles proportional.

@@ -5541,7 +3967,7 @@ PROPOSITION 22.

If four straight lines be proportional, the rectilineal figures similar and similarly described upon them will also be proportional; and, if the rectilineal figures similar and similarly described upon them be proportional, the straight lines will themselves also be proportional.

Let the four straight lines AB, CD, EF, GH be proportional, so that, as AB is to CD, so is EF to GH, and let there be described on AB, CD the similar and similarly situated rectilineal figures KAB, LCD, and on EF, GH the similar and similarly situated rectilineal figures MF, NH; I say that, as KAB is to LCD, so is MF to NH.

-

For let there be taken a third proportional O to AB, CD, and a third proportional P to EF, GH. [VI. 11]

+

For let there be taken a third proportional O to AB, CD, and a third proportional P to EF, GH. [VI. 11]

Then since, as AB is to CD, so is EF to GH, and, as CD is to O, so is GH to P, therefore, ex aequali, as AB is to O, so is EF to P. [V. 22]

But, as AB is to O, so is KAB to LCD, [VI. 19, Por.] and, as EF is to P, so is MF to NH; therefore also, as KAB is to LCD, so is MF to NH. [V. 11]

@@ -5553,7 +3979,7 @@

Therefore MF has the same ratio to each of the figures NH, SR; therefore NH is equal to SR. [V. 9]

But it is also similar and similarly situated to it; therefore GH is equal to QR. -

+

And, since, as AB is to CD, so is EF to QR, while QR is equal to GH, therefore, as AB is to CD, so is EF to GH.

Therefore etc. Q. E. D.

@@ -5564,7 +3990,7 @@

For let them be placed so that BC is in a straight line with CG; therefore DC is also in a straight line with CE.

-

Let the parallelogram DG be completed; let a straight line K be set out, and let it be contrived that, as BC is to CG, so is K to L, and, as DC is to CE, so is L to M. [VI. 12]

+

Let the parallelogram DG be completed; let a straight line K be set out, and let it be contrived that, as BC is to CG, so is K to L, and, as DC is to CE, so is L to M. [VI. 12]

Then the ratios of K to L and of L to M are the same as the ratios of the sides, namely of BC to CG and of DC to CE.

But the ratio of K to M is compounded of the ratio of K to L and of that of L to M; so that K has also to M the ratio compounded of the ratios of the sides.

Now since, as BC is to CG, so is the parallelogram AC to the parallelogram CH, [VI. 1] while, as BC is to CG, so is K to L, therefore also, as K is to L, so is AC to CH. [V. 11]

@@ -5595,7 +4021,7 @@

But it was proved that, as CF is to FA, so also is BE to EA; therefore also, as BE is to EA, so is DG to GA, and therefore, componendo, as BA is to AE, so is DA to AG, [V. 18] and, alternately, as BA is to AD, so is EA to AG. [V. 16]

Therefore in the parallelograms ABCD, EG, the sides about the common angle BAD are proportional.

-

And, since GF is parallel to DC, the angle AFG is equal to the angle DCA; and the angle DAC is common to the two triangles ADC, AGF; therefore the triangle ADC is equiangular with the triangle AGF. +

And, since GF is parallel to DC, the angle AFG is equal to the angle DCA; and the angle DAC is common to the two triangles ADC, AGF; therefore the triangle ADC is equiangular with the triangle AGF.

For the same reason the triangle ACB is also equiangular with the triangle AFE, and the whole parallelogram ABCD is equiangular with the parallelogram EG.

Therefore, proportionally, as AD is to DC, so is AG to GF, as DC is to CA, so is GF to FA, as AC is to CB, so is AF to FE, and further, as CB is to BA, so is FE to EA.

@@ -5610,7 +4036,7 @@

To construct one and the same figure similar to a given rectilineal figure and equal to another given rectilineal figure.

Let ABC be the given rectilineal figure to which the figure to be constructed must be similar, and D that to which it must be equal; thus it is required to construct one and the same figure similar to ABC and equal to D.

-

Let there be applied to BC the parallelogram BE equal to the triangle ABC [I. 44], and to CE the parallelogram CM equal to D in the angle FCE which is equal to the angle CBL. [I. 45]

+

Let there be applied to BC the parallelogram BE equal to the triangle ABC [I. 44], and to CE the parallelogram CM equal to D in the angle FCE which is equal to the angle CBL. [I. 45]

Therefore BC is in a straight line with CF, and LE with EM.

Now let GH be taken a mean proportional to BC, CF [VI. 13], and on GH let KGH be described similar and similarly situated to ABC. [VI. 18]

Then, since, as BC is to GH, so is GH to CF, and, if three straight lines be proportional, as the first is to the third, so is the figure on the first to the similar and similarly situated figure described on the second, [VI. 19, Por.] therefore, as BC is to CF, so is the triangle ABC to the triangle KGH.

@@ -5633,13 +4059,13 @@

For suppose it is not, but, if possible, let AHC be the diameter < of ABCD >, let GF be produced and carried through to H, and let HK be drawn through H parallel to either of the straight lines AD, BC. [I. 31]

Since, then, ABCD is about the same diameter with KG, therefore, as DA is to AB, so is GA to AK. [VI. 24]

But also, because of the similarity of ABCD, EG, as DA is to AB, so is GA to AE; therefore also, as GA is to AK, so is GA to AE. [V. 11]

-

Therefore GA has the same ratio to each of the straight lines AK, AE.

+

Therefore GA has the same ratio to each of the straight lines AK, AE.

Therefore AE is equal to AK [V. 9], the less to the greater : which is impossible.

Therefore ABCD cannot but be about the same diameter with AF; therefore the parallelogram ABCD is about the same diameter with the parallelogram AF.

Therefore etc. Q. E. D.

- + PROPOSITION 27.

Of all the parallelograms applied to the same straight line and deficient by parallelogrammic figures similar and similarly situated to that described on the half of the straight line, that parallelogram is greatest which is applied to the half of the straight line and is similar to the defect.

Let AB be a straight line and let it be bisected at C; let there be applied to the straight line AB the parallelogram AD deficient by the parallelogrammic figure DB described on the half of AB, that is, CB;

I say that, of all the parallelograms applied to AB and deficient by parallelogrammic figures similar and similarly situated to DB, AD is greatest.

@@ -5656,7 +4082,7 @@ PROPOSITION 28.

To a given straight line to apply a parallelogram equal to a given rectilineal figure and deficient by a parallelogrammic figure similar to a given one : thus the given rectilineal figure must not be greater than the parallelogram described on the half of the straight line and similar to the defect.

-

Let AB be the given straight line, C the given rectilineal figure to which the figure to be applied to AB is required to be equal, not being greater than the parallelogram described on the half of AB and similar to the defect, and D the parallelogram to which the defect is required to be similar; thus it is required to apply to the given straight line AB a parallelogram equal to the given rectilineal figure C and deficient by a parallelogrammic figure which is similar to D.

+

Let AB be the given straight line, C the given rectilineal figure to which the figure to be applied to AB is required to be equal, not being greater than the parallelogram described on the half of AB and similar to the defect, and D the parallelogram to which the defect is required to be similar; thus it is required to apply to the given straight line AB a parallelogram equal to the given rectilineal figure C and deficient by a parallelogrammic figure which is similar to D.

Let AB be bisected at the point E, and on EB let EBFG be described similar and similarly situated to D; [VI. 18] let the parallelogram AG be completed.

If then AG is equal to C, that which was enjoined will have been done; for there has been applied to the given straight line AB the parallelogram AG equal to the given rectilineal figure C and deficient by a parallelogrammic figure GB which is similar to D.

@@ -5672,7 +4098,7 @@

Let GO be made equal to KL, and GP equal to LM; and let the parallelogram OGPQ be completed; therefore it is equal and similar to KM.

-

Therefore GQ is also similar to GB; [VI. 21] therefore GQ is about the same diameter with GB. [VI. 26]

+

Therefore GQ is also similar to GB; [VI. 21] therefore GQ is about the same diameter with GB. [VI. 26]

Let GQB be their diameter, and let the figure be described.

Then, since BG is equal to C, KM, and in them GQ is equal to KM, therefore the remainder, the gnomon UWV, is equal to the remainder C.

And, since PR is equal to OS, let QB be added to each; therefore the whole PB is equal to the whole OB.

@@ -5691,7 +4117,7 @@

Let AB be bisected at E; let there be described on EB the parallelogram BF similar and similarly situated to D; and let GH be constructed at once equal to the sum of BF, C and similar and similarly situated to D. [VI. 25]

Let KH correspond to FL and KG to FE.

-

Now, since GH is greater than FB, therefore KH is also greater than FL, and KG than FE.

+

Now, since GH is greater than FB, therefore KH is also greater than FL, and KG than FE.

Let FL, FE be produced, let FLM be equal to KH, and FEN to KG, and let MN be completed; therefore MN is both equal and similar to GH.

But GH is similar to EL; therefore MN is also similar to EL; [VI. 21] therefore EL is about the same diameter with MN. [VI. 26]

@@ -5717,7 +4143,7 @@

And, since BC is equal to CD, let CE be subtracted from each; therefore the remainder BF is equal to the remainder AD. -

+

But it is also equiangular with it; therefore in BF, AD the sides about the equal angles are reciprocally proportional; [VI. 14] therefore, as FE is to ED, so is AE to EB.

But FE is equal to AB, and ED to AE.

@@ -5731,7 +4157,7 @@

In right-angled triangles the figure on the side subtending the right angle is equal to the similar and similarly described figures on the sides containing the right angle.

Let ABC be a right-angled triangle having the angle BAC right; I say that the figure on BC is equal to the similar and similarly described figures on BA, AC.

Let AD be drawn perpendicular.

-

Then since, in the right-angled triangle ABC, AD has been drawn from the right angle at A perpendicular to the base BC, the triangles ABD, ADC adjoining the perpendicular are similar both to the whole ABC and to one another. [VI. 8]

+

Then since, in the right-angled triangle ABC, AD has been drawn from the right angle at A perpendicular to the base BC, the triangles ABD, ADC adjoining the perpendicular are similar both to the whole ABC and to one another. [VI. 8]

And, since ABC is similar to ABD, therefore, as CB is to BA, so is AB to BD. [VI. Def. 1]

And, since three straight lines are proportional, as the first is to the third, so is the figure on the first to the similar and similarly described figure on the second. [VI. 19, Por.]

@@ -5743,7 +4169,7 @@ PROPOSITION 32.

If two triangles having two sides proportional to two sides be placed together at one angle so that their corresponding sides are also parallel, the remaining sides of the triangles will be in a straight line.

-

Let ABC, DCE be two triangles having the two sides BA, AC proportional to the two sides DC, DE, so that, as AB is to AC, so is DC to DE, and AB parallel to DC, and AC to DE; I say that BC is in a straight line with CE.

+

Let ABC, DCE be two triangles having the two sides BA, AC proportional to the two sides DC, DE, so that, as AB is to AC, so is DC to DE, and AB parallel to DC, and AC to DE; I say that BC is in a straight line with CE.

For, since AB is parallel to DC, and the straight line AC has fallen upon them, the alternate angles BAC, ACD are equal to one another. [I. 29]

For the same reason the angle CDE is also equal to the angle ACD; so that the angle BAC is equal to the angle CDE.

@@ -5764,7 +4190,7 @@

Let ABC, DEF be equal circles, and let the angles BGC, EHF be angles at their centres G, H, and the angles BAC, EDF angles at the circumferences; I say that, as the circumference BC is to the circumference EF, so is the angle BGC to the angle EHF, and the angle BAC to the angle EDF.

For let any number of consecutive circumferences CK, KL be made equal to the circumference BC, and any number of consecutive circumferences FM, MN equal to the circumference EF; and let GK, GL, HM, HN be joined.

-

Then, since the circumferences BC, CK, KL are equal to one another, the angles BGC, CGK, KGL are also equal to one another; [III. 27] therefore, whatever multiple the circumference BL is of BC, that multiple also is the angle BGL of the angle BGC.

+

Then, since the circumferences BC, CK, KL are equal to one another, the angles BGC, CGK, KGL are also equal to one another; [III. 27] therefore, whatever multiple the circumference BL is of BC, that multiple also is the angle BGL of the angle BGC.

For the same reason also, whatever multiple the circumference NE is of EF, that multiple also is the angle NHE of the angle EHF.

If then the circumference BL is equal to the circumference EN, the angle BGL is also equal to the angle EHN; [III. 27] if the circumference BL is greater than the circumference EN, the angle BGL is also greater than the angle EHN; and, if less, less.

There being then four magnitudes, two circumferences BC, EF, and two angles BGC, EHF, there have been taken, of the circumference BC and the angle BGC equimultiples, namely the circumference BL and the angle BGL, and of the circumference EF and the angle EHF equimultiples, namely the circumference EN and the angle EHN.

@@ -5775,9 +4201,9 @@

Therefore etc. Q. E. D.

- - - +
+ +
BOOK VII. DEFINITIONS. @@ -5823,7 +4249,7 @@ 11 - +

A prime number is that which is measured by an unit alone.

 @@ -5872,7 +4298,7 @@

A perfect number is that which is equal to its own parts.

 - + BOOK VII. PROPOSITIONS. @@ -5886,13 +4312,13 @@

Let a number measure them, and let it be E; let CD, measuring BF, leave FA less than itself, let AF, measuring DG, leave GC less than itself, and let GC, measuring FH, leave an unit HA.

Since, then, E measures CD, and CD measures BF, therefore E also measures BF.

But it also measures the whole BA; therefore it will also measure the remainder AF.

-

But AF measures DG; therefore E also measures DG.

+

But AF measures DG; therefore E also measures DG.

But it also measures the whole DC therefore it will also measure the remainder CG.

But CG measures FH; therefore E also measures FH.

But it also measures the whole FA; therefore it will also measure the remainder, the unit AH, though it is a number: which is impossible.

Therefore no number will measure the numbers AB, CD; therefore AB, CD are prime to one another. [VII. Def. 12] Q. E. D.

 - + PROPOSITION 2.

@@ -5912,7 +4338,7 @@

But CD measures BE; therefore CF also measures BE.

But it also measures EA; therefore it will also measure the whole BA.

But it also measures CD; therefore CF measures AB, CD.

-

Therefore CF is a common measure of AB, CD.

+

Therefore CF is a common measure of AB, CD.

I say next that it is also the greatest.

For, if CF is not the greatest common measure of AB, CD, some number which is greater than CF will measure the numbers AB, CD.

Let such a number measure them, and let it be G.

@@ -5938,7 +4364,7 @@

For let the greatest common measure, D, of the two numbers A, B be taken; [VII. 2] then D either measures, or does not measure, C.

First, let it measure it.

But it measures A, B also; therefore D measures A, B, C; therefore D is a common measure of A, B, C.

-

I say that it is also the greatest.

+

I say that it is also the greatest.

For, if D is not the greatest common measure of A, B, C, some number which is greater than D will measure the numbers A, B, C.

Let such a number measure them, and let it be E.

Since then E measures A, B, C, it will also measure A, B; therefore it will also measure the greatest common measure of A, B. [VII. 2, Por.]

@@ -5954,7 +4380,7 @@

But it measures C also; therefore E measures A, B, C; therefore E is a common measure of A, B, C.

I say next that it is also the greatest.

For, if E is not the greatest common measure of A, B, C, some number which is greater than E will measure the numbers A, B, C.

-

Let such a number measure them, and let it be F.

+

Let such a number measure them, and let it be F.

Now, since F measures A, B, C, it also measures A, B; therefore it will also measure the greatest common measure of A, B. [VII. 2, Por.]

But the greatest common measure of A, B is D; therefore F measures D.

And it measures C also; therefore F measures D, C; therefore it will also measure the greatest common measure of D, C. [VII. 2, Por.]

@@ -5978,7 +4404,7 @@

But D is equal to each of the numbers BE, EF, FC; therefore each of the numbers BE, EF, FC is also a part of A; so that BC is parts of A.

Therefore etc. Q. E. D.

 - + PROPOSITION 5.

@@ -5994,7 +4420,7 @@

Therefore, whatever multiple BC is of A, the same multiple also is the sum of BC, EF of the sum of A, D.

Therefore, whatever part A is of BC, the same part also is the sum of A, D of the sum of BC, EF. Q. E. D.

 - + PROPOSITION 6.

@@ -6020,7 +4446,7 @@

But, whatever part AE is of CF, the same part also, by hypothesis, is AB of CD; therefore, whatever part AB is of GF, the same part is it of CD also; therefore GF is equal to CD.

Let CF be subtracted from each; therefore the remainder GC is equal to the remainder FD.

Now since, whatever part AE is of CF, the same part also is EB of GC, while GC is equal to FD, therefore, whatever part AE is of CF, the same part also is EB of FD.

-

But, whatever part AE is of CF, the same part also is AB of CD; therefore also the remainder EB is the same part of the remainder FD that the whole AB is of the whole CD. Q. E. D.

+

But, whatever part AE is of CF, the same part also is AB of CD; therefore also the remainder EB is the same part of the remainder FD that the whole AB is of the whole CD. Q. E. D.

 PROPOSITION 8. @@ -6031,7 +4457,7 @@

For let GH be made equal to AB.

Therefore, whatever parts GH is of CD, the same parts also is AE of CF.

-

Let GH be divided into the parts of CD, namely GK, KH, and AE into the parts of CF, namely AL, LE; thus the multitude of GK, KH will be equal to the multitude of AL, LE.

+

Let GH be divided into the parts of CD, namely GK, KH, and AE into the parts of CF, namely AL, LE; thus the multitude of GK, KH will be equal to the multitude of AL, LE.

Now since, whatever part GK is of CD, the same part also is AL of CF, while. CD is greater than CF, therefore GK is also greater than AL.

Let GM be made equal to AL.

Therefore, whatever part GK is of CD, the same part also is GM of CF; therefore also the remainder MK is the same part of the remainder FD that the whole GK is of the whole CD. [VII. 7]

@@ -6050,7 +4476,7 @@

For since, whatever part A is of BC, the same part also is D of EF, therefore, as many numbers as there are in BC equal to A, so many also are there in EF equal to D.

Let BC be divided into the numbers equal to A, namely BG, GC, and EF into those equal to D, namely EH, HF; thus the multitude of BG, GC will be equal to the multitude of EH, HF.

-

Now, since the numbers BG, GC are equal to one another, and the numbers EH, HF are also equal to one another, while the multitude of BG, GC is equal to the multitude of EH, HF, therefore, whatever part or parts BG is of EH, the same part or the same parts is GC of HF also; so that, in addition, whatever part or parts BG is of EH, the same part also, or the same parts, is the sum BC of the sum EF. [VII. 5, 6]

+

Now, since the numbers BG, GC are equal to one another, and the numbers EH, HF are also equal to one another, while the multitude of BG, GC is equal to the multitude of EH, HF, therefore, whatever part or parts BG is of EH, the same part or the same parts is GC of HF also; so that, in addition, whatever part or parts BG is of EH, the same part also, or the same parts, is the sum BC of the sum EF. [VII. 5, 6]

But BG is equal to A, and EH to D; therefore, whatever part or parts A is of D, the same part or the same parts is BC of EF also. Q. E. D.

 @@ -6063,7 +4489,7 @@

For since, whatever parts AB is of C, the same parts also is DE of F, therefore, as many parts of C as there are in AB, so many parts also of F are there in DE.

Let AB be divided into the parts of C, namely AG, GB, and DE into the parts of F, namely DH, HE; thus the multitude of AG, GB will be equal to the multitude of DH, HE.

Now since, whatever part AG is of C, the same part also is DH of F, alternately also, whatever part or parts AG is of DH, the same part or the same parts is C of F also. [VII. 9]

-

For the same reason also, whatever part or parts GB is of HE, the same part or the same parts is C of F also; so that, in addition, whatever parts or part AB is of DE, the same parts also, or the same part, is C of F. [VII. 5, 6] Q. E. D.

+

For the same reason also, whatever part or parts GB is of HE, the same part or the same parts is C of F also; so that, in addition, whatever parts or part AB is of DE, the same parts also, or the same part, is C of F. [VII. 5, 6] Q. E. D.

 PROPOSITION 11. @@ -6087,7 +4513,7 @@

Therefore also the sum of A, C is the same part or the same parts of the sum of B, D that A is of B. [VII. 5, 6]

Therefore, as A is to B, so are A, C to B, D. [VII. Def. 20]

 - + PROPOSITION 13.

@@ -6107,7 +4533,7 @@

For, since, as A is to B, so is D to E, therefore, alternately, as A is to D, so is B to E. [VII. 13] -

+

Again, since, as B is to C, so is E to F, therefore, alternately, as B is to E, so is C to F. [VII. 13]

But, as B is to E, so is A to D; therefore also, as A is to D, so is C to F.

@@ -6117,7 +4543,7 @@ PROPOSITION 15.

If an unit measure any number, and another number measure any other number the same number of times, alternately also, the unit will measure the third number the same number of times that the second measures the fourth. - +

For let the unit A measure any number BC, and let another number D measure any other number EF the same number of times; I say that, alternately also, the unit A measures the number D the same number of times that BC measures EF.

@@ -6130,7 +4556,7 @@

Therefore, as the unit A is to the number D, so is BC to EF.

Therefore the unit A measures the number D the same number of times that BC measures EF. Q. E. D.

- + PROPOSITION 16.

@@ -6166,7 +4592,7 @@

For the same reason, as the unit F is to the number A, so also is C to E; therefore also, as B is to D, so is C to E.

Therefore, alternately, as B is to C, so is D to E. [VII. 13] Q. E. D.

 - + PROPOSITION 18.

@@ -6184,7 +4610,7 @@ If four numbers be proportional, the number produced from the first and fourth will be equal to the number produced from the second and third; and, if the number produced from the first and fourth be equal to that produced from the second and third, the four numbers will be proportional.

Let A, B, C, D be four numbers in proportion, so that, as A is to B, so is C to D; and let A by multiplying D make E, and let B by multiplying C make F; I say that E is equal to F.

-

For let A by multiplying C make G.

+

For let A by multiplying C make G.

Since, then, A by multiplying C has made G, and by multiplying D has made E, the number A by multiplying the two numbers C, D has made G, E.

Therefore, as C is to D, so is G to E. [VII. 17]

But, as C is to D, so is A to B; therefore also, as A is to B, so is G to E.

@@ -6210,7 +4636,7 @@

For, if possible, let it be so; therefore EF is also the same parts of B that CD is of A. [VII. 13 and Def. 20]

Therefore, as many parts of A as there are in CD, so many parts of B are there also in EF.

-

Let CD be divided into the parts of A, namely CG, GD, and EF into the parts of B, namely EH, HF; thus the multitude of CG, GD will be equal to the multitude of EH, HF.

+

Let CD be divided into the parts of A, namely CG, GD, and EF into the parts of B, namely EH, HF; thus the multitude of CG, GD will be equal to the multitude of EH, HF.

Now, since the numbers CG, GD are equal to one another, and the numbers EH, HF are also equal to one another, while the multitude of CG, GD is equal to the multitude of EH, HF, therefore, as CG is to EH, so is GD to HF.

Therefore also, as one of the antecedents is to one of the consequents, so will all the antecedents be to all the consequents. [VII. 12]

Therefore, as CG is to EH, so is CD to EF.

@@ -6229,7 +4655,7 @@

Let them be C, D.

Since, then, the least numbers of those which have the same ratio measure those which have the same ratio the same number of times, the greater the greater and the less the less, that is, the antecedent the antecedent and the consequent the consequent, [VII. 20] therefore C measures A the same number of times that D measures B.

Now, as many times as C measures A, so many units let there be in E.

-

Therefore D also measures B according to the units in E.

+

Therefore D also measures B according to the units in E.

And, since C measures A according to the units in E, therefore E also measures A according to the units in C. [VII. 16]

For the same reason E also measures B according to the units in D. [VII. 16]

Therefore E measures A, B which are prime to one another: which is impossible. [VII. Def. 12]

@@ -6244,7 +4670,7 @@

Let A, B be the least numbers of those which have the same ratio with them; I say that A, B are prime to one another.

For, if they are not prime to one another, some number will measure them.

-

Let some number measure them, and let it be C.

+

Let some number measure them, and let it be C.

And, as many times as C measures A, so many units let there be in D, and, as many times as C measures B, so many units let there be in E

Since C measures A according to the units in D, therefore C by multiplying D has made A. [VII. Def. 15]

@@ -6263,7 +4689,7 @@

Let a number measure them, and let it be D.

Since D measures C, and C measures A, therefore D also measures A.

-

But it also measures B; therefore D measures A, B which are prime to one another: which is impossible. [VII. Def. 12]

+

But it also measures B; therefore D measures A, B which are prime to one another: which is impossible. [VII. Def. 12]

Therefore no number will measure the numbers C, B.

Therefore C, B are prime to one another. Q. E. D.

@@ -6281,7 +4707,7 @@

Therefore E by multiplying F has made D. [VII. Def. 15]

But, further, A by multiplying B has also made D; therefore the product of E, F is equal to the product of A, B.

But, if the product of the extremes be equal to that of the means, the four numbers are proportional; [VII. 19] therefore, as E is to A, so is B to F.

-

But A, E are prime to one another, numbers which are prime to one another are also the least of those which have the same ratio, [VII. 21] and the least numbers of those which have the same ratio with them measure those which have the same ratio the same number of times, the greater the greater and the less the less, that is, the antecedent the antecedent and the consequent the consequent; [VII. 20] therefore E measures B.

+

But A, E are prime to one another, numbers which are prime to one another are also the least of those which have the same ratio, [VII. 21] and the least numbers of those which have the same ratio with them measure those which have the same ratio the same number of times, the greater the greater and the less the less, that is, the antecedent the antecedent and the consequent the consequent; [VII. 20] therefore E measures B.

But it also measures C; therefore E measures B, C which are prime to one another: which is impossible. [VII. Def. 12]

Therefore no number will measure the numbers C, D.

Therefore C, D are prime to one another. Q. E. D. @@ -6298,7 +4724,7 @@

For let D be made equal to A.

Since A, B are prime to one another, and A is equal to D, therefore D, B are also prime to one another.

-

Therefore each of the two numbers D, A is prime to B; therefore the product of D, A will also be prime to B. [VII. 24]

+

Therefore each of the two numbers D, A is prime to B; therefore the product of D, A will also be prime to B. [VII. 24]

But the number which is the product of D, A is C.

Therefore C, B are prime to one another. Q. E. D. The Greek, o( e)k tou= e(no\s au)tw=n geno/menos, literally the number produced from the one of them, @@ -6320,7 +4746,7 @@

But the product of C, D is F.

Therefore E, F are prime to one another. Q. E. D.

- + PROPOSITION 27.

If two numbers be prime to one another, and each by multiplying itself make a certain number, the products will be prime to one another; and, if the original numbers by multiplying the products make certain numbers, the latter will also be prime to one another [and this is always the case with the extremes].

@@ -6348,7 +4774,7 @@

For the same reason AC, CB are also prime to one another.

Therefore CA is prime to each of the numbers AB, BC.

Again, let CA, AB be prime to one another; I say that AB, BC are also prime to one another.

-

For, if AB, BC are not prime to one another, some number will measure AB, BC.

+

For, if AB, BC are not prime to one another, some number will measure AB, BC.

Let a number measure them, and let it be D.

Now, since D measures each of the numbers AB, BC, it will also measure the whole CA.

But it also measures AB; therefore D measures CA, AB which are prime to one another: which is impossible. [VII. Def. 12]

@@ -6365,7 +4791,7 @@

Let C measure them.

Since C measures B, and A does not measure B, therefore C is not the same with A.

-

Now, since C measures B, A, therefore it also measures A which is prime, though it is not the same with it: which is impossible.

+

Now, since C measures B, A, therefore it also measures A which is prime, though it is not the same with it: which is impossible.

Therefore no number will measure B, A.

Therefore A, B are prime to one another. Q. E. D.

 @@ -6384,7 +4810,7 @@

Therefore, as D is to A, so is B to E. [VII. 19]

But D, A are prime to one another, primes are also least, [VII. 21] and the least measure the numbers which have the same ratio the same number of times, the greater the greater and the less the less, that is, the antecedent the antecedent and the consequent the consequent; [VII. 20] therefore D measures B.

Similarly we can also show that, if D do not measure B, it will measure A.

-

Therefore D measures one of the numbers A, B. Q. E. D.

+

Therefore D measures one of the numbers A, B. Q. E. D.

PROPOSITION 31. @@ -6404,7 +4830,7 @@

Thus, if the investigation be continued in this way, some prime number will be found which will measure the number before it, which will also measure A.

For, if it is not found, an infinite series of numbers will measure the number A, each of which is less than the other: which is impossible in numbers.

Therefore some prime number will be found which will measure the one before it, which will also measure A.

-

Therefore any composite number is measured by some prime number. +

Therefore any composite number is measured by some prime number. i.e. the implied problem of finding a prime number which measures A. In the Greek the sentence stops here, but it is necessary to add the words the number before it, which will also measure A, which are found a few lines further down. It is possible that the words may have fallen out of P here by a simple mistake due to o(moiote/leuton (Heiberg).

@@ -6429,7 +4855,7 @@

A, B, C are either prime to one another or not.

Now, if A, B, C are prime to one another, they are the least of those which have the same ratio with them. [VII. 21]

-

But, if not, let D the greatest common measure of A, B, C be taken, [VII. 3] and, as many times as D measures the numbers A, B, C +

But, if not, let D the greatest common measure of A, B, C be taken, [VII. 3] and, as many times as D measures the numbers A, B, C respectively, so many units let there be in the numbers E, F, G respectively.

Therefore the numbers E, F, G measure the numbers A, B, C respectively according to the units in D. [VII. 16]

Therefore E, F, G measure A, B, C the same number of times; therefore E, F, G are in the same ratio with A, B, C. [VII. Def. 20]

@@ -6445,13 +4871,13 @@

Therefore the product of E, D is equal to the product of H, M.

Therefore, as E is to H, so is M to D. [VII. 19]

But E is greater than H; therefore M is also greater than D.

-

And it measures A, B, C: which is impossible, for by hypothesis D is the greatest common measure of A, B, C.

+

And it measures A, B, C: which is impossible, for by hypothesis D is the greatest common measure of A, B, C.

Therefore there cannot be any numbers less than E, F, G which are in the same ratio with A, B, C.

Therefore E, F, G are the least of those which have the same ratio with A, B, C. Q. E. D. literally (as usual) each of the numbers E, F, G measures each of the numbers A, B, C.

- + PROPOSITION 34.

@@ -6471,7 +4897,7 @@ E. [VII. 19]

But A, B are prime, primes are also least, [VII. 21] and the least measure the numbers which have the same ratio the same number of times, the greater the greater and the less the less; [VII. 20] therefore B measures E, as consequent consequent.

And, since A by multiplying B, E has made C, D, therefore, as B is to E, so is C to D. [VII. 17]

-

But B measures E; therefore C also measures D, the greater the less: which is impossible.

+

But B measures E; therefore C also measures D, the greater the less: which is impossible.

Therefore A, B do not measure any number less than C; therefore C is the least that is measured by A, B.

Next, let A, B not be prime to one another, and let F, E, the least numbers of those which have the same ratio with A, B, be taken; [VII. 33] therefore the product of A, E is equal to the product of B, F. [VII. 19]

And let A by multiplying E make C; therefore also B by multiplying F has made C; therefore A, B measure C.

@@ -6486,11 +4912,11 @@

Therefore also, as F is to E, so is H to G.

But F, E are least, and the least measure the numbers which have the same ratio the same number of times, the greater the greater and the less the less; [VII. 20] therefore E measures G.

And, since A by multiplying E, G has made C, D, therefore, as E is to G, so is C to D. [VII. 17]

-

But E measures G; therefore C also measures D, the greater the less: which is impossible.

+

But E measures G; therefore C also measures D, the greater the less: which is impossible.

Therefore A, B will not measure any number which is less than C.

Therefore C is the least that is measured by A, B. Q. E. D.

- + PROPOSITION 35.

@@ -6512,7 +4938,7 @@

Let D, the least number measured by the two numbers A, B, be taken. [VII. 34]

Then C either measures, or does not measure, D.

-

First, let it measure it.

+

First, let it measure it.

But A, B also measure D; therefore A, B, C measure D.

I say next that it is also the least that they measure.

For, if not, A, B, C will measure some number which is less than D.

@@ -6531,7 +4957,7 @@

Let them measure F.

Since A, B, C measure F, therefore also A, B measure F; therefore the least number measured by A, B will also measure F. [VII. 35]

But D is the least number measured by A, B; therefore D measures F.

-

But C also measures F; therefore D, C measure F, so that the least number measured by D, C will also measure F.

+

But C also measures F; therefore D, C measure F, so that the least number measured by D, C will also measure F.

But E is the least number measured by C, D; therefore E measures F, the greater the less: which is impossible.

Therefore A, B, C will not measure any number which is less than E.

Therefore E is the least that is measured by A, B, C. Q. E. D.

@@ -6544,7 +4970,7 @@

For let the number A be measured by any number B; I say that A has a part called by the same name as B.

For, as many times as B measures A, so many units let there be in C.

-

Since B measures A according to the units in C, and the unit D also measures the number C according to the units in it, therefore the unit D measures the number C the same number of times as B measures A.

+

Since B measures A according to the units in C, and the unit D also measures the number C according to the units in it, therefore the unit D measures the number C the same number of times as B measures A.

Therefore, alternately, the unit D measures the number B the same number of times as C measures A; [VII. 15] therefore, whatever part the unit D is of the number B, the same part is C of A also.

But the unit D is a part of the number B called by the same name as it; therefore C is also a part of A called by the same name as B, so that A has a part C which is called by the same name as B. Q. E. D.

@@ -6555,7 +4981,7 @@

For let the number A have any part whatever, B, and let C be a number called by the same name as the part B; I say that C measures A.

-

For, since B is a part of A called by the same name as C, and the unit D is also a part of C called by the same name as it, therefore, whatever part the unit D is of the number C, the same part is B of A also; therefore the unit D measures the number C the same number of times that B measures A.

+

For, since B is a part of A called by the same name as C, and the unit D is also a part of C called by the same name as it, therefore, whatever part the unit D is of the number C, the same part is B of A also; therefore the unit D measures the number C the same number of times that B measures A.

Therefore, alternately, the unit D measures the number B the same number of times that C measures A. [VII. 15]

Therefore C measures A. Q. E. D.

@@ -6569,7 +4995,7 @@

Let D, E, F be numbers called by the same name as the parts A, B, C, and let G, the least number measured by D, E, F, be taken. [VII. 36]

Therefore G has parts called by the same name as D, E, F. [VII. 37]

But A, B, C are parts called by the same name as D, E, F; therefore G has the parts A, B, C.

-

I say next that it is also the least number that has.

+

I say next that it is also the least number that has.

For, if not, there will be some number less than G which will have the parts A, B, C.

Let it be H.

Since H has the parts A, B, C, therefore H will be measured by numbers called by the same name as the parts A, B, C. [VII. 38]

@@ -6578,9 +5004,9 @@

Therefore there will be no number less than G that will have the parts A, B, C. Q. E. D.

- - - +
+
+ BOOK VIII. PROPOSITIONS @@ -6594,7 +5020,7 @@

For, if not, let E, F, G, H be less than A, B, C, D, and in the same ratio with them.

Now, since A, B, C, D are in the same ratio with E, F, G, H, and the multitude of the numbers A, B, C, D is equal to the multitude of the numbers E, F, G, H, therefore, ex aequali, as A is to D, so is E to H. [VII. 14]

-

But A, D are prime, primes are also least, [VII. 21] and the least numbers measure those which have the same ratio the same number of times, the greater the greater and the less the less, that is, the antecedent the antecedent and the consequent the consequent. [VII. 20]

+

But A, D are prime, primes are also least, [VII. 21] and the least numbers measure those which have the same ratio the same number of times, the greater the greater and the less the less, that is, the antecedent the antecedent and the consequent the consequent. [VII. 20]

Therefore A measures E, the greater the less: which is impossible.

Therefore E, F, G, H which are less than A, B, C, D are not in the same ratio with them.

Therefore A, B, C, D are the least of those which have the same ratio with them. Q. E. D.

@@ -6606,7 +5032,7 @@

Let the ratio of A to B be the given ratio in least numbers; thus it is required to find numbers in continued proportion, as many as may be prescribed, and the least that are in the ratio of A to B.

-

Let four be prescribed; let A by multiplying itself make C, and by multiplying B let it make D; let B by multiplying itself make E; further, let A by multiplying C, D, E make F, G, H, and let B by multiplying E make K.

+

Let four be prescribed; let A by multiplying itself make C, and by multiplying B let it make D; let B by multiplying itself make E; further, let A by multiplying C, D, E make F, G, H, and let B by multiplying E make K.

Now, since A by multiplying itself has made C, and by multiplying B has made D, therefore, as A is to B, so is C to D. [VII. 17]

Again, since A by multiplying B has made D, and B by multiplying itself has made E, therefore the numbers A, B by multiplying B have made the numbers D, E respectively.

Therefore, as A is to B, so is D to E. [VII. 18]

@@ -6622,7 +5048,7 @@

I say next that they are the least numbers that are so.

For, since A, B are the least of those which have the same ratio with them, and the least of those which have the same ratio are prime to one another, [VII. 22] therefore A, B are prime to one another.

And the numbers A, B by multiplying themselves respectively have made the numbers C, E, and by multiplying the numbers C, E respectively have made the numbers F, K; therefore C, E and F, K are prime to one another respectively. [VII. 27]

-

But, if there be as many numbers as we please in continued proportion, and the extremes of them be prime to one another, they are the least of those which have the same ratio with them. [VIII. 1]

+

But, if there be as many numbers as we please in continued proportion, and the extremes of them be prime to one another, they are the least of those which have the same ratio with them. [VIII. 1]

Therefore C, D, E and F, G, H, K are the least of those which have the same ratio with A, B. Q. E. D.

PORISM. @@ -6633,7 +5059,7 @@ PROPOSITION 3.

If as many numbers as we please in continued proportion be the least of those which have the same ratio with them, the extremes of them are prime to one another.

Let as many numbers as we please, A, B, C, D, in continued proportion be the least of those which have the same ratio with them;

- I say that the extremes of them A, D are prime to one another.

+ I say that the extremes of them A, D are prime to one another.

For let two numbers E, F, the least that are in the ratio of A, B, C, D, be taken, [VII. 33] then three others G, H, K with the same property; and others, more by one continually, [VIII. 2] until the multitude taken becomes equal to the multitude of the numbers A, B, C, D.

Let them be taken, and let them be L, M, N, O.

Now, since E, F are the least of those which have the same ratio with them, they are prime to one another. [VII. 22]

@@ -6642,7 +5068,7 @@

And L, O are prime to one another.

Therefore A, D are also prime to one another. Q. E. D.

- + PROPOSITION 4.

@@ -6657,7 +5083,7 @@

And, as many times as E measures K, so many times let F measure L also.

Now, since A measures H the same number of times that B measures G, therefore, as A is to B, so is H to G. [VII. Def. 20, VII. 13]

For the same reason also, as C is to D, so is G to K, and further, as E is to F, so is K to L; therefore H, G, K, L are continuously proportional in the ratio of A to B, in the ratio of C to D, and in the ratio of E to F.

-

I say next that they are also the least that have this property.

+

I say next that they are also the least that have this property.

For, if H, G, K, L are not the least numbers continuously proportional in the ratios of A to B, of C to D, and of E to F, let them be N, O, M, P.

Then since, as A is to B, so is N to O, while A, B are least, and the least numbers measure those which have the same ratio the same number of times, the greater the greater and the less the less, that is, the antecedent the antecedent and the consequent the consequent; therefore B measures O. [VII. 20]

For the same reason C also measures O; therefore B, C measure O; therefore the least number measured by B, C will also measure O. [VII. 35]

@@ -6668,7 +5094,7 @@

Let M, the least number measured by E, K, be taken.

And, as many times as K measures M, so many times let H, G measure N, O respectively, and, as many times as E measures M, so many times let F measure P also.

-

Since H measures N the same number of times that G measures O, therefore, as H is to G, so is N to O. [VII. 13 and Def. 20]

+

Since H measures N the same number of times that G measures O, therefore, as H is to G, so is N to O. [VII. 13 and Def. 20]

But, as H is to G, so is A to B; therefore also, as A is to B, so is N to O.

For the same reason also, as C is to D, so is O to M.

@@ -6684,11 +5110,11 @@

But E also measures S; therefore E, K measure S.

Therefore the least number measured by E, K will also measure S. [VII. 35]

But M is the least number measured by E, K; therefore M measures S, the greater the less: which is impossible.

-

Therefore there will not be any numbers less than N, O, M, P continuously proportional in the ratios of A to B, of C to D, and of E to F; therefore N, O, M, P are the least numbers continuously proportional in the ratios A : B, C : D, E : F. Q. E. D. +

Therefore there will not be any numbers less than N, O, M, P continuously proportional in the ratios of A to B, of C to D, and of E to F; therefore N, O, M, P are the least numbers continuously proportional in the ratios A : B, C : D, E : F. Q. E. D. 69, 71, 99. the ratios A : B, C : D, E : F. This abbreviated expression is in the Greek o) *a*b, *g*d, *e*z lo/goi.

 - + PROPOSITION 5.

@@ -6719,7 +5145,7 @@

I say, then, that neither will any other measure any other.

For, if possible, let A measure C.

And, however many A, B, C are, let as many numbers F, G, H, the least of those which have the same ratio with A, B, C, be taken. [VII. 33]

-

Now, since F, G, H are in the same ratio with A, B, C, and the multitude of the numbers A, B, C is equal to the multitude of the numbers F, G, H, therefore, ex aequali, as A is to C, so is F to H. [VII. 14]

+

Now, since F, G, H are in the same ratio with A, B, C, and the multitude of the numbers A, B, C is equal to the multitude of the numbers F, G, H, therefore, ex aequali, as A is to C, so is F to H. [VII. 14]

And since, as A is to B, so is F to G, while A does not measure B, therefore neither does F measure G; [VII. Def. 20] therefore F is not an unit, for the unit measures any number.

Now F, H are prime to one another. [VIII. 3]

And, as F is to H, so is A to C; therefore neither does A measure C.

@@ -6736,7 +5162,7 @@

But A measures D.

Therefore A also measures B. Q. E. D.

 - + PROPOSITION 8.

@@ -6747,7 +5173,7 @@

For, as many as A, B, C, D are in multitude, let so many numbers G, H, K, L, the least of those which have the same ratio with A, C, D, B, be taken; [VII. 33] therefore the extremes of them G, L are prime to one another. [VIII. 3]

Now, since A, C, D, B are in the same ratio with G, H, K, L, and the multitude of the numbers A, C, D, B is equal to the multitude of the numbers G, H, K, L, therefore, ex aequali, as A is to B, so is G to L. [VII. 14]

But, as A is to B, so is E to F; therefore also, as G is to L, so is E to F.

-

But G, L are prime, primes are also least, [VII. 21] and the least numbers measure those which have the same ratio the same number of times, the greater the greater and the less the less, that is, the antecedent the antecedent and the consequent the consequent. [VII. 20]

+

But G, L are prime, primes are also least, [VII. 21] and the least numbers measure those which have the same ratio the same number of times, the greater the greater and the less the less, that is, the antecedent the antecedent and the consequent the consequent. [VII. 20]

Therefore G measures E the same number of times as L measures F.

Next, as many times as G measures E, so many times let H, K also measure M, N respectively; therefore G, H, K, L measure E, M, N, F the same number of times.

Therefore G, H, K, L are in the same ratio with E, M, N, F. [VII. Def. 20]

@@ -6763,7 +5189,7 @@

If two numbers be prime to one another, and numbers fall between them in continued proportion, then, however many numbers fall between them in continued proportion, so many will also fall between each of them and an unit in continued proportion.

-

Let A, B be two numbers prime to one another, and let C, D fall between them in continued proportion, and let the unit E be set out; I say that, as many numbers as fall between A, B in continued proportion, so many will also fall between either of the numbers A, B and the unit in continued proportion.

+

Let A, B be two numbers prime to one another, and let C, D fall between them in continued proportion, and let the unit E be set out; I say that, as many numbers as fall between A, B in continued proportion, so many will also fall between either of the numbers A, B and the unit in continued proportion.

For let two numbers F, G, the least that are in the ratio of A, C, D, B, be taken, three numbers H, K, L with the same property, and others more by one continually, until their multitude is equal to the multitude of A, C, D, B. [VIII. 2]

Let them be taken, and let them be M, N, O, P.

@@ -6772,7 +5198,7 @@

Now, since F by multiplying itself has made H, therefore F measures H according to the units in F.

But the unit E also measures F according to the units in it; therefore the unit E measures the number F the same number of times as F measures H.

Therefore, as the unit E is to the number F, so is F to H. [VII. Def. 20]

-

Again, since F by multiplying H has made M, therefore H measures M according to the units in F.

+

Again, since F by multiplying H has made M, therefore H measures M according to the units in F.

But the unit E also measures the number F according to the units in it; therefore the unit E measures the number F the same number of times as H measures M.

Therefore, as the unit E is to the number F, so is H to M.

But it was also proved that, as the unit E is to the number F, so is F to H; therefore also, as the unit E is to the number F, so is F to H, and H to M.

@@ -6784,7 +5210,7 @@ PROPOSITION 10.

If numbers fall between each of two numbers and an unit in continued proportion, however many numbers fall between each of them and an unit in continued proportion, so many also will fall between the numbers themselves in continued proportion. - +

For let the numbers D, E and F, G respectively fall between the two numbers A, B and the unit C in continued proportion; I say that, as many numbers as have fallen between each of the numbers A, B and the unit C in continued proportion, so many numbers will also fall between A, B in continued proportion.

For let D by multiplying F make H, and let the numbers D, F by multiplying H make K, L respectively.

@@ -6794,7 +5220,7 @@

Again, since, as C is to the number D, so is E to A, therefore the unit C measures the number D the same number of times as E measures A.

But the unit C measures the number D according to the units in D; therefore E also measures A according to the units in D; therefore D by multiplying E has made A.

For the same reason also F by multiplying itself has made G, and by multiplying G has made B.

-

And, since D by multiplying itself has made E and by multiplying F has made H, therefore, as D is to F, so is E to H. [VII. 17]

+

And, since D by multiplying itself has made E and by multiplying F has made H, therefore, as D is to F, so is E to H. [VII. 17]

For the same reason also, as D is to F, so is H to G. [VII. 18]

Therefore also, as E is to H, so is H to G.

@@ -6838,7 +5264,7 @@

For let C by multiplying itself make E, and by multiplying D let it make F; let D by multiplying itself make G, and let the numbers C, D by multiplying F make H, K respectively.

Now, since A is a cube, and C its side, and C by multiplying itself has made E, therefore C by multiplying itself has made E and by multiplying E has made A.

For the same reason also D by multiplying itself has made G and by multiplying G has made B.

-

And, since C by multiplying the numbers C, D has made E, F respectively, therefore, as C is to D, so is E to F. [VII. 17]

+

And, since C by multiplying the numbers C, D has made E, F respectively, therefore, as C is to D, so is E to F. [VII. 17]

For the same reason also, as C is to D, so is F to G. [VII. 18]

Again, since C by multiplying the numbers E, F has made A, H respectively, therefore, as E is to F, so is A to H. [VII. 17]

@@ -6856,7 +5282,7 @@ PROPOSITION 13.

If there be as many numbers as we please in continued proportion, and each by multiplying itself make some number, the products will be proportional; and, if the original numbers by multiplying the products make certain numbers, the latter will also be proportional. - +

Let there be as many numbers as we please, A, B, C, in continued proportion, so that, as A is to B, so is B to C; let A, B, C by multiplying themselves make D, E, F, and by multiplying D, E, F let them make G, H, K; I say that D, E, F and G, H, K are in continued proportion.

@@ -6864,7 +5290,7 @@

And again let B by multiplying C make O, and let the numbers B, C by multiplying O make P, Q respectively.

Then, in manner similar to the foregoing, we can prove that D, L, E and G, M, N, H are continuously proportional in the ratio of A to B, and further E, O, F and H, P, Q, K are continuously proportional in the ratio of B to C.

Now, as A is to B, so is B to C; therefore D, L, E are also in the same ratio with E, O, F, and further G, M, N, H in the same ratio with H, P, Q, K.

-

And the multitude of D, L, E is equal to the multitude of E, O, F, and that of G, M, N, H to that of H, P, Q, K; therefore, ex acquali, as D is to E, so is E to F, and, as G is to H, so is H to K. [VII. 14] Q. E. D.

+

And the multitude of D, L, E is equal to the multitude of E, O, F, and that of G, M, N, H to that of H, P, Q, K; therefore, ex acquali, as D is to E, so is E to F, and, as G is to H, so is H to K. [VII. 14] Q. E. D.

PROPOSITION 14. @@ -6874,7 +5300,7 @@

Let A, B be square numbers, let C, D be their sides, and let A measure B; I say that C also measures D.

For let C by multiplying D make E; therefore A, E, B are continuously proportional in the ratio of C to D. [VIII. 11]

-

And, since A, E, B are continuously proportional, and A measures B, therefore A also measures E. [VIII. 7]

+

And, since A, E, B are continuously proportional, and A measures B, therefore A also measures E. [VIII. 7]

And, as A is to E, so is C to D; therefore also C measures D. [VII. Def. 20]

Again, let C measure D; I say that A also measures B.

For, with the same construction, we can in a similar manner prove that A, E, B are continuously proportional in the ratio of C to D.

@@ -6887,7 +5313,7 @@

If a cube number measure a cube number, the side will also measure the side; and, if the side measure the side, the cube will also measure the cube.

-

For let the cube number A measure the cube B, and let C be the side of A and D of B; I say that C measures D.

+

For let the cube number A measure the cube B, and let C be the side of A and D of B; I say that C measures D.

For let C by multiplying itself make E, and let D by multiplying itself make G; further, let C by multiplying D make F, and let C, D by multiplying F make H, K respectively.

Now it is manifest that E, F, G and A, H, K, B are continuously proportional in the ratio of C to D. [VIII. 11, 12]

@@ -6897,7 +5323,7 @@

For, with the same construction, we can prove in a similar manner that A, H, K, B are continuously proportional in the ratio of C to D.

And, since C measures D, and, as C is to D, so is A to H, therefore A also measures H, [VII. Def. 20] so that A measures B also. Q. E. D.

- + PROPOSITION 16.

@@ -6920,7 +5346,7 @@

For if C measures D, A will also measure B. [VIII. 15]

But A does not measure B; therefore neither does C measure D.

-

Again, let C not measure D; I say that neither will A measure B.

+

Again, let C not measure D; I say that neither will A measure B.

For, if A measures B, C will also measure D. [VIII. 15]

But C does not measure D; therefore neither will A measure B. Q. E. D.

 @@ -6937,7 +5363,7 @@

And, since A is plane, and C, D are its sides, therefore D by multiplying C has made A.

For the same reason also E by multiplying F has made B.

Now let D by multiplying E make G.

-

Then, since D by multiplying C has made A, and by multiplying E has made G, therefore, as C is to E, so is A to G. [VII. 17]

+

Then, since D by multiplying C has made A, and by multiplying E has made G, therefore, as C is to E, so is A to G. [VII. 17]

But, as C is to E, so is D to F; therefore also, as D is to F, so is A to G.

Again, since E by multiplying D has made G, and by multiplying F has made B, therefore, as D is to F, so is G to B. [VII. 17]

But it was also proved that, as D is to F, so is A to G; therefore also, as A is to G, so is G to B.

@@ -6948,7 +5374,7 @@

And, as A is to G, so is C to E, and so is D to F.

Therefore A also has to B the ratio duplicate of that which C has to E or D to F. Q. E. D.

- + PROPOSITION 19.

@@ -6966,7 +5392,7 @@

Therefore M is the product of D, F, as was proved in the theorem preceding this. [VIII. 18]

Now, since D by multiplying C has made K, and by multiplying F has made M, therefore, as C is to F, so is K to M. [VII. 17]

But, as K is to M, so is M to L.

-

Therefore K, M, L are continuously proportional in the ratio of C to F.

+

Therefore K, M, L are continuously proportional in the ratio of C to F.

And since, as C is to D, so is F to G, alternately therefore, as C is to F, so is D to G. [VII. 13]

For the same reason also, as D is to G, so is E to H.

@@ -6984,7 +5410,7 @@

But, as M is to L, so is C to F, D to G, and E to H.

Therefore also, as C is to F, D to G, and E to H, so not only is O to B, but also A to N and N to O.

Therefore A, N, O, B are continuously proportional in the aforesaid ratios of the sides.

-

I say that A also has to B the ratio triplicate of that which the corresponding side has to the corresponding side, that is, of the ratio which the number C has to F, or D to G, and also E to H.

+

I say that A also has to B the ratio triplicate of that which the corresponding side has to the corresponding side, that is, of the ratio which the number C has to F, or D to G, and also E to H.

For, since A, N, O, B are four numbers in continued proportion, therefore A has to B the ratio triplicate of that which A has to N. [V. Def. 10]

But, as A is to N, so it was proved that C is to F, D to G, and also E to H.

Therefore A also has to B the ratio triplicate of that which the corresponding side has to the corresponding side, that is, of the ratio which the number C has to F, D to G, and also E to H. Q. E. D.

@@ -6996,7 +5422,7 @@

For let one mean proportional number C fall between the two numbers A, B; I say that A, B are similar plane numbers.

Let D, E, the least numbers of those which have the same ratio with A, C, be taken; [VII. 33] therefore D measures A the same number of times that E measures C. [VII. 20]

-

Now, as many times as D measures A, so many units let there be in F; therefore F by multiplying D has made A, so that A is plane, and D, F are its sides.

+

Now, as many times as D measures A, so many units let there be in F; therefore F by multiplying D has made A, so that A is plane, and D, F are its sides.

Again, since D, E are the least of the numbers which have the same ratio with C, B, therefore D measures C the same number of times that E measures B. [VII. 20]

As many times, then, as E measures B, so many units let there be in G; therefore E measures B according to the units in G; therefore G by multiplying E has made B.

@@ -7027,7 +5453,7 @@

Let, then, H, K be the sides of E, and L, M of G.

Therefore it is manifest from the theorem before this that E, F, G are continuously proportional in the ratio of H to L and that of K to M.

Now, since E, F, G are the least of the numbers which have the same ratio with A, C, D, and the multitude of the numbers E, F, G is equal to the multitude of the numbers A, C, D, therefore, ex aequali, as E is to G, so is A to D. [VII. 14]

-

But E, G are prime, primes are also least, [VII. 21] and the least measure those which have the same ratio with them the same number of times, the greater the greater and the less the less, that is, the antecedent the antecedent and the consequent the consequent; [VII. 20] therefore E measures A the same number of times that G measures D.

+

But E, G are prime, primes are also least, [VII. 21] and the least measure those which have the same ratio with them the same number of times, the greater the greater and the less the less, that is, the antecedent the antecedent and the consequent the consequent; [VII. 20] therefore E measures A the same number of times that G measures D.

Now, as many times as E measures A, so many units let there be in N.

Therefore N by multiplying E has made A.

But E is the product of H, K; therefore N by multiplying the product of H, K has made A.

@@ -7060,7 +5486,7 @@

Let A, B, C, D be four numbers in continued proportion, and let A be cube; I say that D is also cube.

-

For, since between A, D there are two mean proportional numbers B, C, therefore A, D are similar solid numbers. [VIII. 21]

+

For, since between A, D there are two mean proportional numbers B, C, therefore A, D are similar solid numbers. [VIII. 21]

But A is cube; therefore D is also cube. Q. E. D.

@@ -7080,7 +5506,7 @@

If two numbers have to one another the ratio which a cube number has to a cube number, and the first be cube, the second will also be cube.

-

For let the two numbers A, B have to one another the ratio which the cube number C has to the cube number D, and let A be cube; I say that B is also cube.

+

For let the two numbers A, B have to one another the ratio which the cube number C has to the cube number D, and let A be cube; I say that B is also cube.

For, since C, D are cube, C, D are similar solid numbers.

Therefore two mean proportional numbers fall between C, D. [VIII. 19]

@@ -7095,7 +5521,7 @@

Let A, B be similar plane numbers; I say that A has to B the ratio which a square number has to a square number.

-

For, since A, B are similar plane numbers, therefore one mean proportional number falls between A, B. [VIII. 18]

+

For, since A, B are similar plane numbers, therefore one mean proportional number falls between A, B. [VIII. 18]

Let it so fall, and let it be C; and let D, E, F, the least numbers of those which have the same ratio with A, C, B, be taken; [VII. 33 or VIII. 2] therefore the extremes of them D, F are square. [VIII. 2, Por.]

And since, as D is to F, so is A to B, and D, F are square, therefore A has to B the ratio which a square number has to a square number. Q. E. D.

@@ -7108,12 +5534,12 @@

For, since A, B are similar solid numbers, therefore two mean proportional numbers fall between A, B. [VIII. 19]

Let C, D so fall, and let E, F, G, H, the least numbers of those which have the same ratio with A, C, D, B, and equal with them in multitude, be taken; [VII. 33 or VIII. 2] therefore the extremes of them E, H are cube. [VIII. 2, Por.]

-

And, as E is to H, so is A to B; therefore A also has to B the ratio which a cube number has to a cube number. Q. E. D.

+

And, as E is to H, so is A to B; therefore A also has to B the ratio which a cube number has to a cube number. Q. E. D.

- - - +
+
+ BOOK IX. PROPOSITIONS @@ -7131,7 +5557,7 @@

But, if numbers fall between two numbers in continued proportion, as many as fall between them, so many also fall between those which have the same ratio; [VIII. 8] so that one mean proportional number falls between D, C also.

And D is square; therefore C is also square. [VIII. 22] Q. E. D.

- + PROPOSITION 2.

@@ -7151,7 +5577,7 @@

If a cube number by multiplying itself make some number, the product will be cube.

-

For let the cube number A by multiplying itself make B; I say that B is cube.

+

For let the cube number A by multiplying itself make B; I say that B is cube.

For let C, the side of A, be taken, and let C by multiplying itself make D.

It is then manifest that C by multiplying D has made A.

@@ -7167,7 +5593,7 @@

But, if two mean proportional numbers fall between two numbers, and the first be cube, the second will also be cube. [VIII. 23]

And A is cube; therefore B is also cube. Q. E. D.

- + PROPOSITION 4.

@@ -7188,7 +5614,7 @@

For let the cube number A by multiplying any number B make the cube number C; I say that B is cube.

-

For let A by multiplying itself make D; therefore D is cube. [IX. 3]

+

For let A by multiplying itself make D; therefore D is cube. [IX. 3]

Now, since A by multiplying itself has made D, and by multiplying B has made C, therefore, as A is to B, so is D to C. [VII. 17]

And since D, C are cube, they are similar solid numbers.

Therefore two mean proportional numbers fall between D, C. [VIII. 19]

@@ -7208,7 +5634,7 @@

But the unit also measures A according to the units in it.

Therefore, as the unit is to A, so is A to B. [VII. Def. 20]

And, since A by multiplying B has made C, therefore B measures C according to the units in A.

-

But the unit also measures A according to the units in it.

+

But the unit also measures A according to the units in it.

Therefore, as the unit is to A, so is B to C. [VII. Def. 20]

But, as the unit is to A, so is A to B; therefore also, as A is to B, so is B to C.

And, since B, C are cube, they are similar solid numbers.

@@ -7225,7 +5651,7 @@

For let the composite number A by multiplying any number B make C; I say that C is solid.

For, since A is composite, it will be measured by some number. [VII. Def. 13]

-

Let it be measured by D; and, as many times as D measures A, so many units let there be in E.

+

Let it be measured by D; and, as many times as D measures A, so many units let there be in E.

Since then D measures A according to the units in E, therefore E by multiplying D has made A. [VII. Def. 15]

And, since A by multiplying B has made C, and A is the product of D, E, therefore the product of D, E by multiplying B has made C.

Therefore C is solid, and D, E, B are its sides. Q. E. D.

@@ -7240,7 +5666,7 @@

For since, as the unit is to A, so is A to B, therefore the unit measures the number A the same number of times that A measures B. [VII. Def. 20]

But the unit measures the number A according to the units in it; therefore A also measures B according to the units in A.

Therefore A by multiplying itself has made B; therefore B is square.

-

And, since B, C, D are in continued proportion, and B is square, therefore D is also square. [VIII. 22]

+

And, since B, C, D are in continued proportion, and B is square, therefore D is also square. [VIII. 22]

For the same reason F is also square.

Similarly we can prove that all those which leave out one are square.

I say next that C, the fourth from the unit, is cube, as are also all those which leave out two.

@@ -7266,7 +5692,7 @@

Next, let A be cube; I say that all the rest are also cube.

Now it has been proved that C, the fourth from the unit, is cube, as also are all those which leave out two; [IX. 8] I say that all the rest are also cube.

For, since, as the unit is to A, so is A to B, therefore the unit measures A the same number of times as A measures B.

-

But the unit measures A according to the units in it; therefore A also measures B according to the units in itself; therefore A by multiplying itself has made B.

+

But the unit measures A according to the units in it; therefore A also measures B according to the units in itself; therefore A by multiplying itself has made B.

And A is cube.

But, if a cube number by multiplying itself make some number, the product is cube. [IX. 3]

Therefore B is also cube.

@@ -7278,7 +5704,7 @@

If as many numbers as we please beginning from an unit be in continued proportion, and the number after the unit be not square, neither will any other be square except the third from the unit and all those which leave out one. And, if the number after the unit be not cube, neither will any other be cube except the fourth from the unit and all those which leave out two.

-

Let there be as many numbers as we please, A, B, C, D, E, F, beginning from an unit and in continued proportion, and let A, the number after the unit, not be square; I say that neither will any other be square except the third from the unit <and those which leave out one>.

+

Let there be as many numbers as we please, A, B, C, D, E, F, beginning from an unit and in continued proportion, and let A, the number after the unit, not be square; I say that neither will any other be square except the third from the unit <and those which leave out one>.

For, if possible, let C be square.

But B is also square; [IX. 8] [therefore B, C have to one another the ratio which a square number has to a square number].

@@ -7295,14 +5721,14 @@

And since, as the unit is to A, so is A to B, and the unit measures A according to the units in it, therefore A also measures B according to the units in itself; therefore A by multiplying itself has made the cube number B.

But, if a number by multiplying itself make a cube number, it is also itself cube. [IX. 6]

Therefore A is also cube: which is contrary to the hypothesis.

-

Therefore D is not cube.

+

Therefore D is not cube.

Similarly we can prove that neither is any other of the numbers cube except the fourth from the unit and those which leave out two. Q. E. D.

PROPOSITION II.

If as many numbers as we please beginning from an unit be in continued proportion, the less measures the greater according to some one of the numbers which have place among the proportional numbers. - +

Let there be as many numbers as we please, B, C, D, E, beginning from the unit A and in continued proportion; I say that B, the least of the numbers B, C, D, E, measures E according to some one of the numbers C, D.

@@ -7314,7 +5740,7 @@ Q. E. D. - + PROPOSITION 12.

@@ -7330,7 +5756,7 @@

Therefore, as A is to E, so is F to C. [VII. 19]

But A, E are prime, primes are also least, [VII. 21] and the least measure those which have the same ratio the same number of times, the antecedent the antecedent and the consequent the consequent; [VII. 20] therefore E measures C.

Let it measure it according to G; therefore E by multiplying G has made C.

-

But, further, by the theorem before this, A has also by multiplying B made C. [IX. 11 and Por.]

+

But, further, by the theorem before this, A has also by multiplying B made C. [IX. 11 and Por.]

Therefore the product of A, B is equal to the product of E, G.

Therefore, as A is to E, so is G to B. [VII. 19]

But A, E are prime, primes are also least, [VII. 21] and the least numbers measure those which have the same ratio with them the same number of times, the antecedent the antecedent and the consequent the consequent: [VII. 20] therefore E measures B.

@@ -7353,7 +5779,7 @@

For, if possible, let it be measured by E, and let E not be the same with any of the numbers A, B, C.

It is then manifest that E is not prime.

-

For, if E is prime and measures D, it will also measure A [IX. 12], which is prime, though it is not the same with it: which is impossible.

+

For, if E is prime and measures D, it will also measure A [IX. 12], which is prime, though it is not the same with it: which is impossible.

Therefore E is not prime.

Therefore it is composite.

But any composite number is measured by some prime number; [VII. 31] therefore E is measured by some prime number.

@@ -7368,7 +5794,7 @@

Similarly we can prove that F is measured by A, by proving again that F is not prime.

For, if it is, and measures D, it will also measure A [IX. 12], which is prime, though it is not the same with it: which is impossible; therefore F is not prime.

Therefore it is composite.

-

But any composite number is measured by some prime number; [VII. 31] therefore F is measured by some prime number.

+

But any composite number is measured by some prime number; [VII. 31] therefore F is measured by some prime number.

I say next that it will not be measured by any other prime except A.

For, if any other prime number measures F, and F measures D, that other will also measure D; so that it will also measure A [IX. 12], which is prime, though it is not the same with it: which is impossible.

Therefore A measures F.

@@ -7386,7 +5812,7 @@

Similarly then we can prove that H is not the same with A.

And, since G measures B according to H, therefore G by multiplying H has made B.

But further A has also by multiplying itself made B; [IX. 8] therefore the product of H, G is equal to the square on A.

-

Therefore, as H is to A, so is A to G. [VII. 19]

+

Therefore, as H is to A, so is A to G. [VII. 19]

But A measures G;< therefore H also measures A, which is prime, though it is not the same with it: which is absurd.

Therefore D the greatest will not be measured by any other number except A, B, C. Q. E. D.

 @@ -7395,7 +5821,7 @@

If a number be the least that is measured by prime numbers, it will not be measured by any other prime number except those originally measuring it.

-

For let the number A be the least that is measured by the prime numbers B, C, D; I say that A will not be measured by any other prime number except B, C, D.

+

For let the number A be the least that is measured by the prime numbers B, C, D; I say that A will not be measured by any other prime number except B, C, D.

For, if possible, let it be measured by the prime number E, and let E not be the same with any one of the numbers B, C, D.

Now, since E measures A, let it measure it according to F; therefore E by multiplying F has made A. @@ -7406,7 +5832,7 @@

Therefore they will measure F, which is less than A: which is impossible, for A is by hypothesis the least number measured by B, C, D.

Therefore no prime number will measure A except B, C, D. Q. E. D.

- + PROPOSITION 15.

@@ -7421,7 +5847,7 @@

But further DE is also prime to EF; therefore DF, DE are prime to EF.

But, if two numbers be prime to any number, their product is also prime to the other; [VII. 24] so that the product of FD, DE is prime to EF; hence the product of FD, DE is also prime to the square on EF. [VII. 25]

But the product of FD, DE is the square on DE together with the product of DE, EF; [II. 3] therefore the square on DE together with the product of DE, EF is prime to the square on EF.

-

And the square on DE is A, the product of DE, EF is B, and the square on EF is C; therefore A, B added together are prime to C.

+

And the square on DE is A, the product of DE, EF is B, and the square on EF is C; therefore A, B added together are prime to C.

Similarly we can prove that B, C added together are prime to A.

I say next that A, C added together are also prime to B.

For, since DF is prime to each of the numbers DE, EF, the square on DF is also prime to the product of DE, EF. [VII. 24, 25]

@@ -7431,7 +5857,7 @@

And the square on DE is A, the product of DE, EF is B, and the square on EF is C.

Therefore A, C added together are prime to B. Q. E. D.

 - + PROPOSITION 16.

@@ -7451,7 +5877,7 @@

For let there be as many numbers as we please, A, B, C, D, in continued proportion, and let the extremes of them, A, D, be prime to one another; I say that D is not to any other number as A is to B.

-

For, if possible, as A is to B, so let D be to E; therefore, alternately, as A is to D, so is B to E. [VII. 13]

+

For, if possible, as A is to B, so let D be to E; therefore, alternately, as A is to D, so is B to E. [VII. 13]

But A, D are prime, primes are also least, [VII. 21] and the least numbers measure those which have the same ratio the same number of times, the antecedent the antecedent and the consequent the consequent. [VII. 20]

Therefore A measures B.

And, as A is to B, so is B to C.

@@ -7468,7 +5894,7 @@

Let A, B be the given two numbers, and let it be required to investigate whether it is possible to find a third proportional to them.

Now A, B are either prime to one another or not.

-

And, if they are prime to one another, it has been proved that it is impossible to find a third proportional to them. [IX. 16]

+

And, if they are prime to one another, it has been proved that it is impossible to find a third proportional to them. [IX. 16]

Next, let A, B not be prime to one another, and let B by multiplying itself make C.

Then A either measures C or does not measure it.

@@ -7483,7 +5909,7 @@

But, by hypothesis, it also does not measure it: which is absurd.

Therefore it is not possible to find a third proportional number to A, B when A does not measure C. Q. E. D.

- + PROPOSITION 19.

@@ -7497,7 +5923,7 @@

For, if possible, let D have been found, so that, as A is to B, so is C to D, and let it be contrived that, as B is to C, so is D to E.

Now, since, as A is to B, so is C to D, and, as B is to C, so is D to E, therefore, ex aequali, as A is to C, so is C to E. [VII. 14]

But A, C are prime, primes are also least, [VII. 21] and the least numbers measure those which have the same ratio, the antecedent the antecedent and the consequent the consequent. [VII. 20]

-

Therefore A measures C as antecedent antecedent.

+

Therefore A measures C as antecedent antecedent.

But it also measures itself; therefore A measures A, C which are prime to one another: which is impossible.

Therefore it is not possible to find a fourth proportional to A, B, C.†

Next, let A, B, C be again in continued proportion, but let A, C not be prime to one another.

@@ -7534,7 +5960,7 @@

Therefore G, being a number, will measure the remainder, the unit DF: which is absurd.

Therefore G is not the same with any one of the numbers A, B, C.

And by hypothesis it is prime.

-

Therefore the prime numbers A, B, C, G have been found which are more than the assigned multitude of A, B, C. Q. E. D.

+

Therefore the prime numbers A, B, C, G have been found which are more than the assigned multitude of A, B, C. Q. E. D.

 PROPOSITION 21. @@ -7551,7 +5977,7 @@

If as many odd numbers as we please be added together, and their multitude be even, the whole will be even.

-

For let as many odd numbers as we please, AB, BC, CD, DE, even in multitude, be added together; I say that the whole AE is even.

+

For let as many odd numbers as we please, AB, BC, CD, DE, even in multitude, be added together; I say that the whole AE is even.

For, since each of the numbers AB, BC, CD, DE is odd, if an unit be subtracted from each, each of the remainders will be even; [VII. Def. 7] so that the sum of them will be even. [IX. 21]

But the multitude of the units is also even.

@@ -7578,7 +6004,7 @@

For from the even number AB let the even number BC be subtracted: I say that the remainder CA is even.

-

For, since AB is even, it has a half part. [VII. Def. 6]

+

For, since AB is even, it has a half part. [VII. Def. 6]

For the same reason BC also has a half part; so that the remainder [CA also has a half part, and] AC is therefore even. Q. E. D.

@@ -7602,7 +6028,7 @@

For, since AB is odd, let the unit BD be subtracted; therefore the remainder AD is even. [VII. Def. 7]

For the same reason CD is also even; [VII. Def. 7] so that the remainder CA is also even. [IX. 24] Q. E. D.

 - + PROPOSITION 27.

@@ -7635,7 +6061,7 @@

For let the odd number A by multiplying the odd number B make C; I say that C is odd.

-

For, since A by multiplying B has made C, therefore C is made up of as many numbers equal to B as there are units in A. [VII. Def. 15]

+

For, since A by multiplying B has made C, therefore C is made up of as many numbers equal to B as there are units in A. [VII. Def. 15]

And each of the numbers A, B is odd; therefore C is made up of odd numbers the multitude of which is odd.

Thus C is odd. [IX. 23] Q. E. D.

@@ -7662,7 +6088,7 @@

For let the odd number A be prime to any number B, and let C be double of B; I say that A is prime to C.

-

For, if they are not prime to one another, some number will measure them.

+

For, if they are not prime to one another, some number will measure them.

Let a number measure them, and let it be D.

Now A is odd; therefore D is also odd.

And since D which is odd measures C, and C is even, therefore [D] will measure the half of C also. [IX. 30]

@@ -7683,7 +6109,7 @@

For let an unit be set out.

Since then as many numbers as we please beginning from an unit are in continued proportion, and the number A after the unit is prime, therefore D, the greatest of the numbers A, B, C, D, will not be measured by any other number except A, B, C. [IX. 13]

And each of the numbers A, B, C is even; therefore D is even-times even only. [VII. Def. 8]

-

Similarly we can prove that each of the numbers B, C is even-times even only. Q. E. D.

+

Similarly we can prove that each of the numbers B, C is even-times even only. Q. E. D.

PROPOSITION 33. @@ -7705,7 +6131,7 @@

Now that A is even-times even is manifest; for it has not its half odd. [VII. Def. 8]

I say next that it is also even-times odd.

For, if we bisect A, then bisect its half, and do this continually, we shall come upon some odd number which will measure A according to an even number.

-

For, if not, we shall come upon a dyad, and A will be among those which are doubled from a dyad: which is contrary to the hypothesis.

+

For, if not, we shall come upon a dyad, and A will be among those which are doubled from a dyad: which is contrary to the hypothesis.

Thus A is even-times odd.

But it was also proved even-times even.

Therefore A is both even-times even and even-times odd. Q. E. D.

@@ -7723,7 +6149,7 @@

Separando, as EL is to LF, so is LK to FK, and KH to FH. [VII. 11, 13]

Therefore also, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents; [VII. 12] therefore, as KH is to FH, so are EL, LK. KH to LF, FK, HF.

But KH is equal to CG, FH to A, and LF, FK, HF to D, BC, A; therefore, as CG is to A, so is EH to D, BC, A.

-

Therefore, as the excess of the second is to the first, so is the excess of the last to all those before it. Q. E. D.

+

Therefore, as the excess of the second is to the first, so is the excess of the last to all those before it. Q. E. D.

 PROPOSITION 36. @@ -7731,7 +6157,7 @@ If as many numbers as we please beginning from an unit be set out continuously in double proportion, until the sum of all becomes prime, and if the sum multiplied into the last make some number, the product will be perfect.

For let as many numbers as we please, A, B, C, D, beginning from an unit be set out in double proportion, until the sum of all becomes prime, let E be equal to the sum, and let E by multiplying D make FG; I say that FG is perfect.

-

For, however many A, B, C, D are in multitude, let so many E, HK, L, M be taken in double proportion beginning from E; therefore, ex aequali, as A is to D, so is E to M. [VII. 14]

+

For, however many A, B, C, D are in multitude, let so many E, HK, L, M be taken in double proportion beginning from E; therefore, ex aequali, as A is to D, so is E to M. [VII. 14]

Therefore the product of E, D is equal to the product of A, M. [VII. 19]

And the product of E, D is FG; therefore the product of A, M is also FG.

Therefore A by multiplying M has made FG; therefore M measures FG according to the units in A.

@@ -7745,7 +6171,7 @@

Therefore the whole FG is equal to E, HK, L, M and A, B, C, D and the unit; and it is measured by them.

I say also that FG will not be measured by any other number except A, B, C, D, E, HK, L, M and the unit.

For, if possible, let some number P measure FG, and let P not be the same with any of the numbers A, B, C, D, E, HK, L, M.

-

And, as many times as P measures FG, so many units let there be in Q; therefore Q by multiplying P has made FG.

+

And, as many times as P measures FG, so many units let there be in Q; therefore Q by multiplying P has made FG.

But, further, E has also by multiplying D made FG; therefore, as E is to Q, so is P to D. [VII. 19]

And, since A, B, C, D are continuously proportional beginning from an unit, therefore D will not be measured by any other number except A, B, C. [IX. 13]

And, by hypothesis, P is not the same with any of the numbers A, B, C; therefore P will not measure D.

@@ -7760,15 +6186,15 @@

Therefore the product of B, L is equal to the product of D, E. [VII. 19]

But the product of D, E is equal to the product of Q, P; therefore the product of Q, P is also equal to the product of B, L.

Therefore, as Q is to B, so is L to P. [VII. 19]

-

And Q is the same with B; therefore L is also the same with P; which is impossible, for by hypothesis P is not the same with any of the numbers set out.

+

And Q is the same with B; therefore L is also the same with P; which is impossible, for by hypothesis P is not the same with any of the numbers set out.

Therefore no number will measure FG except A, B, C, D, E, HK, L, M and the unit.

And FG was proved equal to A, B, C, D, E, HK, L, M and the unit; and a perfect number is that which is equal to its own parts; [VII. Def. 22] therefore FG is perfect. Q. E. D.

 - +
- - +
+ BOOK X. DEFINITIONS I. @@ -7789,8 +6215,8 @@

And let the square on the assigned straight line be called rational and those areas which are commensurable with it rational, but those which are incommensurable with it irrational, and the straight lines which produce them irrational, that is, in case the areas are squares, the sides themselves, but in case they are any other rectilineal figures, the straight lines on which are described squares equal to them.

 - - + + PROPOSITIONS I—47. @@ -7801,14 +6227,14 @@

For C if multiplied will sometime be greater than AB. [cf. v. Def. 4]

Let it be multiplied, and let DE be a multiple of C, and greater than. AB; let DE be divided into the parts DF, FG, GE equal to C, from AB let there be subtracted BH greater than its half, and, from AH, HK greater than its half, and let this process be repeated continually until the divisions in AB are equal in multitude with the divisions in DE.

Let, then, AK, KH, HB be divisions which are equal in multitude with DF, FG, GE.

-

Now, since DE is greater than AB, and from DE there has been subtracted EG less than its half, and, from AB, BH greater than its half, therefore the remainder GD is greater than the remainder HA.

+

Now, since DE is greater than AB, and from DE there has been subtracted EG less than its half, and, from AB, BH greater than its half, therefore the remainder GD is greater than the remainder HA.

And, since GD is greater than HA, and there has been subtracted, from GD, the half GF, and, from HA, HK greater than its half, therefore the remainder DF is greater than the remainder AK.

But DF is equal to C; therefore C is also greater than AK.

Therefore AK is less than C.

Therefore there is left of the magnitude AB the magnitude AK which is less than the lesser magnitude set out, namely C. Q. E. D.

And the theorem can be similarly proved even if the parts subtracted be halves.

 - + PROPOSITION 2.

@@ -7824,7 +6250,7 @@

But CF measures BG; therefore E also measures BG.

But it measures the whole AB also; therefore it will also measure the remainder AG, the greater the less: which is impossible.

Therefore no magnitude will measure the magnitudes AB, CD; therefore the magnitudes AB, CD are incommensurable.

-

Therefore etc. [X. Def. 1]

+

Therefore etc. [X. Def. 1]

 PROPOSITION 3. @@ -7837,7 +6263,7 @@

Next, let AB not measure CD.

Then, if the less be continually subtracted in turn from the greater, that which is left over will sometime measure the one before it, because AB, CD are not incommensurable; [cf. X. 2] let AB, measuring ED, leave EC less than itself, let EC, measuring FB, leave AF less than itself, and let AF measure CE.

Since, then, AF measures CE, while CE measures FB, therefore AF will also measure FB.

-

But it measures itself also; therefore AF will also measure the whole AB.

+

But it measures itself also; therefore AF will also measure the whole AB.

But AB measures DE; therefore AF will also measure ED.

But it measures CE also; therefore it also measures the whole CD.

Therefore AF is a common measure of AB, CD.

@@ -7867,7 +6293,7 @@

Next, let D not measure C.

I say first that C, D are commensurable.

For, since A, B, C are commensurable, some magnitude will measure them, and this will of course measure A, B also; so that it will also measure the greatest common measure of A, B, namely D. [X. 3, Por.]

-

But it also measures C; so that the said magnitude will measure C, D; therefore C, D are commensurable.

+

But it also measures C; so that the said magnitude will measure C, D; therefore C, D are commensurable.

Now let their greatest common measure be taken, and let it be E. [X. 3]

Since then E measures D, while D measures A, B, therefore E will also measure A, B.

But it measures C also; therefore E measures A, B, C; therefore E is a common measure of A, B, C.

@@ -7884,7 +6310,7 @@

From this it is manifest that, if a magnitude measure three magnitudes, it will also measure their greatest common measure.

Similarly too, with more magnitudes, the greatest common measure can be found, and the porism can be extended. Q. E. D.

- + PROPOSITION 5. @@ -7895,12 +6321,12 @@

And, as many times as C measures A, so many units let there be in D; and, as many times as C measures B, so many units let there be in E.

Since then C measures A according to the units in D, while the unit also measures D according to the units in it, therefore the unit measures the number D the same number of times as the magnitude C measures A; therefore, as C is to A, so is the unit to D; [VII. Def. 20] therefore, inversely, as A is to C, so is D to the unit. [cf. V. 7, Por.]

-

Again, since C measures B according to the units in E, while the unit also measures E according to the units in it, therefore the unit measures E the same number of times as C measures B; therefore, as C is to B, so is the unit to E.

+

Again, since C measures B according to the units in E, while the unit also measures E according to the units in it, therefore the unit measures E the same number of times as C measures B; therefore, as C is to B, so is the unit to E.

But it was also proved that, as A is to C, so is D to the unit; therefore, ex aequali, as A is to B, so is the number D to E. [V. 22]

Therefore the commensurable magnitudes A, B have to one another the ratio which the number D has to the number E. Q. E. D.

 - + PROPOSITION 6.

If two magnitudes have to one another the ratio which a number has to a number, the magnitudes will be commensurable.

@@ -7916,7 +6342,7 @@

But, as D is to E, so is A to B; therefore also, as A is to B, so is it to F also. [V. 11]

Therefore A has the same ratio to each of the magnitudes B, F; therefore B is equal to F. [V. 9]

But C measures F; therefore it measures B also.

-

Further it measures A also; therefore C measures A, B.

+

Further it measures A also; therefore C measures A, B.

Therefore A is commensurable with B.

Therefore etc.

@@ -7928,7 +6354,7 @@ 15. But the unit measures the number D; therefore C also measures A. These words are redundant, though they are apparently found in all the MSS. - + PROPOSITION 7.

Incommensurable magnitudes have not to one another the ratio which a number has to a number.

@@ -7949,7 +6375,7 @@ PROPOSITION 9. -

The squares on straight lines commensurable in length have to one another the ratio which a square number has to a square number; and squares which have to one another the ratio which a square number has to a square number will also have their sides commensurable in length. But the squares on straight lines incommensurable in length have not to one another the ratio which a square number has to a square number; and squares which have not to one another the ratio which a square number has to a square number will not have their sides commensurable in length either.

+

The squares on straight lines commensurable in length have to one another the ratio which a square number has to a square number; and squares which have to one another the ratio which a square number has to a square number will also have their sides commensurable in length. But the squares on straight lines incommensurable in length have not to one another the ratio which a square number has to a square number; and squares which have not to one another the ratio which a square number has to a square number will not have their sides commensurable in length either.

For let A, B be commensurable in length; I say that the square on A has to the square on B the ratio which a square number has to a square number.

For, since A is commensurable in length with B, therefore A has to B the ratio which a number has to a number. [X. 5]

@@ -7959,7 +6385,7 @@

For since, as the square on A is to the square on B, so is the square on C to the square on D, while the ratio of the square on A to the square on B is duplicate of the ratio of A to B, and the ratio of the square on C to the square on D is duplicate of the ratio of C to D, therefore also, as A is to B, so is C to D.

Therefore A has to B the ratio which the number C has to the number D; therefore A is commensurable in length with B. [X. 6]

Next, let A be incommensurable in length with B; I say that the square on A has not to the square on B the ratio which a square number has to a square number.

-

For, if the square on A has to the square on B the ratio which a square number has to a square number, A will be commensurable with B.

+

For, if the square on A has to the square on B the ratio which a square number has to a square number, A will be commensurable with B.

But it is not; therefore the square on A has not to the square on B the ratio which a square number has to a square number.

Again, let the square on A not have to the square on B the ratio which a square number has to a square number; I say that A is incommensurable in length with B.

For, if A is commensurable with B, the square on A will have to the square on B the ratio which a square number has to a square number.

@@ -7981,7 +6407,7 @@ [PROPOSITION 10.

To find two straight lines incommensurable, the one in length only, and the other in square also, with an assigned straight line.

Let A be the assigned straight line; thus it is required to find two straight lines incommensurable, the one in length only, and the other in square also, with A.

-

Let two numbers B, C be set out which have not to one another the ratio which a square number has to a square number, that is, which are not similar plane numbers; and let it be contrived that, as B is to C, so is the square on A to the square on D +

Let two numbers B, C be set out which have not to one another the ratio which a square number has to a square number, that is, which are not similar plane numbers; and let it be contrived that, as B is to C, so is the square on A to the square on D

—for we have learnt how to do this— [X. 6, Por.] therefore the square on A is commensurable with the square on D. [X. 6]

And, since B has not to C the ratio which a square number has to a square number, therefore neither has the square on A to the square on D the ratio which a square number has to a square number; therefore A is incommensurable in length with D. [X. 9]

Let E be taken a mean proportional between A, D; therefore, as A is to D, so is the square on A to the square on E. [V. Def. 9]

@@ -7995,7 +6421,7 @@

For, since A is commensurable with B, therefore A has to B the ratio which a number has to a number. [X. 5]

And, as A is to B, so is C to D; therefore C also has to D the ratio which a number has to a number; therefore C is commensurable with D. [X. 6]

Next, let A be incommensurable with B; I say that C will also be incommensurable with D.

-

For, since A is incommensurable with B, therefore A has not to B the ratio which a number has to a number. [X. 7]

+

For, since A is incommensurable with B, therefore A has not to B the ratio which a number has to a number. [X. 7]

And, as A is to B, so is C to D; therefore neither has C to D the ratio which a number has to a number; therefore C is incommensurable with D. [X. 8]

Therefore etc.

@@ -8004,7 +6430,7 @@

Magnitudes commensurable with the same magnitude are commensurable with one another also.

For let each of the magnitudes A, B be commensurable with C; I say that A is also commensurable with B.

-

For, since A is commensurable with C, therefore A has to C the ratio which a number has to a number. [X. 5]

+

For, since A is commensurable with C, therefore A has to C the ratio which a number has to a number. [X. 5]

Let it have the ratio which D has to E.

Again, since C is commensurable with B, therefore C has to B the ratio which a number has to a number. [X. 5]

Let it have the ratio which F has to G.

@@ -8016,7 +6442,7 @@

Therefore A has to B the ratio which a number has to a number; therefore A is commensurable with B. [X. 6]

Therefore etc. Q. E. D.

- + PROPOSITION 13.

If two magnitudes be commensurable, and the one of them be incommensurable with any magnitude, the remaining one will also be incommensurable with the same.

@@ -8033,7 +6459,7 @@

Let the semicircle ADB be described on AB, and let AD be fitted into it equal to C; [IV. 1] let DB be joined.

It is then manifest that the angle ADB is right, [III. 31] and that the square on AB is greater than the square on AD, that is, C, by the square on DB. [I. 47]

-

Similarly also, if two straight lines be given, the straight line the square on which is equal to the sum of the squares on them is found in this manner.

+

Similarly also, if two straight lines be given, the straight line the square on which is equal to the sum of the squares on them is found in this manner.

Let AD, DB be the given two straight lines, and let it be required to find the straight line the square on which is equal to the sum of the squares on them.

Let them be placed so as to contain a right angle, that formed by AD, DB; and let AB be joined.

It is again manifest that the straight line the square on which is equal to the sum of the squares on AD, DB is AB. [I. 47] Q. E. D.

@@ -8048,7 +6474,7 @@

Let A, B, C, D be four straight lines in proportion, so that, as A is to B, so is C to D; and let the square on A be greater than the square on B by the square on E, and let the square on C be greater than the square on D by the square on F; I say that, if A is commensurable with E, C is also commensurable with F, and, if A is incommensurable with E, C is also incommensurable with F.

For since, as A is to B, so is C to D, therefore also, as the square on A is to the square on B, so is the square on C to the square on D. [VI. 22]

-

But the squares on E, B are equal to the square on A, and the squares on D, F are equal to the square on C.

+

But the squares on E, B are equal to the square on A, and the squares on D, F are equal to the square on C.

Therefore, as the squares on E, B are to the square on B, so are the squares on D, F to the square on D; therefore, separando, as the square on E is to the square on B, so is the square on F to the square on D; [V. 17] therefore also, as E is to B, so is F to D; [VI. 22] therefore, inversely, as B is to E, so is D to F.

But, as A is to B, so also is C to D; therefore, ex aequali, as A is to E, so is C to F. [V. 22]

Therefore, if A is commensurable with E, C is also commensurable with F, and, if A is incommensurable with E, C is also incommensurable with F. [X. 11]

@@ -8058,7 +6484,7 @@ or whatever it may be, for itself in these cases.

- + PROPOSITION 15.

If two commensurable magnitudes be added together, the whole will also be commensurable with each of them; and, if the whole be commensurable with one of them, the original magnitudes will also be commensurable.

@@ -8075,7 +6501,7 @@

But it measures AB also; therefore D will measure AB, BC; therefore AB, BC are commensurable. [X. Def. 1]

Therefore etc.

 - + PROPOSITION 16.

If two incommensurable magnitudes be added together, the whole will also be incommensurable with each of them; and, if the whole be incommensurable with one of them, the original magnitudes will also be incommensurable.

@@ -8094,7 +6520,7 @@

For, if they are commensurable, some magnitude will measure them.

Let it measure them, and let it be D.

Since then D measures AB, BC. therefore it will also measure the whole AC.

-

But it measures AB also; therefore D measures CA, AB.

+

But it measures AB also; therefore D measures CA, AB.

Therefore CA, AB are commensurable; but they were also, by hypothesis, incommensurable: which is impossible.

Therefore no magnitude will measure AB, BC; therefore AB, BC are incommensurable. [X. Def. 1]

Therefore etc.

@@ -8115,7 +6541,7 @@ If there be two unequal straight lines, and to the greater there be applied a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, and if it divide it into parts which are commensurable in length, then the square on the greater will be greater than the square on the less by the square on a straight line commensurable with the greater.

And, if the square on the greater be greater than the square on the less by the square on a straight line commensurable with - the greater, and if there be applied to the greater a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, it will divide it into parts which are commensurable in length.

+ the greater, and if there be applied to the greater a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, it will divide it into parts which are commensurable in length.

Let A, BC be two unequal straight lines, of which BC is the greater, and let there be applied to BC a parallelogram equal to the fourth part of the square on the less, A, that is, equal to the square on the half of A, and deficient by a square figure. Let this be the rectangle BD, DC, [cf. Lemma]

and let BD be commensurable in length with DC; I say that the square on BC is greater than the square on A by the square on a straight line commensurable with BC.

For let BC be bisected at the point E, and let EF be made equal to DE.

Therefore the remainder DC is equal to BF.

@@ -8125,7 +6551,7 @@

And the square on BC is equal to four times the square on EC, for again BC is double of CE.

Therefore the squares on A, DF are equal to the square on BC, so that the square on BC is greater than the square on A by the square on DF.

It is to be proved that BC is also commensurable with DF.

-

Since BD is commensurable in length with DC, therefore BC is also commensurable in length with CD. [X. 15]

+

Since BD is commensurable in length with DC, therefore BC is also commensurable in length with CD. [X. 15]

But CD is commensurable in length with CD, BF, for CD is equal to BF. [X. 6]

Therefore BC is also commensurable in length with BF, CD, [X. 12] so that BC is also commensurable in length with the remainder FD; [X. 15] therefore the square on BC is greater than the square on A by the square on a straight line commensurable with BC.

Next, let the square on BC be greater than the square on A by the square on a straight line commensurable with BC, let a parallelogram be applied to BC equal to the fourth part of the square on A and deficient by a square figure, and let it be the rectangle BD, DC.

@@ -8146,7 +6572,7 @@

And, if the square on the greater be greater than the square on the less by the square on a straight line incommensurable with the greater, and if there be applied to the greater a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, it divides it into parts which are incommensurable.

Let A, BC be two unequal straight lines, of which BC is the greater, and to BC let there be applied a parallelogram equal to the fourth part of the square on the less, A, and deficient by a square figure. Let this be the rectangle BD, DC, [cf. Lemma before X. 17]

and let BD be incommensurable in length with DC; I say that the square on BC is greater than the square on A by the square on a straight line incommensurable with BC.

For, with the same construction as before, we can prove similarly that the square on BC is greater than the square on A by the square on FD.

-

It is to be proved that BC is incommensurable in length with DF.

+

It is to be proved that BC is incommensurable in length with DF.

Since BD is incommensurable in length with DC, therefore BC is also incommensurable in length with CD. [X. 16]

But DC is commensurable with the sum of BF, DC; [X. 6] therefore BC is also incommensurable with the sum of BF, DC; [X. 13] so that BC is also incommensurable in length with the remainder FD. [X. 16]

And the square on BC is greater than the square on A by the square on FD; therefore the square on BC is greater than the square on A by the square on a straight line incommensurable with BC.

@@ -8169,14 +6595,14 @@

For let the rectangle AC be contained by the rational straight lines AB, BC commensurable in length; I say that AC is rational.

For on AB let the square AD be described; therefore AD is rational. [X. Def. 4]

-

And, since AB is commensurable in length with BC, while AB is equal to BD, therefore BD is commensurable in length with BC.

+

And, since AB is commensurable in length with BC, while AB is equal to BD, therefore BD is commensurable in length with BC.

And, as BD is to BC, so is DA to AC. [VI. 1]

Therefore DA is commensurable with AC. [X. 11]

But DA is rational; therefore AC is also rational. [X. Def. 4]

Therefore etc.

- + PROPOSITION 20.

If a rational area be applied to a rational straight line, it produces as breadth a straight line rational and commensurable in length with the straight line to which it is applied.

For let the rational area AC be applied to AB, a straight line once more rational in any of the aforesaid ways, producing BC as breadth; I say that BC is rational and commensurable in length with BA. For on AB let the square AD be described; therefore AD is rational. [X. Def. 4]

@@ -8193,7 +6619,7 @@ The rectangle contained by rational straight lines commensurable in square only is irrational, and the side of the square equal to it is irrational. Let the latter be called medial.

-

For let the rectangle AC be contained by the rational straight lines AB, BC commensurable in square only; I say that AC is irrational, and the side of the square equal to it is irrational; and let the latter be called medial. +

For let the rectangle AC be contained by the rational straight lines AB, BC commensurable in square only; I say that AC is irrational, and the side of the square equal to it is irrational; and let the latter be called medial.

For on AB let the square AD be described; therefore AD is rational. [X. Def. 4]

@@ -8206,7 +6632,7 @@

If there be two straight lines, then, as the first is to the second, so is the square on the first to the rectangle contained by the two straight lines.

Let FE, EG be two straight lines.

-

I say that, as FE is to EG, so is the square on FE to the rectangle FE, EG.

+

I say that, as FE is to EG, so is the square on FE to the rectangle FE, EG.

For on FE let the square DF be described, and let GD be completed.

Since then, as FE is to EG, so is FD to DG, [VI. 1] and FD is the square on FE, and DG the rectangle DE, EG, that is, the rectangle FE, EG, therefore, as FE is to EG, so is the square on FE to the rectangle FE, EG.

Similarly also, as the rectangle GE, EF is to the square on EF, that is, as GD is to FD, so is GE to EF. Q. E. D.

@@ -8221,7 +6647,7 @@

Let the square on it be equal to GF.

But the square on it is also equal to BD; therefore BD is equal to GF.

But it is also equiangular with it; and in equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional; [VI. 14] therefore, proportionally, as BC is to EG, so is EF to CD.

-

Therefore also, as the square on BC is to the square on EG, so is the square on EF to the square on CD. [VI. 22]

+

Therefore also, as the square on BC is to the square on EG, so is the square on EF to the square on CD. [VI. 22]

But the square on CB is commensurable with the square on EG, for each of these straight lines is rational; therefore the square on EF is also commensurable with the square on CD. [X. 11]

But the square on EF is rational; therefore the square on CD is also rational; [X. Def. 4] therefore CD is rational.

And, since EF is incommensurable in length with EG, for they are commensurable in square only, and, as EF is to EG, so is the square on EF to the rectangle FE, EG, [Lemma] therefore the square on EF is incommensurable with the rectangle FE, EG. [X. 11]

@@ -8239,7 +6665,7 @@

But EC is equal to the square on A, and CF is equal to the square on B; therefore EC is commensurable with CF.

And, as EC is to CF, so is ED to DF; [VI. 1] therefore ED is commensurable in length with DF. [X. 11]

But ED is rational and incommensurable in length with DC; therefore DF is also rational [X. Def. 3] and incommensurable in length with DC. [X. 13]

-

Therefore CD, DF are rational and commensurable in square only.

+

Therefore CD, DF are rational and commensurable in square only.

But the straight line the square on which is equal to the rectangle contained by rational straight lines commensurable in square only is medial; [X. 21] therefore the side of the square equal to the rectangle CD, DF is medial.

And B is the side of the square equal to the rectangle CD, DF; therefore B is medial.

@@ -8268,7 +6694,7 @@

Since then each of the squares AD, BE is medial, and AD is equal to GH, and BE to NL, therefore each of the rectangles GH, NL is also medial.

And they are applied to the rational straight line FG; therefore each of the straight lines FH, KL is rational and incommensurable in length with FG. [X. 22]

And, since AD is commensurable with BE, therefore GH is also commensurable with NL.

-

And, as GH is to NL, so is FH to KL; [VI. 1] therefore FH is commensurable in length with KL. [X. 11]

+

And, as GH is to NL, so is FH to KL; [VI. 1] therefore FH is commensurable in length with KL. [X. 11]

Therefore FH, KL are rational straight lines commensurable in length; therefore the rectangle FH, KL is rational. [X. 19]

And, since DB is equal to BA, and OB to BC, therefore, as DB is to BC, so is AB to BO.

But, as DB is to BC, so is DA to AC, [VI. 1] and, as AB is to BO, so is AC to CO; [id.] therefore, as DA is to AC, so is AC to CO.

@@ -8287,7 +6713,7 @@

But DB is rational; therefore KH is also rational.

Since, then, each of the rectangles AB, AC is medial, and AB is equal to FH, and AC to FG, therefore each of the rectangles FH, FG is also medial.

And they are applied to the rational straight line EF; therefore each of the straight lines HE, EG is rational and incommensurable in length with EF. [X. 22]

-

And, since [DB is rational and is equal to KH, therefore] KH is [also] rational; and it is applied to the rational straight line EF; therefore GH is rational and commensurable in length with EF. [X. 20]

+

And, since [DB is rational and is equal to KH, therefore] KH is [also] rational; and it is applied to the rational straight line EF; therefore GH is rational and commensurable in length with EF. [X. 20]

But EG is also rational, and is incommensurable in length with EF; therefore EG is incommensurable in length with GH. [X. 13]

And, as EG is to GH, so is the square on EG to the rectangle EG, GH; therefore the square on EG is incommensurable with the rectangle EG, GH. [X. 11]

But the squares on EG, GH are commensurable with the square on EG, for both are rational; and twice the rectangle EG, GH is commensurable with the rectangle EG, GH, for it is double of it; [X. 6] therefore the squares on EG, GH are incommensurable with twice the rectangle EG, GH; [X. 13] therefore also the sum of the squares on EG, GH and twice the rectangle EG, GH, that is, the square on EH [II. 4], is incommensurable with the squares on EG, GH. [X. 16]

@@ -8309,7 +6735,7 @@

Therefore C, D are medial and commensurable in square only.

I say that they also contain a rational rectangle.

For since, as A is to B, so is C to D, therefore, alternately, as A is to C, so is B to D. [V. 16]

-

But, as A is to C, so is C to B; therefore also, as C is to B, so is B to D; therefore the rectangle C, D is equal to the square on B.

+

But, as A is to C, so is C to B; therefore also, as C is to B, so is B to D; therefore the rectangle C, D is equal to the square on B.

But the square on B is rational; therefore the rectangle C, D is also rational.

Therefore medial straight lines commensurable in square only have been found which contain a rational rectangle. Q. E. D.

@@ -8319,7 +6745,7 @@

Let the rational straight lines A, B, C commensurable in square only be set out; let D be taken a mean proportional between A, B, [VI. 13] and let it be contrived that, as B is to C, so is D to E. [VI. 12]

-

Since A, B are rational straight lines commensurable in square only, therefore the rectangle A, B, that is, the square on D [VI. 17], is medial. [X. 21]

+

Since A, B are rational straight lines commensurable in square only, therefore the rectangle A, B, that is, the square on D [VI. 17], is medial. [X. 21]

Therefore D is medial. [X. 21]

And since B, C are commensurable in square only, and, as B is to C, so is D to E, therefore D, E are also commensurable in square only. [X. 11]

But D is medial; therefore E is also medial. [X. 23, addition]

@@ -8357,7 +6783,7 @@

First, if possible, let the product of AB, BC together with the square on CE be equal to the square on BE, and let GA be double of the unit DE.

Since then the whole AC is double of the whole CD, and in them AG is double of DE, therefore the remainder GC is also double of the remainder EC; therefore GC is bisected by E.

Therefore the product of GB, BC together with the square on CE is equal to the square on BE. [II. 6]

-

But the product of AB, BC together with the square on CE is also, by hypothesis, equal to the square on BE; therefore the product of GB, BC together with the square on CE is equal to the product of AB, BC together with the square on CE.

+

But the product of AB, BC together with the square on CE is also, by hypothesis, equal to the square on BE; therefore the product of GB, BC together with the square on CE is equal to the product of AB, BC together with the square on CE.

And, if the common square on CE be subtracted, it follows that AB is equal to GB: which is absurd.

Therefore the product of AB, BC together with the square on CE is not equal to the square on BE.

I say next that neither is it less than the square on BE.

@@ -8373,7 +6799,7 @@ PROPOSITION 29.

To find two rational straight lines commensurable in square only and such that the square on the greater is greater than the square on the less by the square on a straight line commensurable in length with the greater.

-

For let there be set out any rational straight line AB, and two square numbers CD, DE such that their difference CE is not square; [Lemma 1] let there be described on AB the semicircle AFB, and let it be contrived that, as DC is to CE, so is the square on BA to the square on AF. [X. 6, Por.] +

For let there be set out any rational straight line AB, and two square numbers CD, DE such that their difference CE is not square; [Lemma 1] let there be described on AB the semicircle AFB, and let it be contrived that, as DC is to CE, so is the square on BA to the square on AF. [X. 6, Por.]

Let FB be joined.

Since, as the square on BA is to the square on AF, so is DC to CE, therefore the square on BA has to the square on AF the ratio which the number DC has to the number CE; therefore the square on BA is commensurable with the square on AF. [X. 6]

@@ -8384,7 +6810,7 @@

And since, as DC is to CE, so is the square on BA to the square on AF, therefore, convertendo, as CD is to DE, so is the square on AB to the square on BF. [V. 19, Por., III. 31, I. 47]

But CD has to DE the ratio which a square number has to a square number: therefore also the square on AB has to the square on BF the ratio which a square number has to a square number; therefore AB is commensurable in length with BF. [X. 9]

And the square on AB is equal to the squares on AF, FB; therefore the square on AB is greater than the square on AF by the square on BF commensurable with AB.

-

Therefore there have been found two rational straight lines BA, AF commensurable in square only and such that the square on the greater AB is greater than the square on the less AF by the square on BF commensurable in length with AB.

+

Therefore there have been found two rational straight lines BA, AF commensurable in square only and such that the square on the greater AB is greater than the square on the less AF by the square on BF commensurable in length with AB.

PROPOSITION 30. @@ -8392,7 +6818,7 @@

Let there be set out a rational straight line AB, and two square numbers CE, ED such that their sum CD is not square; [Lemma 2]

let there be described on AB the semicircle AFB, let it be contrived that, as DC is to CE, so is the square on BA to the square on AF, [X. 6, Por.] and let FB be joined.

Then, in a similar manner to the preceding, we can prove that BA, AF are rational straight lines commensurable in square only.

And since, as DC is to CE, so is the square on BA to the square on AF, therefore, convertendo, as CD is to DE, so is the square on AB to the square on BF. [V. 19, Por., III. 31, I. 47]

-

But CD has not to DE the ratio which a square number has to a square number; therefore neither has the square on AB to the square on BF the ratio which a square number has to a square number; therefore AB is incommensurable in length with BF. [X. 9]

+

But CD has not to DE the ratio which a square number has to a square number; therefore neither has the square on AB to the square on BF the ratio which a square number has to a square number; therefore AB is incommensurable in length with BF. [X. 9]

And the square on AB is greater than the square on AF by the square on FB incommensurable with AB.

Therefore AB, AF are rational straight lines commensurable in square only, and the square on AB is greater than the square on AF by the square on FB incommensurable in length with AB. Q. E. D.

@@ -8404,7 +6830,7 @@

Let there be set out two rational straight lines A, B commensurable in square only and such that the square on A, being the greater, is greater than the square on B the less by the square on a straight line commensurable in length with A. [X. 29]

And let the square on C be equal to the rectangle A, B.

-

Now the rectangle A, B is medial; [X. 21] therefore the square on C is also medial; therefore C is also medial. [X. 21]

+

Now the rectangle A, B is medial; [X. 21] therefore the square on C is also medial; therefore C is also medial. [X. 21]

Let the rectangle C, D be equal to the square on B.

Now the square on B is rational; therefore the rectangle C, D is also rational.

And since, as A is to B, so is the rectangle A, B to the square on B, while the square on C is equal to the rectangle A, B, and the rectangle C, D is equal to the square on B, therefore, as A is to B, so is the square on C to the rectangle C, D.

@@ -8419,7 +6845,7 @@ PROPOSITION 32.

To find two medial straight lines commensurable in square only, containing a medial rectangle, and such that the square on the greater is greater than the square on the less by the square on a straight line commensurable with the greater. - +

Let there be set out three rational straight lines A, B, C commensurable in square only, and such that the square on A is greater than the square on C by the square on a straight line commensurable with A, [X. 29] and let the square on D be equal to the rectangle A, B.

@@ -8432,14 +6858,14 @@

And, since, as A is to C, so is D to E, while the square on A is greater than the square on C by the square on a straight line commensurable with A, therefore also the square on D will be greater than the square on E by the square on a straight line commensurable with D.[X. 14]

I say next that the rectangle D, E is also medial.

For, since the rectangle B, C is equal to the rectangle D, E, while the rectangle B, C is medial, [X. 21] therefore the rectangle D, E is also medial.

-

Therefore two medial straight lines D, E, commensurable in square only, and containing a medial rectangle, have been found such that the square on the greater is greater than the square on the less by the square on a straight line commensurable with the greater.

+

Therefore two medial straight lines D, E, commensurable in square only, and containing a medial rectangle, have been found such that the square on the greater is greater than the square on the less by the square on a straight line commensurable with the greater.

Similarly again it can be proved that the square on D is greater than the square on E by the square on a straight line incommensurable with D, when the square on A is greater than the square on C by the square on a straight line incommensurable with A. [X. 30]

 LEMMA.

Let ABC be a right-angled triangle having the angle A right, and let the perpendicular AD be drawn; I say that the rectangle CB, BD is equal to the square on BA, the rectangle BC, CD equal to the square on CA, the rectangle BD, DC equal to the square on AD, and, further, the rectangle BC, AD equal to the rectangle BA, AC.

-

+

And first that the rectangle CB, BD is equal to the square on BA.

For, since in a right-angled triangle AD has been drawn from the right angle perpendicular to the base, therefore the triangles ABD, ADC are similar both to the whole ABC and to one another. [VI. 8]

And since the triangle ABC is similar to the triangle ABD, therefore, as CB is to BA, so is BA to BD; [VI. 4] therefore the rectangle CB, BD is equal to the square on AB. [VI. 17]

@@ -8454,12 +6880,12 @@ PROPOSITION 33.

To find two straight lines incommensurable in square which make the sum of the squares on them rational but the rectangle contained by them medial.

Let there be set out two rational straight lines AB, BC commensurable in square only and such that the square on the greater AB is greater than the square on the less BC by the square on a straight line incommensurable with AB, [X. 30]

- let BC be bisected at D, let there be applied to AB a parallelogram equal to the square on either of the straight lines BD, DC and deficient by a square figure, and let it be the rectangle AE, EB; [VI. 28] let the semicircle AFB be described on AB, let EF be drawn at right angles to AB, and let AF, FB be joined.

+ let BC be bisected at D, let there be applied to AB a parallelogram equal to the square on either of the straight lines BD, DC and deficient by a square figure, and let it be the rectangle AE, EB; [VI. 28] let the semicircle AFB be described on AB, let EF be drawn at right angles to AB, and let AF, FB be joined.

Then, since AB, BC are unequal straight lines, and the square on AB is greater than the square on BC by the square on a straight line incommensurable with AB, while there has been applied to AB a parallelogram equal to the fourth part of the square on BC, that is, to the square on half of it, and deficient by a square figure, making the rectangle AE, EB, therefore AE is incommensurable with EB. [X. 18]

And, as AE is to EB, so is the rectangle BA, AE to the rectangle AB, BE, while the rectangle BA, AE is equal to the square on AF, and the rectangle AB, BE to the square on BF; therefore the square on AF is incommensurable with the square on FB; therefore AF, FB are incommensurable in square.

And, since AB is rational, therefore the square on AB is also rational; so that the sum of the squares on AF, FB is also rational. [I. 47]

And since, again, the rectangle AE, EB is equal to the square on EF, and, by hypothesis, the rectangle AE, EB is also equal to the square on BD, therefore FE is equal to BD; therefore BC is double of FE, so that the rectangle AB, BC is also commensurable with the rectangle AB, EF.

-

But the rectangle AB, BC is medial; [X. 21] therefore the rectangle AB, EF is also medial. [X. 23, Por.]

+

But the rectangle AB, BC is medial; [X. 21] therefore the rectangle AB, EF is also medial. [X. 23, Por.]

But the rectangle AB, EF is equal to the rectangle AF, FB; [Lemma] therefore the rectangle AF, FB is also medial.

But it was also proved that the sum of the squares on these straight lines is rational.

Therefore two straight lines AF, FB incommensurable in square have been found which make the sum of the squares on them rational, but the rectangle contained by them medial. Q. E. D.

@@ -8470,7 +6896,7 @@ To find two straight lines incommensurable in square which make the sum of the squares on them medial but the rectangle contained by them rational.

Let there be set out two medial straight lines AB, BC, commensurable in square only, such that the rectangle which they contain is rational, and the square on AB is greater than the square on BC by the square on a straight line incommensurable with AB; [X. 31, ad fin.]

let the semicircle ADB be described on AB, let BC be bisected at E, let there be applied to AB a parallelogram equal to the square on BE and deficient by a square figure, namely the rectangle AF, FB; [VI. 28] therefore AF is incommensurable in length with FB. [X. 18]

-

Let FD be drawn from F at right angles to AB, and let AD, DB be joined.

+

Let FD be drawn from F at right angles to AB, and let AD, DB be joined.

Since AF is incommensurable in length with FB, therefore the rectangle BA, AF is also incommensurable with the rectangle AB, BF. [X. 11]

But the rectangle BA, AF is equal to the square on AD, and the rectangle AB, BF to the square on DB; therefore the square on AD is also incommensurable with the square on DB.

And, since the square on AB is medial, therefore the sum of the squares on AD, DB is also medial. [III. 31, I. 47]

@@ -8483,7 +6909,7 @@ PROPOSITION 35.

To find two straight lines incommensurable in square which make the sum of the squares on them medial and the rectangle contained by them medial and moreover incommensurable with the sum of the squares on them.

Let there be set out two medial straight lines AB, BC commensurable in square only, containing a medial rectangle, and such that the square on AB is greater than the square on BC by the square on a straight line incommensurable with AB; [X. 32 -, ad fin.] let the semicircle ADB be described on AB, and let the rest of the construction be as above.

+, ad fin.] let the semicircle ADB be described on AB, and let the rest of the construction be as above.

Then, since AF is incommensurable in length with FB, [X. 18 ] AD is also incommensurable in square with DB. [X. 11 @@ -8499,7 +6925,7 @@ ] so that the square on AB is also incommensurable with the rectangle AB, BE. [X. 11 ]

But the squares on AD, DB are equal to the square on AB, [I. 47 -] and the rectangle AB, FD, that is, the rectangle AD, DB, is equal to the rectangle AB, BE; therefore the sum of the squares on AD, DB is incommensurable with the rectangle AD, DB.

+] and the rectangle AB, FD, that is, the rectangle AD, DB, is equal to the rectangle AB, BE; therefore the sum of the squares on AD, DB is incommensurable with the rectangle AD, DB.

Therefore two straight lines AD, DB incommensurable in square have been found which make the sum of the squares on them medial and the rectangle contained by them medial and moreover incommensurable with the sum of the squares on them. Q. E. D.

@@ -8518,7 +6944,7 @@ ]

But the sum of the squares on AB, BC is rational; therefore the square on AC is irrational, so that AC is also irrational. [X. Def. 4 ]

-

And let it be called binomial.

+

And let it be called binomial.

 PROPOSITION 37. @@ -8526,7 +6952,7 @@ a first bimedial straight line.

For let two medial straight lines AB, BC commensurable in square only and containing a rational rectangle be added together;

I say that the whole AC is irrational.

-

For, since AB is incommensurable in length with BC, therefore the squares on AB, BC are also incommensurable with twice the rectangle AB, BC; [cf. X. 36, ll. 9-20] +

For, since AB is incommensurable in length with BC, therefore the squares on AB, BC are also incommensurable with twice the rectangle AB, BC; [cf. X. 36, ll. 9-20] and, componendo, the squares on AB, BC together with twice the rectangle AB, BC, that is, the square on AC [II. 4], is incommensurable with the rectangle AB, BC. [X. 16 ]

But the rectangle AB, BC is rational, for, by hypothesis, AB, BC are straight lines containing a rational rectangle; therefore the square on AC is irrational; therefore AC is irrational. [X. Def. 4 @@ -8542,7 +6968,7 @@ and, componendo, the squares on AB<

For let a rational straight line DE be set out, and let the parallelogram DF equal to the square on AC be applied to DE, producing DG as breadth. [I. 44 ]

Then, since the square on AC is equal to the squares on AB, BC and twice the rectangle AB, BC, [II. 4 -] let EH, equal to the squares on AB, BC, be applied to DE; therefore the remainder HF is equal to twice the rectangle AB, BC.

+] let EH, equal to the squares on AB, BC, be applied to DE; therefore the remainder HF is equal to twice the rectangle AB, BC.

And, since each of the straight lines AB, BC is medial, therefore the squares on AB, BC are also medial.

But, by hypothesis, twice the rectangle AB, BC is also medial.

And EH is equal to the squares on AB, BC, while FH is equal to twice the rectangle AB, BC; therefore each of the rectangle EH, HF is medial.

@@ -8563,14 +6989,14 @@ and, componendo, the squares on AB< ]

But DE is rational; and the rectangle contained by an irrational and a rational straight line is irrational; [cf. X. 20 ] therefore the area DF is irrational, and the side of the square equal to it is irrational. [X. Def. 4 -]

+]

But AC is the side of the square equal to DF; therefore AC is irrational.

And let it be called a second bimedial straight line. Q. E. D.

PROPOSITION 39.

If two straight lines incommensurable in square which make the sum of the squares on them rational, but the rectangle contained by them medial, be added together, the whole straight line is irrational : and let it be called - major.

+ major.

For let two straight lines AB, BC incommensurable in square, and fulfilling the given conditions [X. 33 ], be added together; I say that AC is irrational.

@@ -8584,7 +7010,7 @@ and, componendo, the squares on AB< PROPOSITION 40.

If two straight lines incommensurable in square which make the sum of the squares on them medial, but the rectangle contained by them rational, be added together, the whole straight line is irrational; and let it be called the - side of a rational plus a medial area.

+ side of a rational plus a medial area.

For let two straight lines AB, BC incommensurable in square, and fulfilling the given conditions [X. 34 ], be added together; I say that AC is irrational.

@@ -8598,7 +7024,7 @@ and, componendo, the squares on AB< PROPOSITION 41.

If two straight lines incommensurable in square which make the sum of the squares on them medial, and the rectangle contained by them medial and also incommensurable with the sum of the squares on them, be added together, the whole straight line is irrational; and let it be called the - side of the sum of two medial areas.

+ side of the sum of two medial areas.

For let two straight lines AB, BC incommensurable in square and satisfying the given conditions [X. 35 ] be added together; I say that AC is irrational.

@@ -8624,7 +7050,7 @@ and, componendo, the squares on AB<

Let the straight line AB be set out, let the whole be cut into unequal parts at each of the points C, D,

and let AC be supposed greater than DB; I say that the squares on AC, CB are greater than the squares on AD, DB.

For let AB be bisected at E.

Then, since AC is greater than DB, let DC be subtracted from each; therefore the remainder AD is greater than the remainder CB.

-

But AE is equal to EB; therefore DE is less than EC; therefore the points C, D are not equidistant from the point of bisection.

+

But AE is equal to EB; therefore DE is less than EC; therefore the points C, D are not equidistant from the point of bisection.

And, since the rectangle AC, CB together with the square on EC is equal to the square on EB, [II. 5 ] and, further, the rectangle AD, DB together with the square on DE is equal to the square on EB, [id.] therefore the rectangle AC, CB together with the square on EC is equal to the rectangle AD, DB together with the square on DE.

And of these the square on DE is less than the square on EC; therefore the remainder, the rectangle AC, CB, is also less than the rectangle AD, DB, so that twice the rectangle AC, CB is also less than twice the rectangle AD, DB.

@@ -8642,7 +7068,7 @@ and, componendo, the squares on AB<

Let AB be a binomial straight line divided into its terms at C; therefore AC, CB are rational straight lines commensurable in square only. [X. 36 ]

-

I say that AB is not divided at another point into two rational straight lines commensurable in square only.

+

I say that AB is not divided at another point into two rational straight lines commensurable in square only.

For, if possible, let it be divided at D also, so that AD, DB are also rational straight lines commensurable in square only.

It is then manifest that AC is not the same with DB.

For, if possible, let it be so.

@@ -8663,7 +7089,7 @@ and, componendo, the squares on AB<

For, if possible, let it be divided at D also, so that AD, DB are also medial straight lines commensurable in square only and containing a rational rectangle.

Since, then, that by which twice the rectangle AD, DB differs from twice the rectangle AC, CB is that by which the squares on AC, CB differ from the squares on AD, DB, while twice the rectangle AD, DB differs from twice the rectangle AC, CB by a rational area—for both are rational— therefore the squares on AC, CB also differ from the squares on AD, DB by a rational area, though they are medial: which is absurd. [x. 26 ]

-

Therefore a first bimedial straight line is not divided into its terms at different points; therefore it is so divided at one point only.

+

Therefore a first bimedial straight line is not divided into its terms at different points; therefore it is so divided at one point only.

PROPOSITION 44. @@ -8676,7 +7102,7 @@ and, componendo, the squares on AB<

Now let a rational straight line EF be set out, let there be applied to EF the rectangular parallelogram EK equal to the square on AB, and let EG equal to the squares on AC, CB be subtracted; therefore the remainder HK is equal to twice the rectangle AC, CB. [II. 4 ]

Again, let there be subtracted EL, equal to the squares on AD, DB, which were proved less than the squares on AC, CB [Lemma -]; therefore the remainder MK is also equal to twice the rectangle AD, DB.

+]; therefore the remainder MK is also equal to twice the rectangle AD, DB.

Now, since the squares on AC, CB are medial, therefore EG is medial.

And it is applied to the rational straight line EF; therefore EH is rational and incommensurable in length with EF. [X. 22 ]

@@ -8697,13 +7123,13 @@ and, componendo, the squares on AB<

But, if two rational straight lines commensurable in square only be added together, the whole is the irrational which is called binomial. [X. 36 ]

Therefore EN is a binomial straight line divided at H.

-

In the same way EM, MN will also be proved to be rational straight lines commensurable in square only; and EN will be a binomial straight line divided at different points, H and M.

+

In the same way EM, MN will also be proved to be rational straight lines commensurable in square only; and EN will be a binomial straight line divided at different points, H and M.

And EH is not the same with MN.

For the squares on AC, CB are greater than the squares on AD, DB.

But the squares on AD, DB are greater than twice the rectangle AD, DB; therefore also the squares on AC, CB, that is, EG, are much greater than twice the rectangle AD, DB, that is, MK, so that EH is also greater than MN.

Therefore EH is not the same with MN. Q. E. D.

 - + PROPOSITION 45.

A major straight line is divided at one and the same point only.

@@ -8714,7 +7140,7 @@ and, componendo, the squares on AB< ]

Therefore a major straight line is not divided at different points; therefore it is only divided at one and the same point. Q. E. D.

 - + PROPOSITION 46.

The side of a rational plus a medial area is divided at one point only.

@@ -8729,7 +7155,7 @@ and, componendo, the squares on AB< PROPOSITION 47.

The side of the sum of two medial areas is divided at one point only.

-

Let AB be divided at C, so that AC, CB are incommensurable in square and make the sum of the squares on AC, CB medial, and the rectangle AC, CB medial and also incommensurable with the sum of the squares on them; I say that AB is not divided at another point so as to fulfil the given conditions.

+

Let AB be divided at C, so that AC, CB are incommensurable in square and make the sum of the squares on AC, CB medial, and the rectangle AC, CB medial and also incommensurable with the sum of the squares on them; I say that AB is not divided at another point so as to fulfil the given conditions.

For, if possible, let it be divided at D, so that again AC is of course not the same as BD, but AC is supposed greater; let a rational straight line EF be set out, and let there be applied to EF the rectangle EG equal to the squares on AC, CB, and the rectangle HK equal to twice the rectangle AC, CB; therefore the whole EK is equal to the square on AB. [II. 4 ]

@@ -8741,7 +7167,7 @@ and, componendo, the squares on AB<

And, since the sum of the squares on AC, CB is incommensurable with twice the rectangle AC, CB, therefore EG is also incommensurable with GN, so that EH is also incommensurable with HN. [VI. 1 , X. 11 ]

-

And they are rational; therefore EH, HN are rational straight lines commensurable in square only; therefore EN is a binomial straight line divided at H. [X. 36 +

And they are rational; therefore EH, HN are rational straight lines commensurable in square only; therefore EN is a binomial straight line divided at H. [X. 36 ]

Similarly we can prove that it is also divided at M.

@@ -8760,7 +7186,7 @@ and, componendo, the squares on AB< 2

but if the lesser term be commensurable in length with the rational straight line set out, let the whole be called a second binomial;

- + 3

and if neither of the terms be commensurable in length with the rational straight line set out, let the whole be called a third binomial.

@@ -8788,7 +7214,7 @@ and, componendo, the squares on AB<

Let it be contrived that, as the number BA is to AC, so is the square on EF to the square on FG. [X. 6, Por.]

But AB has to AC the ratio which a number has to a number; therefore the square on EF also has to the square on FG the ratio which a number has to a number, so that the square on EF is commensurable with the square on FG. [X. 6]

And EF is rational; therefore FG is also rational.

-

And, since BA has not to AC the ratio which a square number has to a square number. neither, therefore, has the square on EF to the square on FG the ratio which a square number has to a square number; therefore EF is incommensurable in length with FG. [X. 9]

+

And, since BA has not to AC the ratio which a square number has to a square number. neither, therefore, has the square on EF to the square on FG the ratio which a square number has to a square number; therefore EF is incommensurable in length with FG. [X. 9]

Therefore EF, FG are rational straight lines commensurable in square only; therefore EG is binomial. [X. 36]

I say that it is also a first binomial straight line.

For since, as the number BA is to AC, so is the square on EF to the square on FG, while BA is greater than AC, therefore the square on EF is also greater than the square on FG.

@@ -8806,7 +7232,7 @@ and, componendo, the squares on AB<

Let it be contrived then that, as the number CA is to AB, so also is the square on EF to the square on FG; [X. 6, Por.] therefore the square on EF is commensurable with the square on FG. [X. 6]

Therefore FG is also rational.

-

Now, since the number CA has not to AB the ratio which a square number has to a square number, neither has the square on EF to the square on FG the ratio which a square number has to a square number.

+

Now, since the number CA has not to AB the ratio which a square number has to a square number, neither has the square on EF to the square on FG the ratio which a square number has to a square number.

Therefore EF is incommensurable in length with FG; [X. 9] therefore EF, FG are rational straight lines commensurable in square only; therefore EG is binomial. [X. 36]

It is next to be proved that it is also a second binomial straight line.

For since, inversely, as the number BA is to AC, so is the square on GF to the square on FE, while BA is greater than AC, therefore the square on GF is greater than the square on FE.

@@ -8823,7 +7249,7 @@ and, componendo, the squares on AB<

Let any other number D, not square, be set out also, and let it not have to either of the numbers BA. AC the ratio which a square number has to a square number.

Let any rational straight line E be set out, and let it be contrived that, as D is to AB, so is the square on E to the square on FG; [X. 6, Por.] therefore the square on E is commensurable with the square on FG. [X. 6]

-

And E is rational; therefore FG is also rational.

+

And E is rational; therefore FG is also rational.

And, since D has not to AB the ratio which a square number has to a square number, neither has the square on E to the square on FG the ratio which a square number has to a square number; therefore E is incommensurable in length with FG. [X. 9]

Next let it be contrived that, as the number BA is to AC, so is the square on FG to the square on GH; [X. 6, Por.] therefore the square on FG is commensurable with the square on GH. [X. 6]

But FG is rational; therefore GH is also rational.

@@ -8833,7 +7259,7 @@ and, componendo, the squares on AB<

For since, as D is to AB, so is the square on E to the square on FG, and, as BA is to AC, so is the square on FG to the square on GH, therefore, ex aequali, as D is to AC, so is the square on E to the square on GH. [V. 22]

But D has not to AC the ratio which a square number has to a square number; therefore neither has the square on E to the square on GH the ratio which a square number has to a square number; therefore E is incommensurable in length with GH. [X. 9]

And since, as BA is to AC, so is the square on FG to the square on GH, therefore the square on FG is greater than the square on GH.

-

Let then the squares on GH, K be equal to the square on FG; therefore, convertendo, as AB is to BC, so is the square on FG to the square on K. [V. 19, Por.]

+

Let then the squares on GH, K be equal to the square on FG; therefore, convertendo, as AB is to BC, so is the square on FG to the square on K. [V. 19, Por.]

But AB has to BC the ratio which a square number has to a square number; therefore the square on FG also has to the square on K the ratio which a square number has to a square number; therefore FG is commensurable in length with K. [X. 9]

Therefore the square on FG is greater than the square on GH by the square on a straight line commensurable with FG.

And FG, GH are rational straight lines commensurable in square only, and neither of them is commensurable in length with E.

@@ -8850,7 +7276,7 @@ and, componendo, the squares on AB<

Therefore EF, FG are rational straight lines commensurable in square only; so that EG is binomial.

I say next that it is also a fourth binomial straight line.

For since, as BA is to AC, so is the square on EF to the square on FG, therefore the square on EF is greater than the square on FG.

-

Let then the squares on FG, H be equal to the square on EF; therefore, convertendo, as the number AB is to BC, so is the square on EF to the square on H. [V. 19, Por.]

+

Let then the squares on FG, H be equal to the square on EF; therefore, convertendo, as the number AB is to BC, so is the square on EF to the square on H. [V. 19, Por.]

But AB has not to BC the ratio which a square number has to a square number; therefore neither has the square on EF to the square on H the ratio which a square number has to a square number.

Therefore EF is incommensurable in length with H; [X. 9] therefore the square on EF is greater than the square on GF by the square on a straight line incommensurable with EF.

And EF, FG are rational straight lines commensurable in square only, and EF is commensurable in length with D.

@@ -8859,7 +7285,7 @@ and, componendo, the squares on AB< PROPOSITION 52.

To find the fifth binomial straight line.

-

Let two numbers AC, CB be set out such that AB has not to either of them the ratio which a square number has to a square number; let any rational straight line D be set out, and let EF be commensurable with D; therefore EF is rational.

+

Let two numbers AC, CB be set out such that AB has not to either of them the ratio which a square number has to a square number; let any rational straight line D be set out, and let EF be commensurable with D; therefore EF is rational.

Let it be contrived that, as CA is to AB, so is the square on EF to the square on FG. [X. 6, Por.]

But CA has not to AB the ratio which a square number has to a square number; therefore neither has the square on EF to the square on FG the ratio which a square number has to a square number.

Therefore EF, FG are rational straight lines commensurable in square only; [X. 9] therefore EG is binomial. [X. 36]

@@ -8881,7 +7307,7 @@ and, componendo, the squares on AB<

Let any rational straight line E be set out, and let it be contrived that, as D is to AB, so is the square on E to the square on FG; [X. 6, Por.] therefore the square on E is commensurable with the square on FG. [X. 6]

And E is rational; therefore FG is also rational.

-

Now, since D has not to AB the ratio which a square number has to a square number, neither has the square on E to the square on FG the ratio which a square number has to a square number; therefore E is incommensurable in length with FG. [X. 9]

+

Now, since D has not to AB the ratio which a square number has to a square number, neither has the square on E to the square on FG the ratio which a square number has to a square number; therefore E is incommensurable in length with FG. [X. 9]

Again, let it be contrived that, as BA is to AC, so is the square on FG to the square on GH. [X. 6, Por.]

Therefore the square on FG is commensurable with the square on HG. [X. 6]

Therefore the square on HG is rational; therefore HG is rational.

@@ -8892,7 +7318,7 @@ and, componendo, the squares on AB<

But D has not to AC the ratio which a square number has to a square number; therefore neither has the square on E to the square on GH the ratio which a square number has to a square number; therefore E is incommensurable in length with GH. [X. 9]

But it was also proved incommensurable with FG; therefore each of the straight lines FG, GH is incommensurable in length with E.

And, since, as BA is to AC, so is the square on FG to the square on GH, therefore the square on FG is greater than the square on GH.

-

Let then the squares on GH, K be equal to the square on FG; therefore, convertendo, as AB is to BC, so is the square on FG to the square on K. [V. 19, Por.]

+

Let then the squares on GH, K be equal to the square on FG; therefore, convertendo, as AB is to BC, so is the square on FG to the square on K. [V. 19, Por.]

But AB has not to BC the ratio which a square number has to a square number; so that neither has the square on FG to the square on K the ratio which a square number has to a square number.

Therefore FG is incommensurable in length with K; [X. 9] therefore the square on FG is greater than the square on GH by the square on a straight line incommensurable with FG.

And FG, GH are rational straight lines commensurable in square only, and neither of them is commensurable in length with the rational straight line E set out.

@@ -8904,7 +7330,7 @@ and, componendo, the squares on AB<

Let the parallelogram AC be completed; I say that AC is a square, that DG is a mean proportional between AB, BC, and further that DC is a mean proportional between AC, CB.

For, since DB is equal to BF, and BE to BG, therefore the whole DE is equal to the whole FG.

-

But DE is equal to each of the straight lines AH, KC, and FG is equal to each of the straight lines AK, HC; [I. 34] therefore each of the straight lines AH, KC is also equal to each of the straight lines AK, HC.

+

But DE is equal to each of the straight lines AH, KC, and FG is equal to each of the straight lines AK, HC; [I. 34] therefore each of the straight lines AH, KC is also equal to each of the straight lines AK, HC.

Therefore the parallelogram AC is equilateral.

And it is also rectangular; therefore AC is a square.

And since, as FB is to BG, so is DB to BE, while, as FB is to BG, so is AB to DG, and, as DB is to BE, so is DG to BC, [VI. 1] therefore also, as AB is to DG, so is DG to BC. [V. 11]

@@ -8920,7 +7346,7 @@ and, componendo, the squares on AB< of the area is the irrational straight line which is called binomial.

For let the area AC be contained by the rational straight line AB and the first binomial AD; I say that the side of the area AC is the irrational straight line which is called binomial.

-

For, since AD is a first binomial straight line, let it be divided into its terms at E, and let AE be the greater term.

+

For, since AD is a first binomial straight line, let it be divided into its terms at E, and let AE be the greater term.

It is then manifest that AE, ED are rational straight lines commensurable in square only, the square on AE is greater than the square on ED by the square on a straight line commensurable with AE, and AE is commensurable in length with the rational straight line AB set out. [X. Deff. II. 1]

Let ED be bisected at the point F.

@@ -8929,7 +7355,7 @@ and, componendo, the squares on AB<

Let GH, EK, FL be drawn from G, E, F parallel to either of the straight lines AB, CD; let the square SN be constructed equal to the parallelogram AH, and the square NQ equal to GK, [II. 14] and let them be placed so that MN is in a straight line with NO; therefore RN is also in a straight line with NP.

And let the parallelogram SQ be completed; therefore SQ is a square. [Lemma]

Now, since the rectangle AG, GE is equal to the square on EF, therefore, as AG is to EF, so is FE to EG; [VI. 17] therefore also, as AH is to EL, so is EL to KG; [VI. 1] therefore EL is a mean proportional between AH, GK.

-

But AH is equal to SN, and GK to NQ; therefore EL is a mean proportional between SN, NQ.

+

But AH is equal to SN, and GK to NQ; therefore EL is a mean proportional between SN, NQ.

But MR is also a mean proportional between the same SN, NQ; [Lemma] therefore EL is equal to MR, so that it is also equal to PO.

But AH, GK are also equal to SN, NQ; therefore the whole AC is equal to the whole SQ, that is, to the square on MO; therefore MO is the side of AC.

@@ -8942,7 +7368,7 @@ and, componendo, the squares on AB<

But AH is equal to SN, and EL to MR; therefore SN is also incommensurable with MR.

But, as SN is to MR, so is PN to NR; [VI. 1] therefore PN is incommensurable with NR. [X. 11]

But PN is equal to MN, and NR to NO; therefore MN is incommensurable with NO.

-

And the square on MN is commensurable with the square on NO, and each is rational; therefore MN, NO are rational straight lines commensurable in square only.

+

And the square on MN is commensurable with the square on NO, and each is rational; therefore MN, NO are rational straight lines commensurable in square only.

Therefore MO is binomial [X. 36] and the side of AC. Q. E. D. 2. side. @@ -8959,7 +7385,7 @@ and, componendo, the squares on AB<

For let the area ABCD be contained by the rational straight line AB and the second binomial AD; I say that the side of the area AC is a first bimedial straight line.

For, since AD is a second binomial straight line, let it be divided into its terms at E, so that AE is the greater term; therefore AE, ED are rational straight lines commensurable in square only, the square on AE is greater than the square on ED by the square on a straight line commensurable with AE, and the lesser term ED is commensurable in length with AB. [X. Deff. II. 2]

-

Let ED be bisected at F, and let there be applied to AE the rectangle AG, GE equal to the square on EF and deficient by a square figure; therefore AG is commensurable in length with GE. [X. 17]

+

Let ED be bisected at F, and let there be applied to AE the rectangle AG, GE equal to the square on EF and deficient by a square figure; therefore AG is commensurable in length with GE. [X. 17]

Through G, E, F let GH, EK, FL be drawn parallel to AB, CD, let the square SN be constructed equal to the parallelogram AH, and the square NQ equal to GK, and let them be placed so that MN is in a straight line with NO; therefore RN is also in a straight line with NP.

Let the square SQ be completed.

@@ -8972,7 +7398,7 @@ and, componendo, the squares on AB<

Therefore BA, AG and BA, GE are pairs of rational straight lines commensurable in square only; so that each of the rectangles AH, GK is medial. [X. 21]

Hence each of the squares SN, NQ is medial.

Therefore MN, NO are also medial.

-

And, since AG is commensurable in length with GE, AH is also commensurable with GK, [VI. 1. X. 11] that is, SN is commensurable with NQ, that is, the square on MN with the square on NO.

+

And, since AG is commensurable in length with GE, AH is also commensurable with GK, [VI. 1. X. 11] that is, SN is commensurable with NQ, that is, the square on MN with the square on NO.

And, since AE is incommensurable in length with ED, while AE is commensurable with AG, and ED is commensurable with EF, therefore AG is incommensurable with EF; [X. 13] so that AH is also incommensurable with EL, that is, SN is incommensurable with MR, that is, PN with NR, [VI. 1, X. 11] that is, MN is incommensurable in length with NO.

But MN, NO were proved to be both medial and commensurable in square; therefore MN, NO are medial straight lines commensurable in square only.

I say next that they also contain a rational rectangle.

@@ -8984,7 +7410,7 @@ and, componendo, the squares on AB< to="DITTO">Therefore BA, AG and BA, GE are pairs of rational straight lines commensurable in square only. The text has Therefore BA, AG, GE are rational straight lines commensurable in square only, which I have altered because it would naturally convey the impression that any two of the three straight lines are commensurable in square only, whereas AG, GE are commensurable in length (I. 18), and it is only the other two pairs which are commensurable in square only.

- + PROPOSITION 56.

@@ -9000,7 +7426,7 @@ and, componendo, the squares on AB< of the area AC, and MN, NO are medial straight lines commensurable in square only; so that MO is bimedial.

It is next to be proved that it is also a second bimedial straight line.

Since DE is incommensurable in length with AB, that is, with EK, and DE is commensurable with EF, therefore EF is incommensurable in length with EK. [X. 13]

-

And they are rational; therefore FE, EK are rational straight lines commensurable in square only.

+

And they are rational; therefore FE, EK are rational straight lines commensurable in square only.

Therefore EL, that is, MR, is medial. [X. 21]

And it is contained by MN, NO; therefore the rectangle MN, NO is medial.

Therefore MO is a second bimedial straight line. [X. 38] Q. E. D.

@@ -9012,7 +7438,7 @@ and, componendo, the squares on AB< of the area is the irrational straight line called major.

For let the area AC be contained by the rational straight line AB and the fourth binomial AD divided into its terms at E, of which terms let AE be the greater; I say that the side - of the area AC is the irrational straight line called major.

+ of the area AC is the irrational straight line called major.

For, since AD is a fourth binomial straight line, therefore AE, ED are rational straight lines commensurable in square only, the square on AE is greater than the square on ED by the square on a straight line incommensurable with AE, and AE is commensurable in length with AB. [X. Deff. II. 4]

Let DE be bisected at F, and let there be applied to AE a parallelogram, the rectangle AG, GE, equal to the square on EF; therefore AG is incommensurable in length with GE. [X. 18]

Let GH, EK, FL be drawn parallel to AB, and let the rest of the construction be as before; it is then manifest that MO is the side @@ -9023,13 +7449,13 @@ and, componendo, the squares on AB<

And, since AE is commensurable with AB, AK is rational; [X. 19] and it is equal to the squares on MN, NO; therefore the sum of the squares on MN, NO is also rational.

And, since DE is incommensurable in length with AB, that is, with EK, while DE is commensurable with EF, therefore EF is incommensurable in length with EK. [X. 13]

Therefore EK, EF are rational straight lines commensurable in square only; therefore LE, that is, MR, is medial. [X. 21]

-

And it is contained by MN, NO; therefore the rectangle MN, NO is medial.

+

And it is contained by MN, NO; therefore the rectangle MN, NO is medial.

And the [sum] of the squares on MN, NO is rational, and MN, NO are incommensurable in square.

But, if two straight lines incommensurable in square and making the sum of the squares on them rational, but the rectangle contained by them medial, be added together, the whole is irrational and is called major. [X. 39]

Therefore MO is the irrational straight line called major and is the side of the area AC. Q. E. D.

 - + PROPOSITION 58.

@@ -9047,13 +7473,13 @@ and, componendo, the squares on AB<

And, since AD is a fifth binomial straight line, and ED the lesser segment, therefore ED is commensurable in length with AB. [X. Deff. II. 5]

But AE is incommensurable with ED; therefore AB is also incommensurable in length with AE. [X. 13]

Therefore AK, that is, the sum of the squares on MN, NO, is medial. [X. 21]

-

And, since DE is commensurable in length with AB, that is, with EK, while DE is commensurable with EF, therefore EF is also commensurable with EK. [X. 12]

+

And, since DE is commensurable in length with AB, that is, with EK, while DE is commensurable with EF, therefore EF is also commensurable with EK. [X. 12]

And EK is rational; therefore EL, that is, MR, that is, the rectangle MN, NO, is also rational. [X. 19]

Therefore MN, NO are straight lines incommensurable in square which make the sum of the squares on them medial, but the rectangle contained by them rational.

Therefore MO is the side of a rational plus a medial area [X. 40] and is the side of the area AC. Q. E. D.

 - + PROPOSITION 59.

@@ -9072,7 +7498,7 @@ and, componendo, the squares on AB<

And, since AE is incommensurable with EF, AK is also incommensurable with EL. [VI. 1, X. 11]

But AK is the sum of the squares on MN, NO, and EL is the rectangle MN, NO; therefore the sum of the squares on MN, NO is incommensurable with the rectangle MN, NO.

-

And each of them is medial, and MN, NO are incommensurable in square.

+

And each of them is medial, and MN, NO are incommensurable in square.

Therefore MO is the side of the sum of two medial areas [X. 41], and is the side of AC. Q. E. D.

@@ -9080,7 +7506,7 @@ and, componendo, the squares on AB< [LEMMA.

If a straight line be cut into unequal parts, the squares on the unequal parts are greater than twice the rectangle contained by the unequal parts.

-

Let AB be a straight line, and let it be cut into unequal parts at C, and let AC be the greater; I say that the squares on AC, CB are greater than twice the rectangle AC, CB.

+

Let AB be a straight line, and let it be cut into unequal parts at C, and let AC be the greater; I say that the squares on AC, CB are greater than twice the rectangle AC, CB.

For let AB be bisected at D.

Since then a straight line has been cut into equal parts at D, and into unequal parts at C, therefore the rectangle AC, CB together with the square on CD is equal to the square on AD, [II. 5] so that the rectangle AC, CB is less than double of the square on AD.

But the squares on AC, CB are double of the squares on AD, DC; [II. 9] therefore the squares on AC, CB are greater than twice the rectangle AC, CB. Q. E. D.]

@@ -9094,7 +7520,7 @@ and, componendo, the squares on AB<

For let there be applied to DE the rectangle DH equal to the square on AC, and KL equal to the square on BC; therefore the remainder, twice the rectangle AC, CB, is equal to MF.

Let MG be bisected at N, and let NO be drawn parallel [to ML or GF].

Therefore each of the rectangles MO, NF is equal to once the rectangle AC, CB.

-

Now, since AB is a binomial divided into its terms at C, therefore AC, CB are rational straight lines commensurable in square only; [X. 36] therefore the squares on AC, CB are rational and commensurable with one another, so that the sum of the squares on AC, CB is also rational. [X. 15]

+

Now, since AB is a binomial divided into its terms at C, therefore AC, CB are rational straight lines commensurable in square only; [X. 36] therefore the squares on AC, CB are rational and commensurable with one another, so that the sum of the squares on AC, CB is also rational. [X. 15]

And it is equal to DL; therefore DL is rational.

And it is applied to the rational straight line DE; therefore DM is rational and commensurable in length with DE. [X. 20]

Again, since AC, CB are rational straight lines commensurable in square only, therefore twice the rectangle AC, CB, that is MF, is medial. [X. 21]

@@ -9104,7 +7530,7 @@ and, componendo, the squares on AB<

It is next to be proved that it is also a first binomial straight line.

Since the rectangle AC, CB is a mean proportional between the squares on AC, CB, [cf. Lemma after X. 53] therefore MO is also a mean proportional between DH, KL.

Therefore, as DH is to MO, so is MO to KL, that is, as DK is to MN, so is MN to MK; [VI. 1] therefore the rectangle DK, KM is equal to the square on MN. [VI. 17]

-

And, since the square on AC is commensurable with the square on CB, DH is also commensurable with KL, so that DK is also commensurable with KM. [VI. 1, X. 11]

+

And, since the square on AC is commensurable with the square on CB, DH is also commensurable with KL, so that DK is also commensurable with KM. [VI. 1, X. 11]

And, since the squares on AC, CB are greater than twice the rectangle AC, CB, [Lemma] therefore DL is also greater than MF, so that DM is also greater than MG. [VI. 1]

And the rectangle DK, KM is equal to the square on MN, that is, to the fourth part of the square on MG, and DK is commensurable with KM.

But, if there be two unequal straight lines, and to the greater there be applied a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, and if it divide it into commensurable parts, the square on the greater is greater than the square on the less by the square on a straight line commensurable with the greater; [X. 17] therefore the square on DM is greater than the square on MG by the square on a straight line commensurable with DM.

@@ -9116,7 +7542,7 @@ and, componendo, the squares on AB<

The square on the first bimedial straight line applied to a rational straight line produces as breadth the second binomial.

Let AB be a first bimedial straight line divided into its medials at C, of which medials AC is the greater; let a rational straight line DE be set out, and let there be applied to DE the parallelogram DF equal to the square on AB, producing DG as its breadth; I say that DG is a second binominal straight line.

-

For let the same construction as before be made.

+

For let the same construction as before be made.

Then, since AB is a first bimedial divided at C, therefore AC, CB are medial straight lines commensurable in square only, and containing a rational rectangle, [X. 37] so that the squares on AC, CB are also medial. [X. 21]

Therefore DL is medial. [X. 15 and 23, Por.]

And it has been applied to the rational straight line DE; therefore MD is rational and incommensurable in length with DE. [X. 22]

@@ -9134,7 +7560,7 @@ and, componendo, the squares on AB< PROPOSITION 62.

The square on the second bimedial straight line applied to a rational straight line produces as breadth the third binomial.

Let AB be a second bimedial straight line divided into its medials at C, so that AC is the greater segment; let DE be any rational straight line, and to DE let there be applied the parallelogram DF equal to the square on AB and producing DG as its breadth;

I say that DG is a third binomial straight line.

-

Let the same construction be made as before shown.

+

Let the same construction be made as before shown.

Then, since AB is a second bimedial divided at C, therefore AC, CB are medial straight lines commensurable in square only and containing a medial rectangle, [X. 38] so that the sum of the squares on AC, CB is also medial. [X. 15 and 23 Por.]

And it is equal to DL; therefore DL is also medial.

And it is applied to the rational straight line DE; therefore MD is also rational and incommensurable in length with DE. [X. 22]

@@ -9146,12 +7572,12 @@ and, componendo, the squares on AB<

In manner similar to the foregoing we may conclude that DM is greater than MG, and that DK is commensurable with KM.

And the rectangle DK, KM is equal to the square on MN; therefore the square on DM is greater than the square on MG by the square on a straight line commensurable with DM.

And neither of the straight lines DM, MG is commensurable in length with DE.

-

Therefore DG is a third binomial straight line. [X. Deff. II. 3] Q. E. D.

+

Therefore DG is a third binomial straight line. [X. Deff. II. 3] Q. E. D.

PROPOSITION 63.

The square on the major straight line applied to a rational straight line produces as breadth the fourth binomial.

-

Let AB be a major straight line divided at C, so that AC is greater than CB; let DE be a rational straight line, and to DE let there be applied the parallelogram DF equal to the square on AB and producing DG as its breadth; I say that DG is a fourth binomial straight line.

+

Let AB be a major straight line divided at C, so that AC is greater than CB; let DE be a rational straight line, and to DE let there be applied the parallelogram DF equal to the square on AB and producing DG as its breadth; I say that DG is a fourth binomial straight line.

Let the same construction be made as before shown.

Then, since AB is a major straight line divided at C, AC, CB are straight lines incommensurable in square which make the sum of the squares on them rational, but the rectangle contained by them medial. [X. 39]

@@ -9161,11 +7587,11 @@ and, componendo, the squares on AB<

It is to be proved that it is also a fourth binomial straight line.

In manner similar to the foregoing we can prove that DM is greater than MG, and that the rectangle DK, KM is equal to the square on MN.

Since then the square on AC is incommensurable with the square on CB, therefore DH is also incommensurable with KL, so that DK is also incommensurable with KM. [VI. 1, X. 11]

-

But, if there be two unequal straight lines, and to the greater there be applied a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, and if it divide it into incommensurable parts, then the square on the greater will be greater than the square on the less by the square on a straight line incommensurable in length with the greater; [X. 18] therefore the square on DM is greater than the square on MG by the square on a straight line incommensurable with DM.

+

But, if there be two unequal straight lines, and to the greater there be applied a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, and if it divide it into incommensurable parts, then the square on the greater will be greater than the square on the less by the square on a straight line incommensurable in length with the greater; [X. 18] therefore the square on DM is greater than the square on MG by the square on a straight line incommensurable with DM.

And DM, MG are rational straight lines commensurable in square only, and DM is commensurable with the rational straight line DE set out.

Therefore DG is a fourth binomial straight line. [X. Deff. II. 4] Q. E. D.

- + PROPOSITION 64.

The square on the side of a rational plus a medial area applied to a rational straight line produces as breadth the fifth binomial.

@@ -9177,14 +7603,14 @@ and, componendo, the squares on AB<

Again, since twice the rectangle AC, CB, that is MF, is rational, therefore MG is rational and commensurable with DE. [X. 20]

Therefore DM is incommensurable with MG; [X. 13] therefore DM, MG are rational straight lines commensurable in square only; therefore DG is binomial. [X. 36]

I say next that it is also a fifth binomial straight line.

-

For it can be proved similarly that the rectangle DK, KM is equal to the square on MN, and that DK is incommensurable in length with KM; therefore the square on DM is greater than the square on MG by the square on a straight line incommensurable with DM. [X. 18]

+

For it can be proved similarly that the rectangle DK, KM is equal to the square on MN, and that DK is incommensurable in length with KM; therefore the square on DM is greater than the square on MG by the square on a straight line incommensurable with DM. [X. 18]

And DM, MG are commensurable in square only, and the less, MG, is commensurable in length with DE.

Therefore DG is a fifth binomial. Q. E. D.

 PROPOSITION 65.

The square on the side of the sum of two medial areas applied to a rational straight line produces as breadth the sixth binomial.

-

Let AB be the side of the sum of two medial areas, divided at C, let DE be a rational straight line, and let there be applied to DE the parallelogram DF equal to the square on AB, producing DG as its breadth; I say that DG is a sixth binomial straight line.

+

Let AB be the side of the sum of two medial areas, divided at C, let DE be a rational straight line, and let there be applied to DE the parallelogram DF equal to the square on AB, producing DG as its breadth; I say that DG is a sixth binomial straight line.

For let the same construction be made as before.

Then, since AB is the side of the sum of two medial areas, divided at C, therefore AC, CB are straight lines incommensurable in square which make the sum of the squares on them medial, the rectangle contained by them medial, and moreover the sum of the squares on them incommensurable with the rectangle contained by them, [X. 41]

so that, in accordance with what was before proved, each of the rectangles DL, MF is medial.

And they are applied to the rational straight line DE; therefore each of the straight lines DM, MG is rational and incommensurable in length with DE. [X. 22]

@@ -9193,13 +7619,13 @@ and, componendo, the squares on AB<

I say next that it is also a sixth binomial straight line.

Similarly again we can prove that the rectangle DK, KM is equal to the square on MN, and that DK is incommensurable in length with KM; and, for the same reason, the square on DM is greater than the square on MG by the square on a straight line incommensurable in length with DM.

And neither of the straight lines DM, MG is commensurable in length with the rational straight line DE set out.

-

Therefore DG is a sixth binomial straight line. Q. E. D.

+

Therefore DG is a sixth binomial straight line. Q. E. D.

PROPOSITION 66.

A straight line commensurable in length with a binomial straight line is itself also binomial and the same in order.

Let AB be binomial, and let CD be commensurable in length with AB;

I say that CD is binomial and the same in order with AB.

-

For, since AB is binomial, let it be divided into its terms at E, and let AE be the greater term; therefore AE, EB are rational straight lines commensurable in square only. [X. 36]

+

For, since AB is binomial, let it be divided into its terms at E, and let AE be the greater term; therefore AE, EB are rational straight lines commensurable in square only. [X. 36]

Let it be contrived that, as AB is to CD, so is AE to CF; [VI. 12] therefore also the remainder EB is to the remainder FD as AB is to CD. [V. 19]

But AB is commensurable in length with CD; therefore AE is also commensurable with CF, and EB with FD. [X. 11]

And AE, EB are rational; therefore CF, FD are also rational.

@@ -9211,7 +7637,7 @@ and, componendo, the squares on AB<

For the square on AE is greater than the square on EB either by the square on a straight line commensurable with AE or by the square on a straight line incommensurable with it.

If then the square on AE is greater than the square on EB by the square on a straight line commensurable with AE, the square on CF will also be greater than the square on FD by the square on a straight line commensurable with CF. [X. 14]

And, if AE is commensurable with the rational straight line set out, CF will also be commensurable with it, [X. 12] and for this reason each of the straight lines AB, CD is a first binomial, that is, the same in order. [X. Deff. II. 1]

-

But, if EB is commensurable with the rational straight line set out, FD is also commensurable with it, [X. 12] and for this reason again CD will be the same in order with AB, for each of them will be a second binomial. [X. Deff. II. 2]

+

But, if EB is commensurable with the rational straight line set out, FD is also commensurable with it, [X. 12] and for this reason again CD will be the same in order with AB, for each of them will be a second binomial. [X. Deff. II. 2]

But, if neither of the straight lines AE, EB is commensurable with the rational straight line set out, neither of the straight lines CF, FD will be commensurable with it, [X. 13] and each of the straight lines AB, CD is a third binomial. [X. Deff. II. 3]

But, if the square on AE is greater than the square on EB by the square on a straight line incommensurable with AE, the square on CF is also greater than the square on FD by the square on a straight line incommensurable with CF. [X. 14]

And, if AE is commensurable with the rational straight line set out, CF is also commensurable with it, and each of the straight lines AB, CD is a fourth binomial. [X. Deff. II. 4]

@@ -9224,7 +7650,7 @@ and, componendo, the squares on AB<

A straight line commensurable in length with a bimedial straight line is itself also bimedial and the same in order.

Let AB be bimedial, and let CD be commensurable in length with AB; I say that CD is bimedial and the same in order with AB.

-

For, since AB is bimedial, let it be divided into its medials at E; therefore AE, EB are medial straight lines commensurable in square only. [X. 37, 38]

+

For, since AB is bimedial, let it be divided into its medials at E; therefore AE, EB are medial straight lines commensurable in square only. [X. 37, 38]

And let it be contrived that, as AB is to CD, so is AE to CF; therefore also the remainder EB is to the remainder FD as AB is to CD. [V. 19]

But AB is commensurable in length with CD; therefore AE, EB are also commensurable with CF, FD respectively. [X. 11]

But AE, EB are medial; therefore CF, FD are also medial. [X. 23]

@@ -9236,7 +7662,7 @@ and, componendo, the squares on AB<

If therefore the rectangle AE, EB is rational, the rectangle CF, FD is also rational, [and for this reason CD is a first bimedial]; [X. 37] but if medial, medial, [X. 23, Por.] and each of the straight lines AB, CD is a second bimedial. [X. 38]

And for this reason CD will be the same in order with AB. Q. E. D.

- + PROPOSITION 68.

A straight line commensurable with a major straight line is itself also major.

@@ -9249,7 +7675,7 @@ and, componendo, the squares on AB<

And since, as AE is to CF, so is EB to FD, alternately also, as AE is to EB, so is CF to FD; [V. 16] therefore also, componendo, as AB is to BE, so is CD to DF; [V. 18] therefore also, as the square on AB is to the square on BE, so is the square on CD to the square on DF. [VI. 20]

Similarly we can prove that, as the square on AB is to the square on AE, so also is the square on CD to the square on CF.

Therefore also, as the square on AB is to the squares on AE, EB, so is the square on CD to the squares on CF, FD; therefore also, alternately, as the square on AB is to the square on CD, so are the squares on AE, EB to the squares on CF, FD. [V. 16]

-

But the square on AB is commensurable with the square on CD; therefore the squares on AE, EB are also commensurable with the squares on CF, FD.

+

But the square on AB is commensurable with the square on CD; therefore the squares on AE, EB are also commensurable with the squares on CF, FD.

And the squares on AE, EB together are rational; therefore the squares on CF, FD together are rational.

Similarly also twice the rectangle AE, EB is commensurable with twice the rectangle CF, FD.

And twice the rectangle AE, EB is medial; therefore twice the rectangle CF, FD is also medial. [X. 23, Por.]

@@ -9266,7 +7692,7 @@ and, componendo, the squares on AB<

We can then prove similarly that CF, FD are incommensurable in square, and the sum of the squares on AE, EB is commensurable with the sum of the squares on CF, FD, and the rectangle AE, EB with the rectangle CF, FD; so that the sum of the squares on CF, FD is also medial, and the rectangle CF, FD rational.

Therefore CD is the side of a rational plus a medial area. Q. E. D.

 - + PROPOSITION 70.

A straight line commensurable with the side of the sum of two medial areas is the side of the sum of two medial areas.

@@ -9280,7 +7706,7 @@ and, componendo, the squares on AB< PROPOSITION 71.

If a rational and a medial area be added together, four irrational straight lines arise, namely a binomial or a first bimedial or a major or a side of a rational plus a medial area.

Let. AB be rational, and CD medial; I say that the side - of the area AD is a binomial or a first bimedial or a major or a side of a rational plus a medial area.

+ of the area AD is a binomial or a first bimedial or a major or a side of a rational plus a medial area.

For AB is either greater or less than CD.

First, let it be greater; let a rational straight line EF be set out, let there be applied to EF the rectangle EG equal to AB, producing EH as breadth, and let HI, equal to DC, be applied to EF, producing HK as breadth.

@@ -9290,7 +7716,7 @@ and, componendo, the squares on AB<

And it is applied to the rational straight line EF, producing HK as breadth; therefore HK is rational and incommensurable in length with EF [X. 22]

And, since CD is medial, while AB is rational, therefore AB is incommensurable with CD, so that EG is also incommensurable with HI.

But, as EG is to HI, so is EH to HK; [VI. 1] therefore EH is also incommensurable in length with HK. [X. 11]

-

And both are rational; therefore EH, HK are rational straight lines commensurable in square only; therefore EK is a binomial straight line, divided at H. [X. 36]

+

And both are rational; therefore EH, HK are rational straight lines commensurable in square only; therefore EK is a binomial straight line, divided at H. [X. 36]

And, since AB is greater than CD, while AB is equal to EG and CD to HI, therefore EG is also greater than HI; therefore EH is also greater than HK.

The square, then, on EH is greater than the square on HK either by the square on a straight line commensurable in length with EH or by the square on a straight line incommensurable with it.

First, let the square on it be greater by the square on a straight line commensurable with itself.

@@ -9307,7 +7733,7 @@ and, componendo, the squares on AB< of the area EI is major; so that the side of the area AD is also major.

Next, let AB be less than CD; therefore EG is also less than HI, so that EH is also less than HK.

-

Now the square on HK is greater than the square on EH either by the square on a straight line commensurable with HK or by the square on a straight line incommensurable with it.

+

Now the square on HK is greater than the square on EH either by the square on a straight line commensurable with HK or by the square on a straight line incommensurable with it.

First, let the square on it be greater by the square on a straight line commensurable in length with itself.

Now the lesser straight line EH is commensurable in length with the rational straight line EF set out; therefore EK is a second binomial. [X. Deff. II. 2]

But EF is rational, and, if an area be contained by a rational straight line and the second binomial, the side of the square equal to it is a first bimedial; [X. 55] therefore the side @@ -9329,7 +7755,7 @@ and, componendo, the squares on AB<

For AB is either greater or less than CD.

First, if it so chance, let AB be greater than CD.

-

Let the rational straight line EF be set out, and to EF let there be applied the rectangle EG equal to AB and producing EH as breadth, and the rectangle HI equal to CD and producing HK as breadth.

+

Let the rational straight line EF be set out, and to EF let there be applied the rectangle EG equal to AB and producing EH as breadth, and the rectangle HI equal to CD and producing HK as breadth.

Now, since each of the areas AB, CD is medial, therefore each of the areas EG, HI is also medial.

And they are applied to the rational straight line FE, producing EH, HK as breadth; therefore each of the straight lines EH, HK is rational and incommensurable in length with EF. [X. 22]

And, since AB is incommensurable with CD, and AB is equal to EG, and CD to HI, therefore EG is also incommensurable with HI.

@@ -9343,7 +7769,7 @@ and, componendo, the squares on AB< of EI, that is, of AD, is a second bimedial.

Next, let the square on EH be greater than the square on HK by the square on a straight line incommensurable in length with EH.

Now each of the straight lines EH, HK is incommensurable in length with EF; therefore EK is a sixth binomial. [X. Deff. II. 6]

-

But, if an area be contained by a rational straight line and the sixth binomial, the side +

But, if an area be contained by a rational straight line and the sixth binomial, the side of the area is the side of the sum of two medial areas; [X. 59] so that the side of the area AD is also the side of the sum of two medial areas.

Therefore etc. Q. E. D.

@@ -9354,7 +7780,7 @@ and, componendo, the squares on AB< an apotome.

For from the rational straight line AB let the rational straight line BC, commensurable with the whole in square only, be subtracted;

I say that the remainder AC is the irrational straight line called apotome.

For, since AB is incommensurable in length with BC, and, as AB is to BC, so is the square on AB to the rectangle AB, BC, therefore the square on AB is incommensurable with the rectangle AB, BC. [X. 11]

-

But the squares on AB, BC are commensurable with the square on AB, [X. 15] and twice the rectangle AB, BC is commensurable with the rectangle AB, BC. [X. 6]

+

But the squares on AB, BC are commensurable with the square on AB, [X. 15] and twice the rectangle AB, BC is commensurable with the rectangle AB, BC. [X. 6]

And, inasmuch as the squares on AB, BC are equal to twice the rectangle AB, BC together with the square on CA, [II. 7] therefore the squares on AB, BC are also incommensurable with the remainder, the square on AC. [X. 13, 16]

But the squares on AB, BC are rational; therefore AC is irrational. [X. Def. 4]

And let it be called an apotome. Q. E. D.

@@ -9363,14 +7789,14 @@ and, componendo, the squares on AB< PROPOSITION 74.

If from a medial straight line there be subtracted a medial straight line which is commensurable with the whole in square only, and which contains with the whole a rational rectangle, the remainder is irrational. And let it be called a first apotome of a medial - straight line.

+ straight line.

For from the medial straight line AB let there be subtracted the medial straight line BC which is commensurable with AB in square only and with AB makes the rectangle AB, BC rational;

I say that the remainder AC is irrational; and let it be called a first apotome of a medial straight line.

For, since AB, BC are medial, the squares on AB, BC are also medial.

But twice the rectangle AB, BC is rational; therefore the squares on AB, BC are incommensurable with twice the rectangle AB, BC; therefore twice the rectangle AB, BC is also incommensurable with the remainder, the square on AC, [Cf. II. 7] since, if the whole is incommensurable with one of the magnitudes, the original magnitudes will also be incommensurable. [X. 16]

But twice the rectangle AB, BC is rational; therefore the square on AC is irrational; therefore AC is irrational. [X. Def. 4]

And let it be called a first apotome of a medial straight line.

- + PROPOSITION 75

If from a medial straight line there be subtracted a medial straight line which is commensurable with the whole in square only, and which contains with the whole a medial rectangle, the remainder is irrational; and let it be called a @@ -9381,7 +7807,7 @@ and, componendo, the squares on AB<

For let a rational straight line DI be set out, let DE equal to the squares on AB, BC be applied to DI, producing DG as breadth, and let DH equal to twice the rectangle AB, BC be applied to DI, producing DF as breadth; therefore the remainder FE is equal to the square on AC. [II. 7]

Now, since the squares on AB, BC are medial and commensurable, therefore DE is also medial. [X. 15 and 23, Por.]

And it is applied to the rational straight line DI, producing DG as breadth; therefore DG is rational and incommensurable in length with DI. [X. 22]

-

Again, since the rectangle AB, BC is medial, therefore twice the rectangle AB, BC is also medial. [X. 23, Por.]

+

Again, since the rectangle AB, BC is medial, therefore twice the rectangle AB, BC is also medial. [X. 23, Por.]

And it is equal to DH; therefore DH is also medial.

And it has been applied to the rational straight line DI, producing DF as breadth; therefore DF is rational and incommensurable in length with DI. [X. 22]

And, since AB, BC are commensurable in square only, therefore AB is incommensurable in length with BC; therefore the square on AB is also incommensurable with the rectangle AB, BC. [X. 11]

@@ -9402,7 +7828,7 @@ and, componendo, the squares on AB<

I say that the remainder AC is the irrational straight line called minor.

For, since the sum of the squares on AB, BC is rational, while twice the rectangle AB, BC is medial, therefore the squares on AB, BC are incommensurable with twice the rectangle AB, BC; and, convertendo, the squares on AB, BC are incommensurable with the remainder, the square on AC. [II. 7, X. 16]

But the squares on AB, BC are rational; therefore the square on AC is irrational; therefore AC is irrational.

-

And let it be called minor.

+

And let it be called minor.

 PROPOSITION 77 @@ -9412,7 +7838,7 @@ and, componendo, the squares on AB<

For, since the sum of the squares on AB, BC is medial, while twice the rectangle AB, BC is rational, therefore the squares on AB, BC are incommensurable with twice the rectangle AB, BC; therefore the remainder also, the square on AC, is incommensurable with twice the rectangle AB, BC. [II. 7, X. 16]

And twice the rectangle AB, BC is rational; therefore the square on AC is irrational; therefore AC is irrational.

-

And let it be called that which produces with a rational area a medial whole. Q. E. D.

+

And let it be called that which produces with a rational area a medial whole. Q. E. D.

PROPOSITION 78 @@ -9421,7 +7847,7 @@ and, componendo, the squares on AB< that which produces with a medial area a medial whole.

For from the straight line AB let there be subtracted the straight line BC incommensurable in square with AB and fulfilling the given conditions; [X. 35]

I say that the remainder AC is the irrational straight line called that which produces with a medial area a medial whole.

-

For let a rational straight line DI be set out, to DI let there be applied DE equal to the squares on AB, BC, producing DG as breadth, and let DH equal to twice the rectangle AB, BC be subtracted.

+

For let a rational straight line DI be set out, to DI let there be applied DE equal to the squares on AB, BC, producing DG as breadth, and let DH equal to twice the rectangle AB, BC be subtracted.

Therefore the remainder FE is equal to the square on AC, [II. 7] so that AC is the side of FE.

Now, since the sum of the squares on AB, BC is medial and is equal to DE, therefore DE is medial.

@@ -9436,7 +7862,7 @@ and, componendo, the squares on AB< is irrational.

And AC is the side of FE; therefore AC is irrational.

-

And let it be called that which produces with a medial area a medial whole. Q. E. D.

+

And let it be called that which produces with a medial area a medial whole. Q. E. D.

PROPOSITION 79 @@ -9445,14 +7871,14 @@ and, componendo, the squares on AB<

I say that no other rational straight line can be annexed to AB which is commensurable with the whole in square only.

For, if possible, let BD be so annexed; therefore AD, DB are also rational straight lines commensurable in square only. [X. 73]

-

Now, since the excess of the squares on AD, DB over twice the rectangle AD, DB is also the excess of the squares on AC, CB over twice the rectangle AC, CB, for both exceed by the same, the square on AB, [II. 7] therefore, alternately, the excess of the squares on AD, DB over the squares on AC, CB is the excess of twice the rectangle AD, DB over twice the rectangle AC, CB.

+

Now, since the excess of the squares on AD, DB over twice the rectangle AD, DB is also the excess of the squares on AC, CB over twice the rectangle AC, CB, for both exceed by the same, the square on AB, [II. 7] therefore, alternately, the excess of the squares on AD, DB over the squares on AC, CB is the excess of twice the rectangle AD, DB over twice the rectangle AC, CB.

But the squares on AD, DB exceed the squares on AC, CB by a rational area, for both are rational; therefore twice the rectangle AD, DB also exceeds twice the rectangle AC, CB by a rational area: which is impossible, for both are medial [X. 21], and a medial area does not exceed a medial by a rational area. [X. 26]

Therefore no other rational straight line can be annexed to AB which is commensurable with the whole in square only.

Therefore only one rational straight line can be annexed to an apotome which is commensurable with the whole in square only. Q. E. D.

 PROPOSITION 80. -

To a first apotome of a medial straight line only one medial straight line can be annexed which is commensurable with the whole in square only and which contains with the whole a rational rectangle.

+

To a first apotome of a medial straight line only one medial straight line can be annexed which is commensurable with the whole in square only and which contains with the whole a rational rectangle.

For let AB be a first apotome of a medial straight line, and let BC be an annex to AB; therefore AC, CB are medial straight lines commensurable in square only and such that the rectangle AC, CB which they contain is rational; [X. 74] I say that no other medial straight line can be annexed to AB which is commensurable with the whole in square only and which contains with the whole a rational area.

For, if possible, let DB also be so annexed; therefore AD, DB are medial straight lines commensurable in square only and such that the rectangle AD, DB which they contain is rational. [X. 74]

@@ -9461,7 +7887,7 @@ and, componendo, the squares on AB<

Therefore the squares on AD, DB also exceed the squares on AC, CB by a rational area. which is impossible, for both are medial [X. 15 and 23, Por.], and a medial area does not exceed a medial by a rational area. [X. 26]

Therefore etc. Q. E. D.

- + PROPOSITION 81.

To a second apotome of a medial straight line only one medial straight line can be annexed which is commensurable with the whole in square only and which contains with the whole a medial rectangle.

@@ -9473,7 +7899,7 @@ and, componendo, the squares on AB< of EL.

Again, let EI equal to the squares on AD, DB be applied to EF, producing EN as breadth.

But EL is also equal to the square on AB; therefore the remainder HI is equal to twice the rectangle AD, DB. [II. 7]

-

Now, since AC, CB are medial straight lines, therefore the squares on AC, CB are also medial.

+

Now, since AC, CB are medial straight lines, therefore the squares on AC, CB are also medial.

And they are equal to EG; therefore EG is also medial. [X. 15 and 23, Por.]

And it is applied to the rational straight line EF, producing EM as breadth; therefore EM is rational and incommensurable in length with EF. [X. 22]

Again, since the rectangle AC, CB is medial, twice the rectangle AC, CB is also medial. [X. 23, Por.]

@@ -9484,14 +7910,14 @@ and, componendo, the squares on AB<

But the squares on AC, CB are commensurable with the square on AC, while twice the rectangle AC, CB is commensurable with the rectangle AC, CB; [X. 6] therefore the squares on AC, CB are incommensurable with twice the rectangle AC, CB. [X. 13]

And EG is equal to the squares on AC, CB, while GH is equal to twice the rectangle AC, CB; therefore EG is incommensurable with HG.

But, as EG is to HG, so is EM to HM; [VI. 1] therefore EM is incommensurable in length with MH. [X. 11]

-

And both are rational; therefore EM, MH are rational straight lines commensurable in square only; therefore EH is an apotome, and HM an annex to it. [X. 73]

+

And both are rational; therefore EM, MH are rational straight lines commensurable in square only; therefore EH is an apotome, and HM an annex to it. [X. 73]

Similarly we can prove that HN is also an annex to it; therefore to an apotome different straight lines are annexed which are commensurable with the wholes in square only: which is impossible. [X. 79]

Therefore etc. Q. E. D.

 PROPOSITION 82.

To a minor straight line only one straight line can be annexed which is incommensurable in square with the whole and which makes, with the whole, the sum of the squares on them rational but twice the rectangle contained by them medial.

-

Let AB be the minor straight line, and let BC be an annex to AB;

therefore AC, CB are straight lines incommensurable in square which make the sum of the squares on them rational, but twice the rectangle contained by them medial. [X. 76]

+

Let AB be the minor straight line, and let BC be an annex to AB;

therefore AC, CB are straight lines incommensurable in square which make the sum of the squares on them rational, but twice the rectangle contained by them medial. [X. 76]

I say that no other straight line can be annexed to AB fulfilling the same conditions.

For, if possible, let BD be so annexed; therefore AD, DB are also straight lines incommensurable in square which fulfil the aforesaid conditions. [X. 76]

Now, since the excess of the squares on AD, DB over the squares on AC, CB is also the excess of twice the rectangle AD, DB over twice the rectangle AC, CB, while the squares on AD, DB exceed the squares on AC, CB by a rational area, for both are rational, therefore twice the rectangle AD, DB also exceeds twice the rectangle AC, CB by a rational area: which is impossible, for both are medial. [X. 26]

@@ -9499,7 +7925,7 @@ and, componendo, the squares on AB< PROPOSITION 83. -

To a straight line which produces with a rational area a medial whole only one straight line can be annexed which is incommensurable in square with the whole straight line and which with the whole straight line makes the sum of the squares on them medial, but twice the rectangle contained by them rational.

+

To a straight line which produces with a rational area a medial whole only one straight line can be annexed which is incommensurable in square with the whole straight line and which with the whole straight line makes the sum of the squares on them medial, but twice the rectangle contained by them rational.

Let AB be the straight line which produces with a rational area a medial whole, and let BC be an annex to AB; therefore AC, CB are straight lines incommensurable in square which fulfil the given conditions. [X. 77]

I say that no other straight line can be annexed to AB which fulfils the same conditions.

@@ -9509,7 +7935,7 @@ and, componendo, the squares on AB< PROPOSITION 84. -

To a straight line which produces with a medial area a medial whole only one straight line can be annexed which is incommensurable in square with the whole straight line and which with the whole straight line makes the sum of the squares on them medial and twice the rectangle contained by them both medial and also incommensurable with the sum of the squares on them.

+

To a straight line which produces with a medial area a medial whole only one straight line can be annexed which is incommensurable in square with the whole straight line and which with the whole straight line makes the sum of the squares on them medial and twice the rectangle contained by them both medial and also incommensurable with the sum of the squares on them.

Let AB be the straight line which produces with a medial area a medial whole, and BC an annex to it; therefore AC, CB are straight lines incommensurable in square which fulfil the aforesaid conditions. [X. 78]

I say that no other straight line can be annexed to AB which fulfils the aforesaid conditions.

@@ -9517,7 +7943,7 @@ and, componendo, the squares on AB<

Let a rational straight line EF be set out, let EG equal to the squares on AC, CB be applied to EF, producing EM as breadth, and let HG equal to twice the rectangle AC, CB be applied to EF, producing HM as breadth; therefore the remainder, the square on AB [II. 7], is equal to EL; therefore AB is the side of EL.

Again, let EI equal to the squares on AD, DB be applied to EF, producing EN as breadth.

-

But the square on AB is also equal to EL; therefore the remainder, twice the rectangle AD, DB [II. 7], is equal to HI.

+

But the square on AB is also equal to EL; therefore the remainder, twice the rectangle AD, DB [II. 7], is equal to HI.

Now, since the sum of the squares on AC, CB is medial and is equal to EG, therefore EG is also medial.

And it is applied to the rational straight line EF, producing EM as breadth; therefore EM is rational and incommensurable in length with EF. [X. 22]

Again, since twice the rectangle AC, CB is medial and is equal to HG, therefore HG is also medial.

@@ -9527,7 +7953,7 @@ and, componendo, the squares on AB<

Similarly we can prove that EH is again an apotome and HN an annex to it.

Therefore to an apotome different rational straight lines are annexed which are commensurable with the wholes in square only: which was proved impossible. [X. 79]

Therefore no other straight line can be so annexed to AB.

-

Therefore to AB only one straight line can be annexed which is incommensurable in square with the whole and which with the whole makes the squares on them added together medial, twice the rectangle contained by them medial, and also the squares on them incommensurable with twice the rectangle contained by them. Q. E. D.

+

Therefore to AB only one straight line can be annexed which is incommensurable in square with the whole and which with the whole makes the squares on them added together medial, twice the rectangle contained by them medial, and also the squares on them incommensurable with twice the rectangle contained by them. Q. E. D.

@@ -9557,7 +7983,7 @@ and, componendo, the squares on AB<

and, if neither, a sixth.

 - + PROPOSITIONS 85—115. @@ -9572,14 +7998,14 @@ and, componendo, the squares on AB<

And both are rational; therefore BG, GC are rational straight lines commensurable in square only; therefore BC is an apotome. [X. 73]

I say next that it is also a first apotome.

For let the square on H be that by which the square on BG is greater than the square on GC.

-

Now since. as ED is to FD, so is the square on BG to the square on GC, therefore also, convertendo, [v. 19, Por.] as DE is to EF, so is the square on GB to the square on H.

+

Now since. as ED is to FD, so is the square on BG to the square on GC, therefore also, convertendo, [v. 19, Por.] as DE is to EF, so is the square on GB to the square on H.

But DE has to EF the ratio which a square number has to a square number, for each is square; therefore the square on GB also has to the square on H the ratio which a square number has to a square number; therefore BG is commensurable in length with H. [X. 9]

And the square on BG is greater than the square on GC by the square on a straight line commensurable in length with BG.

And the whole BG is commensurable in length with the rational straight line A set out.

Therefore BC is a first apotome. [X. Deff. III. 1]

Therefore the first apotome BC has been found. (Being) that which it was required to find.

 - + PROPOSITION 86.

To find the second apotome.

@@ -9596,7 +8022,7 @@ and, componendo, the squares on AB<

Since then, as the square on BG is to the square on GC, so is the number ED to the number DF, therefore, convertendo, as the square on BG is to the square on H, so is DE to EF. [V. 19, Por.]

And each of the numbers DE, EF is square; therefore the square on BG has to the square on H the ratio which a square number has to a square number; therefore BG is commensurable in length with H. [X. 9]

And the square on BG is greater than the square on GC by the square on H; therefore the square on BG is greater than the square on GC - by the square on a straight line commensurable in length with BG.

+ by the square on a straight line commensurable in length with BG.

And CG, the annex, is commensurable with the rational straight line A set out.

Therefore BC is a second apotome. [X. Deff. III. 2]

Therefore the second apotome BC has been found. Q. E. D.

@@ -9606,7 +8032,7 @@ and, componendo, the squares on AB<

To find the third apotome.

Let a rational straight line A be set out, let three numbers E, BC, CD be set out which have not to one another the ratio which a square number has to a square number, but let CB have to BD the ratio which a square number has to a square number.

-

Let it be contrived that, as E is to BC, so is the square on A to the square on FG, and, as BC is to CD, so is the square on FG to the square on GH. [X. 6, Por.]

+

Let it be contrived that, as E is to BC, so is the square on A to the square on FG, and, as BC is to CD, so is the square on FG to the square on GH. [X. 6, Por.]

Since then, as E is to BC, so is the square on A to the square on FG, therefore the square on A is commensurable with the square on FG. [X. 6]

But the square on A is rational; therefore the square on FG is also rational; therefore FG is rational.

And, since E has not to BC the ratio which a square number has to a square number, therefore neither has the square on A to the square on FG the ratio which a square number has to a square number; therefore A is incommensurable in length with FG. [X. 9]

@@ -9615,7 +8041,7 @@ and, componendo, the squares on AB<

And, since BC has not to CD the ratio which a square number has to a square number, therefore neither has the square on FG to the square on GH the ratio which a square number has to a square number; therefore FG is incommensurable in length with GH. [X. 9]

And both are rational; therefore FG, GH are rational straight lines commensurable in square only; therefore FH is an apotome. [X. 73]

I say next that it is also a third apotome.

-

For since, as E is to BC, so is the square on A to the square on FG, and, as BC is to CD, so is the square on FG to the square on HG, therefore, ex aequali, as E is to CD, so is the square on A to the square on HG. [V. 22]

+

For since, as E is to BC, so is the square on A to the square on FG, and, as BC is to CD, so is the square on FG to the square on HG, therefore, ex aequali, as E is to CD, so is the square on A to the square on HG. [V. 22]

But E has not to CD the ratio which a square number has to a square number; therefore neither has the square on A to the square on GH the ratio which a square number has to a square number; therefore A is incommensurable in length with GH. [X. 9]

Therefore neither of the straight lines FG, GH is commensurable in length with the rational straight line A set out.

Now let the square on K be that by which the square on FG is greater than the square on GH.

@@ -9632,7 +8058,7 @@ and, componendo, the squares on AB<

Let two numbers DF, FE be set out such that the whole DE has not to either of the numbers DF, EF the ratio which a square number has to a square number.

Let it be contrived that, as DE is to EF, so is the square on BG to the square on GC; [X. 6, Por.] therefore the square on BG is commensurable with the square on GC. [X. 6]

-

But the square on BG is rational; therefore the square on GC is also rational; therefore GC is rational.

+

But the square on BG is rational; therefore the square on GC is also rational; therefore GC is rational.

Now, since DE has not to EF the ratio which a square number has to a square number, therefore neither has the square on BG to the square on GC the ratio which a square number has to a square number; therefore BG is incommensurable in length with GC. [X. 9]

And both are rational; therefore BG, GC are rational straight lines commensurable in square only; therefore BC is an apotome. [X. 73]

Now let the square on H be that by which the square on BG is greater than the square on GC.

@@ -9651,7 +8077,7 @@ and, componendo, the squares on AB<

Therefore the square on GB is also rational; [X. 6] therefore BG is also rational.

Now since, as DE is to EF, so is the square on BG to the square on GC, while DE has not to EF the ratio which a square number has to a square number, therefore neither has the square on BG to the square on GC the ratio which a square number has to a square number; therefore BG is incommensurable in length with GC. [X. 9]

-

And both are rational; therefore BG, GC are rational straight lines commensurable in square only; therefore BC is an apotome. [X. 73]

+

And both are rational; therefore BG, GC are rational straight lines commensurable in square only; therefore BC is an apotome. [X. 73]

I say next that it is also a fifth apotome.

For let the square on H be that by which the square on BG is greater than the square on GC.

Since then, as the square on BG is to the square on GC, so is DE to EF, therefore, convertendo, as ED is to DF, so is the square on BG to the square on H. [V. 19, Por.]

@@ -9660,7 +8086,7 @@ and, componendo, the squares on AB<

And the annex CG is commensurable in length with the rational straight line A set out; therefore BC is a fifth apotome. [X. Deff. III. 5]

Therefore the fifth apotome BC has been found. Q. E. D.

- + PROPOSITION 90.

To find the sixth apotome.

@@ -9672,7 +8098,7 @@ and, componendo, the squares on AB<

And, since E has not to BC the ratio which a square number has to a square number, therefore neither has the square on A to the square on FG the ratio which a square number has to a square number; therefore A is incommensurable in length with FG. [X. 9]

Again, since, as BC is to CD, so is the square on FG to the square on GH, therefore the square on FG is commensurable with the square on GH. [X. 6]

But the square on FG is rational; therefore the square on GH is also rational; therefore GH is also rational.

-

And, since BC has not to CD the ratio which a square number has to a square number, therefore neither has the square on FG to the square on GH the ratio which a square number has to a square number; therefore FG is incommensurable in length with GH. [X. 9]

+

And, since BC has not to CD the ratio which a square number has to a square number, therefore neither has the square on FG to the square on GH the ratio which a square number has to a square number; therefore FG is incommensurable in length with GH. [X. 9]

And both are rational; therefore FG, GH are rational straight lines commensurable in square only; therefore FH is an apotome. [X. 73]

I say next that it is also a sixth apotome.

For since, as E is to BC, so is the square on A to the square on FG, and, as BC is to CD, so is the square on FG to the square on GH, therefore, ex aequali, as E is to CD, so is the square on A to the square on GH. [v. 22]

@@ -9680,7 +8106,7 @@ and, componendo, the squares on AB<

Now let the square on K be that by which the square on FG is greater than the square on GH.

Since then, as BC is to CD, so is the square on FG to the square on GH, therefore, convertendo, as CB is to BD, so is the square on FG to the square on K. [v. 19, Por.]

But CB has not to BD the ratio which a square number has to a square number; therefore neither has the square on FG to the square on K the ratio which a square number has to a square number; therefore FG is incommensurable in length with K. [X. 9]

-

And the square on FG is greater than the square on GH by the square on K; therefore the square on FG is greater than the square on GH by the square on a straight line incommensurable in length with FG.

+

And the square on FG is greater than the square on GH by the square on K; therefore the square on FG is greater than the square on GH by the square on a straight line incommensurable in length with FG.

And neither of the straight lines FG, GH is commensurable with the rational straight line A set out.

Therefore FH is a sixth apotome. [X. Deff. III. 6]

Therefore the sixth apotome FH has been found. Q. E. D.

@@ -9691,14 +8117,14 @@ and, componendo, the squares on AB< of the area is an apotome.

For let the area AB be contained by the rational straight line AC and the first apotome AD;

I say that the side - of the area AB is an apotome.

+ of the area AB is an apotome.

For, since AD is a first apotome, let DG be its annex; therefore AG, GD are rational straight lines commensurable in square only. [X. 73]

And the whole AG is commensurable with the rational straight line AC set out, and the square on AG is greater than the square on GD by the square on a straight line commensurable in length with AG; [X. Deff. III. 1] if therefore there be applied to AG a parallelogram equal to the fourth part of the square on DG and deficient by a square figure, it divides it into commensurable parts. [X. 17]

Let DG be bisected at E, let there be applied to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF, FG; therefore AF is commensurable with FG.

And through the points E, F, G let EH, FI, GK be drawn parallel to AC.

Now, since AF is commensurable in length with FG, therefore AG is also commensurable in length with each of the straight lines AF, FG. [X. 15]

-

But AG is commensurable with AC; therefore each of the straight lines AF, FG is commensurable in length with AC. [X. 12]

+

But AG is commensurable with AC; therefore each of the straight lines AF, FG is commensurable in length with AC. [X. 12]

And AC is rational; therefore each of the straight lines AF, FG is also rational, so that each of the rectangles AI, FK is also rational. [X. 19]

Now, since DE is commensurable in length with EG, therefore DG is also commensurable in length with each of the straight lines DE, EG. [X. 15]

But DG is rational and incommensurable in length with AC; therefore each of the straight lines DE, EG is also rational and incommensurable in length with AC; [X. 13] therefore each of the rectangles DH, EK is medial. [X. 21]

@@ -9710,7 +8136,7 @@ and, componendo, the squares on AB<

But EK is equal to DH, and MN to LO; therefore DK is equal to the gnomon UVW and NO.

But AK is also equal to the squares LM, NO; therefore the remainder AB is equal to ST.

But ST is the square on LN; therefore the square on LN is equal to AB; therefore LN is the side - of AB.

+ of AB.

I say next that LN is an apotome.

For, since each of the rectangles AI, FK is rational, and they are equal to LM, NO, therefore each of the squares LM, NO, that is, the squares on LP, PN respectively, is also rational; therefore each of the straight lines LP, PN is also rational.

Again, since DH is medial and is equal to LO, therefore LO is also medial.

@@ -9726,14 +8152,14 @@ and, componendo, the squares on AB< PROPOSITION 92.

If an area be contained by a rational straight line and a second apotome, the side of the area is a first apotome of a medial straight line.

-

For let the area AB be contained by the rational straight line AC and the second apotome AD; I say that the side +

For let the area AB be contained by the rational straight line AC and the second apotome AD; I say that the side of the area AB is a first apotome of a medial straight line.

For let DG be the annex to AD; therefore AG, GD are rational straight lines commensurable in square only, [X. 73] and the annex DG is commensurable with the rational straight line AC set out, while the square on the whole AG is greater than the square on the annex GD by the square on a straight line commensurable in length with AG. [X. Deff. III. 2]

Since then the square on AG is greater than the square on GD by the square on a straight line commensurable with AG, therefore, if there be applied to AG a parallelogram equal to the fourth part of the square on GD and deficient by a square figure, it divides it into commensurable parts. [X. 17]

Let then DG be bisected at E, let there be applied to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF, FG; therefore AF is commensurable in length with FG.

Therefore AG is also commensurable in length with each of the straight lines AF, FG. [X. 15]

-

But AG is rational and incommensurable in length with AC; therefore each of the straight lines AF, FG is also rational and incommensurable in length with AC; [X. 13] therefore each of the rectangles AI, FK is medial. [X. 21]

+

But AG is rational and incommensurable in length with AC; therefore each of the straight lines AF, FG is also rational and incommensurable in length with AC; [X. 13] therefore each of the rectangles AI, FK is medial. [X. 21]

Again, since DE is commensurable with EG, therefore DG is also commensurable with each of the straight lines DE, EG. [X. 15]

But DG is commensurable in length with AC.

Therefore each of the rectangles DH, EK is rational. [X. 19]

@@ -9743,7 +8169,7 @@ and, componendo, the squares on AB<

And, since the rectangle AF, FG is equal to the square on EG, therefore, as AF is to EG, so is EG to FG, [VI. 17] while, as AF is to EG, so is AI to EK, and, as EG is to FG, so is EK to FK; [VI. 1] therefore EK is a mean proportional between AI, FK. [V. 11]

But MN is also a mean proportional between the squares LM, NO, and AI is equal to LM, and FK to NO; therefore MN is also equal to EK.

But DH is equal to EK, and LO equal to MN; therefore the whole DK is equal to the gnomon UVW and NO.

-

Since then the whole AK is equal to LM, NO, and, in these, DK is equal to the gnomon UVW and NO, therefore the remainder AB is equal to TS.

+

Since then the whole AK is equal to LM, NO, and, in these, DK is equal to the gnomon UVW and NO, therefore the remainder AB is equal to TS.

But TS is the square on LN; therefore the square on LN is equal to the area AB; therefore LN is the side of the area AB.

I say that LN is a first apotome of a medial straight line.

@@ -9756,7 +8182,7 @@ and, componendo, the squares on AB<

Therefore the side of the area AB is a first apotome of a medial straight line. Q. E. D.

- + PROPOSITION 93.

If an area be contained by a rational straight line and a third apotome, the side @@ -9766,7 +8192,7 @@ and, componendo, the squares on AB<

For let DG be the annex to AD; therefore AG, GD are rational straight lines commensurable in square only, and neither of the straight lines AG, GD is commensurable in length with the rational straight line AC set out, while the square on the whole AG is greater than the square on the annex DG by the square on a straight line commensurable with AG. [X. Deff. III. 3]

Since then the square on AG is greater than the square on GD by the square on a straight line commensurable with AG, therefore, if there be applied to AG a parallelogram equal to the fourth part of the square on DG and deficient by a square figure, it will divide it into commensurable parts. [X. 17]

-

Let then DG be bisected at E, let there be applied to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF, FG.

+

Let then DG be bisected at E, let there be applied to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF, FG.

Let EH, FI, GK be drawn through the points E, F, G parallel to AC.

Therefore AF, FG are commensurable; therefore AI is also commensurable with FK. [VI. 1, X. 11]

And, since AF, FG are commensurable in length, therefore AG is also commensurable in length with each of the straight lines AF, FG. [X. 15]

@@ -9779,7 +8205,7 @@ and, componendo, the squares on AB<

But, as AF is to EG, so is AI to EK; [VI. 1] therefore AI is incommensurable with EK. [X. 11]

Now let the square LM be constructed equal to AI, and let there be subtracted NO equal to FK and being about the same angle with LM; therefore LM, NO are about the same diameter. [VI. 26]

Let PR be their diameter, and let the figure be drawn.

-

Now, since the rectangle AF, FG is equal to the square on EG, therefore, as AF is to EG, so is EG to FG. [VI. 17]

+

Now, since the rectangle AF, FG is equal to the square on EG, therefore, as AF is to EG, so is EG to FG. [VI. 17]

But, as AF is to EG, so is AI to EK, and, as EG is to FG, so is EK to FK; [VI. 1] therefore also, as AI is to EK, so is EK to FK; [V. 11] therefore EK is a mean proportional between AI, FK.

But MN is also a mean proportional between the squares LM, NO, and AI is equal to LM, and FK to NO; therefore EK is also equal to MN.

But MN is equal to LO, and EK equal to DH; therefore the whole DK is also equal to the gnomon UVW and NO.

@@ -9790,13 +8216,13 @@ and, componendo, the squares on AB<

And, since AI is commensurable with FK, [VI. 1, X. 11] therefore the square on LP is also commensurable with the square on PN.

Again, since AI was proved incommensurable with EK, therefore LM is also incommensurable with MN, that is, the square on LP with the rectangle LP, PN; so that LP is also incommensurable in length with PN; [VI. 1, X. 11] therefore LP, PN are medial straight lines commensurable in square only.

I say next that they also contain a medial rectangle.

-

For, since EK was proved medial, and is equal to the rectangle LP, PN, therefore the rectangle LP, PN is also medial, so that LP, PN are medial straight lines commensurable in square only which contain a medial rectangle.

+

For, since EK was proved medial, and is equal to the rectangle LP, PN, therefore the rectangle LP, PN is also medial, so that LP, PN are medial straight lines commensurable in square only which contain a medial rectangle.

Therefore LN is a second apotome of a medial straight line; [X. 75] and it is the side of the area AB.

Therefore the side of the area AB is a second apotome of a medial straight line. Q. E. D.

- + PROPOSITION 94.

If an area be contained by a rational straight line and a fourth apotome, the side @@ -9806,7 +8232,7 @@ and, componendo, the squares on AB<

For let DG be the annex to AD; therefore AG, GD are rational straight lines commensurable in square only, AG is commensurable in length with the rational straight line AC set out, and the square on the whole AG is greater than the square on the annex DG by the square on a straight line incommensurable in length with AG, [X. Deff. III. 4]

Since then the square on AG is greater than the square on GD by the square on a straight line incommensurable in length with AG, therefore, if there be applied to AG a parallelogram equal to the fourth part of the square on DG and deficient by a square figure, it will divide it into incommensurable parts. [X. 18]

-

Let then DG be bisected at E, let there be applied to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF, FG; therefore AF is incommensurable in length with FG.

+

Let then DG be bisected at E, let there be applied to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF, FG; therefore AF is incommensurable in length with FG.

Let EH, FI, GK be drawn through E, F, G parallel to AC, BD.

Since then AG is rational and commensurable in length with AC, therefore the whole AK is rational. [X. 19]

Again, since DG is incommensurable in length with AC, and both are rational, therefore DK is medial. [X. 21]

@@ -9819,7 +8245,7 @@ and, componendo, the squares on AB<

But MN is also a mean proportional between the squares LM, NO, and AI is equal to LM, and FK to NO; therefore EK is also equal to MN.

But DH is equal to EK, and LO is equal to MN; therefore the whole DK is equal to the gnomon UVW and NO.

Since, then, the whole AK is equal to the squares LM, NO, and, in these, DK is equal to the gnomon UVW and the square NO, therefore the remainder AB is equal to ST, that is, to the square on LN; therefore LN is the side - of the area AB.

+ of the area AB.

I say that LN is the irrational straight line called minor.

For, since AK is rational and is equal to the squares on LP, PN, therefore the sum of the squares on LP, PN is rational.

Again, since DK is medial, and DK is equal to twice the rectangle LP, PN, therefore twice the rectangle LP, PN is medial.

@@ -9838,7 +8264,7 @@ and, componendo, the squares on AB<

For let the area AB be contained by the rational straight line AC and the fifth apotome AD; I say that the side of the area AB is a straight line which produces with a rational area a medial whole.

-

For let DG be the annex to AD; therefore AG, GD are rational straight lines commensurable in square only,

the annex GD is commensurable in length with the rational straight line AC set out, and the square on the whole AG is greater than the square on the annex DG by the square on a straight line incommensurable with AG. [X. Deff. III. 5]

+

For let DG be the annex to AD; therefore AG, GD are rational straight lines commensurable in square only,

the annex GD is commensurable in length with the rational straight line AC set out, and the square on the whole AG is greater than the square on the annex DG by the square on a straight line incommensurable with AG. [X. Deff. III. 5]

Therefore, if there be applied to AG a parallelogram equal to the fourth part of the square on DG and deficient by a square figure, it will divide it into incommensurable parts. [X. 18]

Let then DG be bisected at the point E, let there be applied to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF, FG; therefore AF is incommensurable in length with FG.

Now, since AG is incommensurable in length with CA, and both are rational, therefore AK is medial. [X. 21]

@@ -9850,13 +8276,13 @@ and, componendo, the squares on AB<

I say that LN is the straight line which produces with a rational area a medial whole.

For, since AK was proved medial and is equal to the squares on LP, PN, therefore the sum of the squares on LP, PN is medial.

Again, since DK is rational and is equal to twice the rectangle LP, PN, the latter is itself also rational.

-

And, since AI is incommensurable with FK, therefore the square on LP is also incommensurable with the square on PN; therefore LP, PN are straight lines incommensurable in square which make the sum of the squares on them medial but twice the rectangle contained by them rational.

+

And, since AI is incommensurable with FK, therefore the square on LP is also incommensurable with the square on PN; therefore LP, PN are straight lines incommensurable in square which make the sum of the squares on them medial but twice the rectangle contained by them rational.

Therefore the remainder LN is the irrational straight line called that which produces with a rational area a medial whole; [X. 77] and it is the side of the area AB.

Therefore the side of the area AB is a straight line which produces with a rational area a medial whole. Q. E. D.

- + PROPOSITION 96.

If an area be contained by a rational straight line and a sixth apotome, the side @@ -9866,7 +8292,7 @@ and, componendo, the squares on AB<

For let DG be the annex to AD; therefore AG, GD are rational straight lines commensurable in square only, neither of them is commensurable in length with the rational straight line AC set out, and the square on the whole AG is greater than the square on the annex DG by the square on a straight line incommensurable in length with AG. [X. Deff. III. 6]

Since then the square on AG is greater than the square on GD by the square on a straight line incommensurable in length with AG, therefore, if there be applied to AG a parallelogram equal to the fourth part of the square on DG and deficient by a square figure, it will divide it into incommensurable parts. [X. 18]

-

Let then DG be bisected at E, let there be applied to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF, FG; therefore AF is incommensurable in length with FG.

+

Let then DG be bisected at E, let there be applied to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF, FG; therefore AF is incommensurable in length with FG.

But, as AF is to FG, so is AI to FK. [VI. 1] therefore AI is incommensurable with FK. [X. 11]

And, since AG, AC are rational straight lines commensurable in square only, AK is medial. [X. 21]

Again, since AC, DG are rational straight lines and incommensurable in length, DK is also medial. [X. 21]

@@ -9880,14 +8306,14 @@ and, componendo, the squares on AB<

For, since AK was proved medial and is equal to the squares on LP, PN, therefore the sum of the squares on LP, PN is medial.

Again, since DK was proved medial and is equal to twice the rectangle LP, PN, twice the rectangle LP, PN is also medial.

And, since AK was proved incommensurable with DK, the squares on LP, PN are also incommensurable with twice the rectangle LP, PN.

-

And, since AI is incommensurable with FK, therefore the square on LP is also incommensurable with the square on PN; therefore LP, PN are straight lines incommensurable in square which make the sum of the squares on them medial, twice the rectangle contained by them medial, and further the squares on them incommensurable with twice the rectangle contained by them.

+

And, since AI is incommensurable with FK, therefore the square on LP is also incommensurable with the square on PN; therefore LP, PN are straight lines incommensurable in square which make the sum of the squares on them medial, twice the rectangle contained by them medial, and further the squares on them incommensurable with twice the rectangle contained by them.

Therefore LN is the irrational straight line called that which produces with a medial area a medial whole; [X. 78] and it is the side of the area AB.

Therefore the side of the area is a straight line which produces with a medial area a medial whole. Q. E. D.

 - + PROPOSITION 97.

@@ -9901,7 +8327,7 @@ and, componendo, the squares on AB<

Let FM be bisected at the point N, and let NO be drawn through N parallel to CD; therefore each of the rectangles FO, LN is equal to the rectangle AG, GB.

Now, since the squares on AG, GB are rational, and DM is equal to the squares on AG, GB,. therefore DM is rational.

And it has been applied to the rational straight line CD, producing CM as breadth; therefore CM is rational and commensurable in length with CD. [X. 20]

-

Again, since twice the rectangle AG, GB is medial, and FL is equal to twice the rectangle AG, GB, therefore FL is medial.

+

Again, since twice the rectangle AG, GB is medial, and FL is equal to twice the rectangle AG, GB, therefore FL is medial.

And it is applied to the rational straight line CD, producing FM as breadth; therefore FM is rational and incommensurable in length with CD. [X. 22]

And, since the squares on AG, GB are rational, while twice the rectangle AG, GB is medial, therefore the squares on AG, GB are incommensurable with twice the rectangle AG, GB.

And CL is equal to the squares on AG, GB, and FL to twice the rectangle AG, GB; therefore DM is incommensurable with FL.

@@ -9912,7 +8338,7 @@ and, componendo, the squares on AB<

But, as CH is to NL, so is CK to NM, and, as NL is to KL, so is NM to KM; [VI. 1] therefore the rectangle CK, KM is equal to the square on NM [VI. 17], that is, to the fourth part of the square on FM.

And, since the square on AG is commensurable with the square on GB, CH is also commensurable with KL.

But, as CH is to KL, so is CK to KM; [VI. 1] therefore CK is commensurable with KM. [X. 11]

-

Since then CM, MF are two unequal straight lines, and to CM there has been applied the rectangle CK, KM equal to the fourth part of the square on FM and deficient by a square figure, while CK is commensurable with KM, therefore the square on CM is greater than the square on MF by the square on a straight line commensurable in length with CM. [X. 17]

+

Since then CM, MF are two unequal straight lines, and to CM there has been applied the rectangle CK, KM equal to the fourth part of the square on FM and deficient by a square figure, while CK is commensurable with KM, therefore the square on CM is greater than the square on MF by the square on a straight line commensurable in length with CM. [X. 17]

And CM is commensurable in length with the rational straight line CD set out; therefore CF is a first apotome. [X. Deff. III. 1]

Therefore etc. Q. E. D.

 @@ -9925,7 +8351,7 @@ and, componendo, the squares on AB<

For let BG be the annex to AB;. therefore AG, GB are medial straight lines commensurable in square only which contain a rational rectangle. [X. 74]

To CD let there be applied CH equal to the square on AG, producing CK as breadth, and KL equal to the square on GB, producing KM as breadth; therefore the whole CL is equal to the squares on AG, GB; therefore CL is also medial. [X. 15 and 23, Por.]

-

And it is applied to the rational straight line CD, producing CM as breadth; therefore CM is rational and incommensurable in length with CD. [X. 22]

+

And it is applied to the rational straight line CD, producing CM as breadth; therefore CM is rational and incommensurable in length with CD. [X. 22]

Now, since CL is equal to the squares on AG, GB, and, in these, the square on AB is equal to CE, therefore the remainder, twice the rectangle AG, GB, is equal to FL. [II. 7]

But twice the rectangle AG, GB is rational; therefore FL is rational.

And it is applied to the rational straight line FE, producing FM as breadth; therefore FM is also rational and commensurable in length with CD. [X. 20]

@@ -9935,12 +8361,12 @@ and, componendo, the squares on AB<

I say next that it is also a second apotome.

For let FM be bisected at N, and let NO be drawn through N parallel to CD; therefore each of the rectangles FO, NL is equal to the rectangle AG, GB.

Now, since the rectangle AG, GB is a mean proportional between the squares on AG, GB, and the square on AG is equal to CH, the rectangle AG, GB to NL, and the square on BG to KL, therefore NL is also a mean proportional between CH, KL; therefore, as CH is to NL, so is NL to KL.

-

But, as CH is to NL, so is CK to NM, and, as NL is to KL, so is NM to MK; [VI. 1] therefore, as CK is to NM, so is NM, so is KM; [V. 11] therefore the rectangle CK, KM is equal to the square on NM [VI. 17], that is, to the fourth part of the square on FM.

+

But, as CH is to NL, so is CK to NM, and, as NL is to KL, so is NM to MK; [VI. 1] therefore, as CK is to NM, so is NM, so is KM; [V. 11] therefore the rectangle CK, KM is equal to the square on NM [VI. 17], that is, to the fourth part of the square on FM.

Since the CM, MF are two unequal straight lines, and the rectangle CK, KM equal to the fourth part of the square on MF and deficient by a square figure has been applied to the greater, CM, and divides it into commensurable parts, therefore the square on CM is greater than the square on MF by the square on a straight line commensurable in length with CM. [X. 17]

And the annex FM is commensurable in length with the rational straight line CD set out; therefore CF is a second apotome. [X. Deff. III. 2]

Therefore etc. Q. E. D.

- + PROPOSITION 99.

The square on a second apotome of a medial straight line applied to a rational straight line produces as breadth a third apotome.

@@ -9950,7 +8376,7 @@ and, componendo, the squares on AB<

Let CH equal to the square on AG be applied to CD, producing CK as breadth, and let KL equal to the square on BG be applied to KH, producing KM as breadth; therefore the whole CL is equal to the squares on AG, GB; therefore CL is also medial. [X. 15 and 23, Por.]

And it is applied to the rational straight line CD, producing CM as breadth; therefore CM is rational and incommensurable in length with CD. [X. 22]

Now, since the whole CL is equal to the squares on AG, GB, and, in these, CE is equal to the square on AB, therefore the remainder LF is equal to twice the rectangle AG, GB. [II. 7]

-

Let then FM be bisected at the point N, and let NO be drawn parallel to CD; therefore each of the rectangles FO, NL is equal to the rectangle AG, GB.

+

Let then FM be bisected at the point N, and let NO be drawn parallel to CD; therefore each of the rectangles FO, NL is equal to the rectangle AG, GB.

But the rectangle AG, GB is medial; therefore FL is also medial.

And it is applied to the rational straight line EF, producing FM as breadth; therefore FM is also rational and incommensurable in length with CD. [X. 22]

And, since AG, GB are commensurable in square only, therefore AG is incommensurable in length with GB; therefore the square on AG is also incommensurable with the rectangle AG, GB. [VI. 1, X. 11]

@@ -9960,7 +8386,7 @@ and, componendo, the squares on AB<

And both are rational; therefore CM, MF are rational straight lines commensurable in square only; therefore CF is an apotome. [X. 73]

I say next that it is also a third apotome.

For, since the square on AG is commensurable with the square on GB, therefore CH is also commensurable with KL, so that CK is also commensurable with KM. [VI. 1, X. 11]

-

And, since the rectangle AG, GB is a mean proportional between the squares on AG, GB, and CH is equal to the square on AG, KL equal to the square on GB, and NL equal to the rectangle AG, GB, therefore NL is also a mean proportional between CH, KL; therefore, as CH is to NL, so is NL to KL.

+

And, since the rectangle AG, GB is a mean proportional between the squares on AG, GB, and CH is equal to the square on AG, KL equal to the square on GB, and NL equal to the rectangle AG, GB, therefore NL is also a mean proportional between CH, KL; therefore, as CH is to NL, so is NL to KL.

But, as CH is to NL, so is CK to NM, and, as NL is to KL, so is NM to KM; [VI. 1] therefore, as CK is to MN, so is MN to KM; [V. 11] therefore the rectangle CK, KM is equal to [the square on MN, that is, to] the fourth part of the square on FM.

Since then CM, MF are two unequal straight lines, and a parallelogram equal to the fourth part of the square on FM and deficient by a square figure has been applied to CM, and divides it into commensurable parts, therefore the square on CM is greater than the square on MF by the square on a straight line commensurable with CM. [X. 17]

And neither of the straight lines CM, MF is commensurable in length with the rational straight line CD set out; therefore CF is a third apotome. [X. Deff. III. 3]

@@ -9976,7 +8402,7 @@ and, componendo, the squares on AB<

For let BG be the annex to AB; therefore AG, GB are straight lines incommensurable in square which make the sum of the squares on AG, GB rational, but twice the rectangle AG, GB medial. [X. 76]

To CD let there be applied CH equal to the square on AG and producing CK as breadth, and KL equal to the square on BG, producing KM as breadth; therefore the whole CL is equal to the squares on AG, GB.

And the sum of the squares on AG, GB is rational; therefore CL is also rational.

-

And it is applied to the rational straight line CD, producing CM as breadth; therefore CM is also rational and commensurable in length with CD. [X. 20]

+

And it is applied to the rational straight line CD, producing CM as breadth; therefore CM is also rational and commensurable in length with CD. [X. 20]

And, since the whole CL is equal to the squares on AG, GB, and, in these, CE is equal to the square on AB, therefore the remainder FL is equal to twice the rectangle AG, GB. [II. 7]

Let then FM be bisected at the point N, and let NO be drawn through N parallel to either of the straight lines CD, ML; therefore each of the rectangles FO, NL is equal to the rectangle AG, GB.

And, since twice the rectangle AG, GB is medial and is equal to FL, therefore FL is also medial.

@@ -9987,7 +8413,7 @@ and, componendo, the squares on AB<

And both are rational; therefore CM, MF are rational straight lines commensurable in square only; therefore CF is an apotome. [X. 73]

I say that it is also a fourth apotome.

For, since AG, GB are incommensurable in square, therefore the square on AG is also incommensurable with the square on GB.

-

And CH is equal to the square on AG, and KL equal to the square on GB; therefore CH is incommensurable with KL.

+

And CH is equal to the square on AG, and KL equal to the square on GB; therefore CH is incommensurable with KL.

But, as CH is to KL, so is CK to KM; [VI. 1] therefore CK is incommensurable in length with KM. [X. 11]

And, since the rectangle AG, GB is a mean proportional between the squares on AG, GB, and the square on AG is equal to CH, the square on GB to KL, and the rectangle AG, GB to NL, therefore NL is a mean proportional between CH, KL; therefore, as CH is to NL, so is NL to KL.

But, as CH is to NL, so is CK to NM, and, as NL is to KL, so is NM to KM; [VI. 1] therefore, as CK is to MN, so is MN to KM; [V. 11] therefore the rectangle CK, KM is equal to the square on MN [VI. 17], that is, to the fourth part of the square on FM.

@@ -10003,7 +8429,7 @@ and, componendo, the squares on AB<

Let AB be the straight line which produces with a rational area a medial whole, and CD a rational straight line, and to CD let CE be applied equal to the square on AB and producing CF as breadth; I say that CF is a fifth apotome.

For let BG be the annex to AB; therefore AG, GB are straight lines incommensurable in square which make the sum of the squares on them medial but twice the rectangle contained by them rational. [X. 77]

-

To CD let there be applied CH equal to the square on AG, and KL equal to the square on GB; therefore the whole CL is equal to the squares on AG, GB.

+

To CD let there be applied CH equal to the square on AG, and KL equal to the square on GB; therefore the whole CL is equal to the squares on AG, GB.

But the sum of the squares on AG, GB together is medial; therefore CL is medial.

And it is applied to the rational straight line CD, producing CM as breadth; therefore CM is rational and incommensurable with CD. [X. 22]

And, since the whole CL is equal to the squares on AG, GB, and, in these, CE is equal to the square on AB, therefore the remainder FL is equal to twice the rectangle AG, GB. [II. 7]

@@ -10015,12 +8441,12 @@ and, componendo, the squares on AB<

And both are rational; therefore CM, MF are rational straight lines commensurable in square only; therefore CF is an apotome. [X. 73]

I say next that it is also a fifth apotome.

For we can prove similarly that the rectangle CK, KM is equal to the square on NM, that is, to the fourth part of the square on FM.

-

And, since the square on AG is incommensurable with the square on GB, while the square on AG is equal to CH, and the square on GB to KL, therefore CH is incommensurable with KL.

+

And, since the square on AG is incommensurable with the square on GB, while the square on AG is equal to CH, and the square on GB to KL, therefore CH is incommensurable with KL.

But, as CH is to KL, so is CK to KM; [VI. 1] therefore CK is incommensurable in length with KM. [X. 11]

Since then CM, MF are two unequal straight lines, and a parallelogram equal to the fourth part of the square on FM and deficient by a square figure has been applied to CM, and divides it into incommensurable parts, therefore the square on CM is greater than the square on MF by the square on a straight line incommensurable with CM. [X. 18]

And the annex FM is commensurable with the rational straight line CD set out; therefore CF is a fifth apotome. [X. Deff. III. 5] Q. E. D.

- + PROPOSITION 102.

@@ -10032,7 +8458,7 @@ and, componendo, the squares on AB<

Now to CD let there be applied CH equal to the square on AG and producing CK as breadth, and KL equal to the square on BG; therefore the whole CL is equal to the squares on AG, GB; therefore CL is also medial.

And it is applied to the rational straight line CD, producing CM as breadth; therefore CM is rational and incommensurable in length with CD. [X. 22]

Since now CL is equal to the squares on AG, GB, and, in these, CE is equal to the square on AB, therefore the remainder FL is equal to twice the rectangle AG, GB. [II. 7]

-

And twice the rectangle AG, GB is medial; therefore FL is also medial.

+

And twice the rectangle AG, GB is medial; therefore FL is also medial.

And it is applied to the rational straight line FE, producing FM as breadth; therefore FM is rational and incommensurable in length with CD. [X. 22]

And, since the squares on AG, GB are incommensurable with twice the rectangle AG, GB, and CL is equal to the squares on AG, GB, and FL equal to twice the rectangle AG, GB, therefore CL is incommensurable with FL.

But, as CL is to FL, so is CM to MF; [VI. 1] therefore CM is incommensurable in length with MF. [X. 11]

@@ -10043,14 +8469,14 @@ and, componendo, the squares on AB<

And, since AG, GB are incommensurable in square, therefore the square on AG is incommensurable with the square on GB.

But CH is equal to the square on AG, and KL is equal to the square on GB; therefore CH is incommensurable with KL.

But, as CH is to KL, so is CK to KM; [VI. 1] therefore CK is incommensurable with KM. [X. 11]

-

And, since the rectangle AG, GB is a mean proportional between the squares on AG, GB, and CH is equal to the square on AG, KL equal to the square on GB, and NL equal to the rectangle AG, GB, therefore NL is also a mean proportional between CH, KL; therefore, as CH is to NL, so is NL to KL.

+

And, since the rectangle AG, GB is a mean proportional between the squares on AG, GB, and CH is equal to the square on AG, KL equal to the square on GB, and NL equal to the rectangle AG, GB, therefore NL is also a mean proportional between CH, KL; therefore, as CH is to NL, so is NL to KL.

And for the same reason as before the square on CM is greater than the square on MF by the square on a straight line incommensurable with CM. [X. 18]

And neither of them is commensurable with the rational straight line CD set out; therefore CF is a sixth apotome. [X. Deff. III. 6] Q. E. D.

 PROPOSITION 103.

A straight line commensurable in length with an apotome is an apotome and the same in order.

-

Let AB be an apotome, and let CD be commensurable in length with AB;

I say that CD is also an apotome and the same in order with AB.

+

Let AB be an apotome, and let CD be commensurable in length with AB;

I say that CD is also an apotome and the same in order with AB.

For, since AB is an apotome, let BE be the annex to it; therefore AE, EB are rational straight lines commensurable in square only. [X. 73]

Let it be contrived that the ratio of BE to DF is the same as the ratio of AB to CD; [VI. 12] therefore also, as one is to one, so are all to all; [V. 12] therefore also, as the whole AE is to the whole CF, so is AB to CD.

But AB is commensurable in length with CD.

@@ -10060,7 +8486,7 @@ and, componendo, the squares on AB<

And the square on AE is greater than the square on EB either by the square on a straight line commensurable with AE or by the square on a straight line incommensurable with it.

If then the square on AE is greater than the square on EB by the square on a straight line commensurable with AE, the square on CF will also be greater than the square on FD by the square on a straight line commensurable with CF. [X. 14]

And, if AE is commensurable in length with the rational straight line set out, CF is so also, [X. 12] if BE, then DF also, [id.] and, if neither of the straight lines AE, EB, then neither of the straight lines CF, FD. [X. 13]

-

But, if the square on AE is greater than the square on EB by the square on a straight line incommensurable with AE, the square on CF will also be greater than the square on FD by the square on a straight line incommensurable with CF. [X. 14]

+

But, if the square on AE is greater than the square on EB by the square on a straight line incommensurable with AE, the square on CF will also be greater than the square on FD by the square on a straight line incommensurable with CF. [X. 14]

And, if AE is commensurable in length with the rational straight line set out, CF is so also, if BE, then DF also, [X. 12] and, if neither of the straight lines AE, EB, then neither of the straight lines CF, FD. [X. 13]

Therefore CD is an apotome and the same in order with AB. Q. E. D.

@@ -10072,7 +8498,7 @@ and, componendo, the squares on AB<

For, since AB is an apotome of a medial straight line, let EB be the annex to it.

Therefore AE, EB are medial straight lines commensurable in square only. [X. 74, 75]

Let it be contrived that, as AB is to CD, so is BE to DF; [VI. 12] therefore AE is also commensurable with CF, and BE with DF. [V. 12, X. 11]

-

But AE, EB are medial straight lines commensurable in square only; therefore CF, FD are also medial straight lines [X. 23] commensurable in square only; [X. 13] therefore CD is an apotome of a medial straight line. [X. 74, 75]

+

But AE, EB are medial straight lines commensurable in square only; therefore CF, FD are also medial straight lines [X. 23] commensurable in square only; [X. 13] therefore CD is an apotome of a medial straight line. [X. 74, 75]

I say next that it is also the same in order with AB.

Since, as AE is to EB, so is CF to FD, therefore also, as the square on AE is to the rectangle AE, EB, so is the square on CF to the rectangle CF, FD.

But the square on AE is commensurable with the square on CF; therefore the rectangle AE, EB is also commensurable with the rectangle CF, FD. [V. 16, X. 11]

@@ -10088,7 +8514,7 @@ and, componendo, the squares on AB<

Now since, as AE is to EB, so is CF to FD, [V. 12, V. 16] therefore also, as the square on AE is to the square on EB, so is the square on CF to the square on FD. [VI. 22]

Therefore, componendo, as the squares on AE, EB are to the square on EB, so are the squares on CF, FD to the square on FD. [V. 18]

But the square on BE is commensurable with the square on DF; therefore the sum of the squares on AE, EB is also commensurable with the sum of the squares on CF, FD. [V. 16, X. 11]

-

But the sum of the squares on AE, EB is rational; [X. 76] therefore the sum of the squares on CF, FD is also rational. [X. Def. 4]

+

But the sum of the squares on AE, EB is rational; [X. 76] therefore the sum of the squares on CF, FD is also rational. [X. Def. 4]

Again, since, as the square on AE is to the rectangle AE, EB, so is the square on CF to the rectangle CF, FD, while the square on AE is commensurable with the square on CF, therefore the rectangle AE, EB is also commensurable with the rectangle CF, FD.

But the rectangle AE, EB is medial; [X. 76] therefore the rectangle CF, FD is also medial; [X. 23, Por.] therefore CF, FD are straight lines incommensurable in square which make the sum of the squares on them rational, but the rectangle contained by them medial.

Therefore CD is minor. [X. 76] Q. E. D.

@@ -10100,7 +8526,7 @@ and, componendo, the squares on AB<

For let BE be the annex to AB; therefore AE, EB are straight lines incommensurable in square which make the sum of the squares on AE, EB medial, but the rectangle contained by them rational. [X. 77]

Let the same construction be made.

-

Then we can prove, in manner similar to the foregoing, that CF, FD are in the same ratio as AE, EB, the sum of the squares on AE, EB is commensurable with the sum of the squares on CF, FD, and the rectangle AE, EB with the rectangle CF, FD; so that CF, FD are also straight lines incommensurable in square which make the sum of the squares on CF, FD medial, but the rectangle contained by them rational.

+

Then we can prove, in manner similar to the foregoing, that CF, FD are in the same ratio as AE, EB, the sum of the squares on AE, EB is commensurable with the sum of the squares on CF, FD, and the rectangle AE, EB with the rectangle CF, FD; so that CF, FD are also straight lines incommensurable in square which make the sum of the squares on CF, FD medial, but the rectangle contained by them rational.

Therefore CD is a straight line which produces with a rational area a medial whole. [X. 77] Q. E. D.

@@ -10112,7 +8538,7 @@ and, componendo, the squares on AB<

Now, as was proved, AE, EB are commensurable with CF, FD, the sum of the squares on AE, EB with the sum of the squares on CF, FD, and the rectangle AE, EB with the rectangle CF, FD; therefore CF, FD are also straight lines incommensurable in square which make the sum of the squares on them medial, the rectangle contained by them medial, and further the sum of the squares on them incommensurable with the rectangle contained by them.

Therefore CD is a straight line which produces with a medial area a medial whole. [X. 78]

 - + PROPOSITION 108.

If from a rational area a medial area be subtracted, the side @@ -10126,7 +8552,7 @@ and, componendo, the squares on AB<

Therefore FH, FK are rational straight lines commensurable in square only; therefore KH is an apotome [X. 73], and KF the annex to it.

Now the square on HF is greater than the square on FK by the square on a straight line either commensurable with HF or not commensurable.

First, let the square on it be greater by the square on a straight line commensurable with it.

-

Now the whole HF is commensurable in length with the rational straight line FG set out; therefore KH is a first apotome. [X. Deff. III. 1]

+

Now the whole HF is commensurable in length with the rational straight line FG set out; therefore KH is a first apotome. [X. Deff. III. 1]

But the side of the rectangle contained by a rational straight line and a first apotome is an apotome. [X. 91]

Therefore the side @@ -10138,7 +8564,7 @@ and, componendo, the squares on AB< PROPOSITION 109.

If from a medial area a rational area be subtracted, there arise two other irrational straight lines, either a first apotome of a medial straight line or a straight line which produces with a rational area a medial whole.

-

For from the medial area BC let the rational area BD be subtracted.

+

For from the medial area BC let the rational area BD be subtracted.

I say that the side of the remainder EC becomes one of two irrational straight lines, either a first apotome of a medial straight line or a straight line which produces with a rational area a medial whole.

@@ -10149,13 +8575,13 @@ and, componendo, the squares on AB<

But FG is rational; so that the side of LH, that is, of EC, is a first apotome of a medial straight line. [X. 92]

But, if the square on HF is greater than the square on FK by the square on a straight line incommensurable with HF, while the annex FK is commensurable in length with the rational straight line FG set out, KH is a fifth apotome; [X. Deff. III. 5] so that the side - of EC is a straight line which produces with a rational area a medial whole. [X. 95]

+ of EC is a straight line which produces with a rational area a medial whole. [X. 95]

PROPOSITION 110.

If from a medial area there be subtracted a medial area incommensurable with the whole, the two remaining irrational straight lines arise, either a second apotome of a medial straight line or a straight line which produces with a medial area a medial whole.

For, as in the foregoing figures, let there be subtracted from the medial area BC the medial area BD incommensurable with the whole;

I say that the side - of EC is one of two irrational straight lines, either a second apotome of a medial straight line or a straight line which produces with a medial area a medial whole.

+ of EC is one of two irrational straight lines, either a second apotome of a medial straight line or a straight line which produces with a medial area a medial whole.

For, since each of the rectangles BC, BD is medial, and BC is incommensurable with BD, it follows that each of the straight lines FH, FK will be rational and incommensurable in length with FG. [X. 22]

And, since BC is incommensurable with BD, that is, GH with GK, HF is also incommensurable with FK; [VI. 1, X. 11] therefore FH, FK are rational straight lines commensurable in square only; therefore KH is an apotome. [X. 73]

If then the square on FH is greater than the square on FK by the square on a straight line commensurable with FH, while neither of the straight lines FH, FK is commensurable in length with the rational straight line FG set out, KH is a third apotome. [X. Deff. III. 3]

@@ -10177,7 +8603,7 @@ and, componendo, the squares on AB<

Then, since AB is an apotome, DE is a first apotome. [X. 97]

Let EF be the annex to it; therefore DF, FE are rational straight lines commensurable in square only, the square on DF is greater than the square on FE by the square on a straight line commensurable with DF, and DF is commensurable in length with the rational straight line DC set out. [X. Deff. III. 1]

Again, since AB is binomial, therefore DE is a first binomial straight line. [X. 60]

-

Let it be divided into its terms at G, and let DG be the greater term; therefore DG, GE are rational straight lines commensurable in square only, the square on DG is greater than the square on GE by the square on a straight line commensurable with DG, and the greater term DG is commensurable in length with the rational straight line DC set out. [X. Deff. II. 1]

+

Let it be divided into its terms at G, and let DG be the greater term; therefore DG, GE are rational straight lines commensurable in square only, the square on DG is greater than the square on GE by the square on a straight line commensurable with DG, and the greater term DG is commensurable in length with the rational straight line DC set out. [X. Deff. II. 1]

Therefore DF is also commensurable in length with DG; [X. 12] therefore the remainder GF is also commensurable in length with DF. [X. 15]

But DF is incommensurable in length with EF; therefore FG is also incommensurable in length with EF. [X. 13]

Therefore GF, FE are rational straight lines commensurable in square only; therefore EG is an apotome. [X. 73]

@@ -10187,7 +8613,7 @@ and, componendo, the squares on AB< PROPOSITION 112.

The square on a rational straight line applied to the binomial straight line produces as breadth an apotome the terms of which are commensurable with the terms of the binomial and moreover in the same ratio; and further the apotome so arising will have the same order as the binomial straight line.

-

Let A be a rational straight line, let BC be a binomial, and let DC be its greater term; let the rectangle BC, EF be equal to the square on A;

I say that EF is an apotome the terms of which are commensurable with CD, DB, and in the same ratio, and further EF will have the same order as BC.

+

Let A be a rational straight line, let BC be a binomial, and let DC be its greater term; let the rectangle BC, EF be equal to the square on A;

I say that EF is an apotome the terms of which are commensurable with CD, DB, and in the same ratio, and further EF will have the same order as BC.

For again let the rectangle BD, G be equal to the square on A.

Since then the rectangle BC, EF is equal to the rectangle BD, G, therefore, as CB is to BD, so is G to EF. [VI. 16]

But CB is greater than BD; therefore G is also greater than EF. [V. 16, V. 14]

@@ -10198,7 +8624,7 @@ and, componendo, the squares on AB<

And, as the square on HK is to the square on KF, so is HK to KE, since the three straight lines HK, KF, KE are proportional. [V. Def. 9]

Therefore HK is commensurable in length with KE, so that HE is also commensurable in length with EK. [X. 15]

Now, since the square on A is equal to the rectangle EH, BD, while the square on A is rational, therefore the rectangle EH, BD is also rational.

-

And it is applied to the rational straight line BD; therefore EH is rational and commensurable in length with BD; [X. 20] so that EK, being commensurable with it, is also rational and commensurable in length with BD.

+

And it is applied to the rational straight line BD; therefore EH is rational and commensurable in length with BD; [X. 20] so that EK, being commensurable with it, is also rational and commensurable in length with BD.

Since, then, as CD is to DB, so is FK to KE, while CD, DB are straight lines commensurable in square only, therefore FK, KE are also commensurable in square only. [X. 11]

But KE is rational; therefore FK is also rational.

Therefore FK, KE are rational straight lines commensurable in square only; therefore EF is an apotome. [X. 73]

@@ -10206,7 +8632,7 @@ and, componendo, the squares on AB<

If then the square on CD is greater than the square on DB by the square on a straight line commensurable with CD, the square on FK is also greater than the square on KE by the square on a straight line commensurable with FK. [X. 14]

And, if CD is commensurable in length with the rational straight line set out, so also is FK; [X. 11, 12] if BD is so commensurable, so also is KE; [X. 12] but, if neither of the straight lines CD, DB is so commensurable, neither of the straight lines FK, KE is so.

But, if the square on CD is greater than the square on DB by the square on a straight line incommensurable with CD, the square on FK is also greater than the square on KE by the square on a straight line incommensurable with FK. [X. 14]

-

And, if CD is commensurable with the rational straight line set out, so also is FK; if BD is so commensurable, so also is KE; but, if neither of the straight lines CD, DB is so commensurable, neither of the straight lines FK, KE is so; so that FE is an apotome, the terms of which FK, KE are commensurable with the terms CD, DB of the binomial straight line and in the same ratio, and it has the same order as BC. Q. E. D.

+

And, if CD is commensurable with the rational straight line set out, so also is FK; if BD is so commensurable, so also is KE; but, if neither of the straight lines CD, DB is so commensurable, neither of the straight lines FK, KE is so; so that FE is an apotome, the terms of which FK, KE are commensurable with the terms CD, DB of the binomial straight line and in the same ratio, and it has the same order as BC. Q. E. D.

PROPOSITION 113. @@ -10216,7 +8642,7 @@ and, componendo, the squares on AB<

For let DC be the annex to BD; therefore BC, CD are rational straight lines commensurable in square only. [X. 73]

Let the rectangle BC, G be also equal to the square on A.

But the square on A is rational; therefore the rectangle BC, G is also rational.

-

And it has been applied to the rational straight line BC; therefore G is rational and commensurable in length with BC. [X. 20]

+

And it has been applied to the rational straight line BC; therefore G is rational and commensurable in length with BC. [X. 20]

Since now the rectangle BC, G is equal to the rectangle BD, KH, therefore, proportionally, as CB is to BD, so is KH to G. [VI. 16]

But BC is greater than BD; therefore KH is also greater than G. [V. 16, V. 14]

Let KE be made equal to G; therefore KE is commensurable in length with BC.

@@ -10227,7 +8653,7 @@ and, componendo, the squares on AB<

But the square on KF is commensurable with the square on FH, for KF, FH are commensurable in square; therefore KF is also commensurable in length with FE, [X. 11] so that KF is also commensurable in length with KE. [X. 15]

But KE is rational and commensurable in length with BC; therefore KF is also rational and commensurable in length with BC. [X. 12]

And, since, as BC is to CD, so is KF to FH, alternately, as BC is to KF, so is DC to FH. [V. 16]

-

But BC is commensurable with KF; therefore FH is also commensurable in length with CD. [X. 11]

+

But BC is commensurable with KF; therefore FH is also commensurable in length with CD. [X. 11]

But BC, CD are rational straight lines commensurable in square only; therefore KF, FH are also rational straight lines [X. Def. 3] commensurable in square only; therefore KH is binomial. [X. 36]

If now the square on BC is greater than the square on CD by the square on a straight line commensurable with BC, the square on KF will also be greater than the square on FH by the square on a straight line commensurable with KF. [X 14]

And, if BC is commensurable in length with the rational straight line set out, so also is KF; if CD is commensurable in length with the rational straight line set out, so also is FH, but, if neither of the straight lines BC, CD, then neither of the straight lines KF, FH.

@@ -10244,7 +8670,7 @@ and, componendo, the squares on AB<

For let a rational straight line H be set out, and to CD let there be applied a rectangle equal to the square on H and producing KL as breadth.

Therefore KL is an apotome.

-

Let its terms be KM, ML commensurable with the terms CE, ED of the binomial straight line and in the same ratio. [X. 112]

+

Let its terms be KM, ML commensurable with the terms CE, ED of the binomial straight line and in the same ratio. [X. 112]

But CE, ED are also commensurable with AF, FB and in the same ratio; therefore, as AF is to FB, so is KM to ML.

Therefore, alternately, as AF is to KM, so is BF to LM; therefore also the remainder AB is to the remainder KL as AF is to KM. [V. 19]

But AF is commensurable with KM; [X. 12] therefore AB is also commensurable with KL. [X. 11]

@@ -10268,12 +8694,12 @@ and, componendo, the squares on AB<

And it is not the same with any of the preceding; for the square on none of the preceding, if applied to a rational straight line produces as breadth a medial straight line.

Again, let the square on D be equal to the rectangle B, C; therefore the square on D is irrational. [deduction from X. 20]

Therefore D is irrational; [X. Def. 4] and it is not the same with any of the preceding, for the square on none of the preceding, if applied to a rational straight line, produces C as breadth.

-

Similarly, if this arrangement proceeds ad infinitum, it is manifest that from the medial straight line there arise irrational straight lines infinite in number, and none is the same with any of the preceding. Q. E. D.

+

Similarly, if this arrangement proceeds ad infinitum, it is manifest that from the medial straight line there arise irrational straight lines infinite in number, and none is the same with any of the preceding. Q. E. D.

 - - - +
+
+ BOOK XI. DEFINITIONS. @@ -10309,7 +8735,7 @@ and, componendo, the squares on AB< 8

Parallel planes are those which do not meet.

- + 9

Similar solid figures are those contained by similar planes equal in multitude.

@@ -10357,7 +8783,7 @@ and, componendo, the squares on AB< 19

The axis of the cone is the straight line which remains fixed and about which the triangle is turned.

 - + 20

And the base is the circle described by the straight line which is carried round.

@@ -10395,9 +8821,9 @@ and, componendo, the squares on AB<

A dodecahedron is a solid figure contained by twelve equal, equilateral, and equiangular pentagons.

 - - - + + + BOOK XI. PROPOSITIONS. @@ -10424,7 +8850,7 @@ and, componendo, the squares on AB<

For let points F, G be taken at random on EC, EB, let CB, FG be joined, and let FH, GK be drawn across; I say first that the triangle ECB is in one plane.

For, if part of the triangle ECB, either FHC or GBK, is in the plane of reference, and the rest in another, a part also of one of the straight lines EC, EB will be in the plane of reference, and a part in another.

-

But, if the part FCBG of the triangle ECB be in the plane of reference, and the rest in another, a part also of both the straight lines EC, EB will be in the plane of reference and a part in another: which was proved absurd. [XI. 1]

+

But, if the part FCBG of the triangle ECB be in the plane of reference, and the rest in another, a part also of both the straight lines EC, EB will be in the plane of reference and a part in another: which was proved absurd. [XI. 1]

Therefore the triangle ECB is in one plane.

But, in whatever plane the triangle ECB is, in that plane also is each of the straight lines EC, EB, and, in whatever plane each of the straight lines EC, EB is, in that plane are AB, CD also. [XI. 1]

Therefore the straight lines AB, CD are in one plane, and every triangle is in one plane. Q. E. D.

@@ -10444,7 +8870,7 @@ and, componendo, the squares on AB< PROPOSITION 4.

If a straight line be set up at right angles to two straight lines which cut one another, at their common point of section, it will also be at right angles to the plane through them. -

+

For let a straight line EF be set up at right angles to the two straight lines AB, CD, which cut one another at the point E, from E; I say that EF is also at right angles to the plane through AB, CD.

For let AE, EB, CE, ED be cut off equal to one another, and let any straight line GEH be drawn across through E, at random; let AD, CB be joined, and further let FA, FG, FD, FC, FH, FB be joined from the point F taken at random <on EF>.

@@ -10453,7 +8879,7 @@ and, componendo, the squares on AB<

Therefore GE is equal to EH, and AG to BH.

And, since AE is equal to EB, while FE is common and at right angles, therefore the base FA is equal to the base FB. [I. 4]

For the same reason FC is also equal to FD.

-

And, since AD is equal to CB, and FA is also equal to FB, the two sides FA, AD are equal to the two sides FB, BC respectively; and the base FD was proved equal to the base FC; therefore the angle FAD is also equal to the angle FBC. [I. 8]

+

And, since AD is equal to CB, and FA is also equal to FB, the two sides FA, AD are equal to the two sides FB, BC respectively; and the base FD was proved equal to the base FC; therefore the angle FAD is also equal to the angle FBC. [I. 8]

And since, again, AG was proved equal to BH, and further FA also equal to FB, the two sides FA, AG are equal to the two sides FB, BH.

And the angle FAG was proved equal to the angle FBH; therefore the base FG is equal to the base FH. [I. 4]

Now since, again, GE was proved equal to EH, and EF is common, the two sides GE, EF are equal to the two sides HE, EF; and the base FG is equal to the base FH; therefore the angle GEF is equal to the angle HEF. [I. 8]

@@ -10474,7 +8900,7 @@ and, componendo, the squares on AB<

For suppose they are not, but, if possible, let BD, BE be in the plane of reference and BC in one more elevated; let the plane through AB, BC be produced; it will thus make, as common section in the plane of reference, a straight line. [XI. 3]

Let it make BF.

Therefore the three straight lines AB, BC, BF are in one plane, namely that drawn through AB, BC.

-

Now, since AB is at right angles to each of the straight lines BD, BE, therefore AB is also at right angles to the plane through BD, BE. [XI. 4]

+

Now, since AB is at right angles to each of the straight lines BD, BE, therefore AB is also at right angles to the plane through BD, BE. [XI. 4]

But the plane through BD, BE is the plane of reference; therefore AB is at right angles to the plane of reference.

Thus AB will also make right angles with all the straight lines which meet it and are in the plane of reference. [XI. Def. 3]

But BF which is in the plane of reference meets it; therefore the angle ABF is right.

@@ -10494,7 +8920,7 @@ and, componendo, the squares on AB<

But each of the straight lines BD, BE is in the plane of reference and meets AB; therefore each of the angles ABD, ABE is right.

For the same reason each of the angles CDB, CDE is also right.

And, since AB is equal to DE, and BD is common, the two sides AB, BD are equal to the two sides ED, DB; and they include right angles; therefore the base AD is equal to the base BE. [I. 4]

-

And, since AB is equal to DE, while AD is also equal to BE, the two sides AB, BE are equal to the two sides ED, DA; and AE is their common base; therefore the angle ABE is equal to the angle EDA. [I. 8]

+

And, since AB is equal to DE, while AD is also equal to BE, the two sides AB, BE are equal to the two sides ED, DA; and AE is their common base; therefore the angle ABE is equal to the angle EDA. [I. 8]

But the angle ABE is right; therefore the angle EDA is also right; therefore ED is at right angles to DA.

But it is also at right angles to each of the straight lines BD, DC; therefore ED is set up at right angles to the three straight lines BD, DA, DC at their point of meeting; therefore the three straight lines BD, DA, DC are in one plane. [XI. 5]

But, in whatever plane DB, DA are, in that plane is AB also, for every triangle is in one plane; [XI. 2] therefore the straight lines AB, BD, DC are in one plane.

@@ -10507,7 +8933,7 @@ and, componendo, the squares on AB<

Let AB, CD be two parallel straight lines, and let points E, F be taken at random on them respectively; I say that the straight line joining the points E, F is in the same plane with the parallel straight lines.

-

For suppose it is not, but, if possible, let it be in a more elevated plane as EGF, and let a plane be drawn through EGF; it will then make, as section in the plane of reference, a straight line. [XI. 3]

+

For suppose it is not, but, if possible, let it be in a more elevated plane as EGF, and let a plane be drawn through EGF; it will then make, as section in the plane of reference, a straight line. [XI. 3]

Let it make it, as EF; therefore the two straight lines EGF, EF will enclose an area: which is impossible.

Therefore the straight line joined from E to F is not in a plane more elevated; therefore the straight line joined from E to F is in the plane through the parallel straight lines AB, CD.

Therefore etc. Q. E. D.

@@ -10521,7 +8947,7 @@ and, componendo, the squares on AB<

For let AB, CD meet the plane of reference at the points B, D, and let BD be joined; therefore AB, CD, BD are in one plane. [XI. 7]

Let DE be drawn, in the plane of reference, at right angles to BD, let DE be made equal to AB, and let BE, AE, AD be joined.

Now, since AB is at right angles to the plane of reference, therefore AB is also at right angles to all the straight lines which meet it and are in the plane of reference; [XI. Def. 3] therefore each of the angles ABD, ABE is right.

-

And, since the straight line BD has fallen on the parallels AB, CD, therefore the angles ABD, CDB are equal to two right angles. [I. 29]

+

And, since the straight line BD has fallen on the parallels AB, CD, therefore the angles ABD, CDB are equal to two right angles. [I. 29]

But the angle ABD is right; therefore the angle CDB is also right; therefore CD is at right angles to BD.

And, since AB is equal to DE, and BD is common, the two sides AB, BD are equal to the two sides ED, DB; and the angle ABD is equal to the angle EDB, for each is right; therefore the base AD is equal to the base BE.

And, since AB is equal to DE, and BE to AD, the two sides AB, BE are equal to the two sides ED, DA respectively, and AE is their common base; therefore the angle ABE is equal to the angle EDA.

@@ -10531,7 +8957,7 @@ and, componendo, the squares on AB<

But DC is in the plane through BD, DA, inasmuch as AB, BD are in the plane through BD, DA, [XI. 2] and DC is also in the plane in which AB, BD are.

Therefore ED is at right angles to DC, so that CD is also at right angles to DE.

But CD is also at right angles to BD.

-

Therefore CD is set up at right angles to the two straight lines DE, DB which cut one another, from the point of section at D; so that CD is also at right angles to the plane through DE, DB. [XI. 4]

+

Therefore CD is set up at right angles to the two straight lines DE, DB which cut one another, from the point of section at D; so that CD is also at right angles to the plane through DE, DB. [XI. 4]

But the plane through DE, DB is the plane of reference; therefore CD is at right angles to the plane of reference.

Therefore etc. Q. E. D.

@@ -10548,7 +8974,7 @@ and, componendo, the squares on AB<

For the same reason CD is also at right angles to the plane through HG, GK; therefore each of the straight lines AB, CD is at right angles to the plane through HG, GK.

But if two straight lines be at right angles to the same plane, the straight lines are parallel; [XI. 6] therefore AB is parallel to CD.

- + PROPOSITION 10.

If two straight lines meeting one another be parallel to two straight lines meeting one another not in the same plane, they will contain equal angles. @@ -10562,7 +8988,7 @@ and, componendo, the squares on AB<

But straight lines which are parallel to the same straight line and are not in the same plane with it are parallel to one another; [XI. 9] therefore AD is parallel and equal to CF.

And AC, DF join them; therefore AC is also equal and parallel to DF. [I. 33]

Now, since the two sides AB, BC are equal to the two sides DE, EF, and the base AC is equal to the base DF, therefore the angle ABC is equal to the angle DEF. [I. 8]

-

Therefore etc.

+

Therefore etc.

 PROPOSITION 11. @@ -10575,7 +9001,7 @@ and, componendo, the squares on AB<

If then AD is also perpendicular to the plane of reference, that which was enjoined will have been done.

But, if not, let DE be drawn from the point D at right angles to BC and in the plane of reference, [I. 11] let AF be drawn from A perpendicular to DE, [I. 12] and let GH be drawn through the point F parallel to BC. [I. 31]

Now, since BC is at right angles to each of the straight lines DA, DE, therefore BC is also at right angles to the plane through ED, DA. [XI. 4]

-

And GH is parallel to it; but, if two straight lines be parallel, and one of them be at right angles to any plane, the remaining one will also be at right angles to the same plane; [XI. 8] therefore GH is also at right angles to the plane through ED, DA.

+

And GH is parallel to it; but, if two straight lines be parallel, and one of them be at right angles to any plane, the remaining one will also be at right angles to the same plane; [XI. 8] therefore GH is also at right angles to the plane through ED, DA.

Therefore GH is also at right angles to all the straight lines which meet it and are in the plane through ED, DA. [XI. Def. 3]

But AF meets it and is in the plane through ED, DA; therefore GH is at right angles to FA, so that FA is also at right angles to GH.

But AF is also at right angles to DE; therefore AF is at right angles to each of the straight lines GH, DE.

@@ -10594,7 +9020,7 @@ and, componendo, the squares on AB<

Then, since AD, CB are two parallel straight lines, while one of them, BC, is at right angles to the plane of reference, therefore the remaining one, AD, is also at right angles to the plane of reference. [XI. 8]

Therefore AD has been set up at right angles to the given plane from the point A in it.

 - + PROPOSITION 13.

@@ -10609,7 +9035,7 @@ and, componendo, the squares on AB<

And they are in one plane: which is impossible.

Therefore etc. Q. E. D.

 - + PROPOSITION 14.

@@ -10626,7 +9052,7 @@ and, componendo, the squares on AB<

Therefore the planes CD, EF will not meet when produced; therefore the planes CD, EF are parallel. [XI. Def. 8]

Therefore planes to which the same straight line is at right angles are parallel. Q. E. D.

 - + PROPOSITION 15.

If two straight lines meeting one another be parallel to two straight lines meeting one another, not being in the same plane, the planes through them are parallel. @@ -10639,11 +9065,11 @@ and, componendo, the squares on AB<

And, since BA is parallel to GH, [XI. 9] therefore the angles GBA, BGH are equal to two right angles. [I. 29]

But the angle BGH is right; therefore the angle GBA is also right; therefore GB is at right angles to BA.

For the same reason GB is also at right angles to BC.

-

Since then the straight line GB is set up at right angles to the two straight lines BA, BC which cut one another, therefore GB is also at right angles to the plane through BA, BC. [XI. 4]

+

Since then the straight line GB is set up at right angles to the two straight lines BA, BC which cut one another, therefore GB is also at right angles to the plane through BA, BC. [XI. 4]

But planes to which the same straight line is at right angles are parallel; [XI. 14] therefore the plane through AB, BC is parallel to the plane through DE, EF.

Therefore, if two straight lines meeting one another be parallel to two straight lines meeting one another, not in the same plane, the planes through them are parallel. Q. E. D.

 - + PROPOSITION 16.

If two parallel planes be cut by any plane, their common sections are parallel. @@ -10659,7 +9085,7 @@ and, componendo, the squares on AB<

Similarly we can prove that neither will the straight lines EF, GH meet when produced in the direction of E, G.

But straight lines which do not meet in either direction are parallel. [I. Def. 23]

Therefore EF is parallel to GH.

-

Therefore etc. Q. E. D.

+

Therefore etc. Q. E. D.

 PROPOSITION 17. @@ -10667,7 +9093,7 @@ and, componendo, the squares on AB<

For let the two straight lines AB, CD be cut by the parallel planes GH, KL, MN at the points A, E, B and C, F, D; I say that, as the straight line AE is to EB, so is CF to FD.

-

For let AC, BD, AD be joined, let AD meet the plane KL at the point O, and let EO, OF be joined.

+

For let AC, BD, AD be joined, let AD meet the plane KL at the point O, and let EO, OF be joined.

Now, since the two parallel planes KL, MN are cut by the plane EBDO, their common sections EO, BD are parallel. [XI. 16]

For the same reason, since the two parallel planes GH, KL are cut by the plane AOFC, their common sections AC, OF are parallel. [id.]

And, since the straight line EO has been drawn parallel to BD, one of the sides of the triangle ABD, therefore, proportionally, as AE is to EB, so is AO to OD. [VI. 2]

@@ -10682,7 +9108,7 @@ and, componendo, the squares on AB<

For let any straight line AB be at right angles to the plane of reference; I say that all the planes through AB are also at right angles to the plane of reference.

For let the plane DE be drawn through AB, let CE be the common section of the plane DE and the plane of reference, let a point F be taken at random on CE, and from F let FG be drawn in the plane DE at right angles to CE. [I. 11]

-

Now, since AB is at right angles to the plane of reference, AB is also at right angles to all the straight lines which meet it and are in the plane of reference; [XI. Def. 3] so that it is also at right angles to CE; therefore the angle ABF is right.

+

Now, since AB is at right angles to the plane of reference, AB is also at right angles to all the straight lines which meet it and are in the plane of reference; [XI. Def. 3] so that it is also at right angles to CE; therefore the angle ABF is right.

But the angle GFB is also right; therefore AB is parallel to FG. [I. 28]

But AB is at right angles to the plane of reference; therefore FG is also at right angles to the plane of reference. [XI. 8]

Now a plane is at right angles to a plane, when the straight lines drawn, in one of the planes, at right angles to the common section of the planes are at right angles to the remaining plane. [XI. Def. 4]

@@ -10697,7 +9123,7 @@ and, componendo, the squares on AB<

For let the two planes AB, BC be at right angles to the plane of reference, and let BD be their common section; I say that BD is at right angles to the plane of reference.

For suppose it is not, and from the point D let DE be drawn in the plane AB at right angles to the straight line AD, and DF in the plane BC at right angles to CD.

-

Now, since the plane AB is at right angles to the plane of reference, and DE has been drawn in the plane AB at right angles to AD, their common section, therefore DE is at right angles to the plane of reference. [XI. Def. 4]

+

Now, since the plane AB is at right angles to the plane of reference, and DE has been drawn in the plane AB at right angles to AD, their common section, therefore DE is at right angles to the plane of reference. [XI. Def. 4]

Similarly we can prove that DF is also at right angles to the plane of reference.

Therefore from the same point D two straight lines have been set up at right angles to the plane of reference on the same side: which is impossible. [XI. 13]

Therefore no straight line except the common section DB of the planes AB, BC can be set up from the point D at right angles to the plane of reference.

@@ -10710,7 +9136,7 @@ and, componendo, the squares on AB<

For let the solid angle at A be contained by the three plane angles BAC, CAD, DAB; I say that any two of the angles BAC, CAD, DAB, taken together in any manner, are greater than the remaining one.

If now the angles BAC, CAD, DAB are equal to one another, it is manifest that any two are greater than the remaining one.

-

But, if not, let BAC be greater, and on the straight line AB, and at the point A on it, let the angle BAE be constructed, in the plane through BA, AC, equal to the angle DAB; let AE be made equal to AD, and let BEC, drawn across through the point E, cut the straight lines AB, AC at the points B, C; let DB, DC be joined.

+

But, if not, let BAC be greater, and on the straight line AB, and at the point A on it, let the angle BAE be constructed, in the plane through BA, AC, equal to the angle DAB; let AE be made equal to AD, and let BEC, drawn across through the point E, cut the straight lines AB, AC at the points B, C; let DB, DC be joined.

Now, since DA is equal to AE, and AB is common, two sides are equal to two sides; and the angle DAB is equal to the angle BAE; therefore the base DB is equal to the base BE. [I. 4]

And, since the two sides BD, DC are greater than BC, [I. 20] and of these DB was proved equal to BE, therefore the remainder DC is greater than the remainder EC.

Now, since DA is equal to AE, and AC is common, and the base DC is greater than the base EC, therefore the angle DAC is greater than the angle EAC. [I. 25]

@@ -10728,7 +9154,7 @@ and, componendo, the squares on AB<

Now, since the solid angle at B is contained by the three plane angles CBA, ABD, CBD, any two are greater than the remaining one; [XI. 20] therefore the angles CBA, ABD are greater than the angle CBD.

For the same reason the angles BCA, ACD are also greater than the angle BCD, and the angles CDA, ADB are greater than the angle CDB; therefore the six angles CBA, ABD, BCA, ACD, CDA, ADB are greater than the three angles CBD, BCD, CDB.

-

But the three angles CBD, BDC, BCD are equal to two right angles; [I. 32] therefore the six angles CBA, ABD, BCA, ACD, CDA, ADB are greater than two right angles.

+

But the three angles CBD, BDC, BCD are equal to two right angles; [I. 32] therefore the six angles CBA, ABD, BCA, ACD, CDA, ADB are greater than two right angles.

And, since the three angles of each of the triangles ABC, ACD, ADB are equal to two right angles, therefore the nine angles of the three triangles, the angles CBA, ACB, BAC, ACD, CDA, CAD, ADB, DBA, BAD are equal to six right angles; and of them the six angles ABC, BCA, ACD, CDA, ADB, DBA are greater than two right angles; therefore the remaining three angles BAC, CAD, DAB containing the solid angle are less than four right angles.

Therefore etc. Q. E. D.

@@ -10736,13 +9162,13 @@ and, componendo, the squares on AB< PROPOSITION 22.

If there be three plane angles of which two, taken together in any manner, are greater than the remaining one, and they are contained by equal straight lines, it is possible to construct a triangle out of the straight lines joining the extremities of the equal straight lines.

-

Let there be three plane angles ABC, DEF, GHK, of which two, taken together in any manner, are greater than the remaining one, namely the angles ABC, DEF greater than the angle GHK, the angles DEF, GHK greater than the angle ABC, and, further, the angles GHK, ABC greater than the angle DEF; let the straight lines AB, BC, DE, EF, GH, HK be equal, and let AC, DF, GK be joined; I say that it is possible to construct a triangle out of straight lines equal to AC, DF, GK, that is, that any two of the straight lines AC, DF, GK are greater than the remaining one.

+

Let there be three plane angles ABC, DEF, GHK, of which two, taken together in any manner, are greater than the remaining one, namely the angles ABC, DEF greater than the angle GHK, the angles DEF, GHK greater than the angle ABC, and, further, the angles GHK, ABC greater than the angle DEF; let the straight lines AB, BC, DE, EF, GH, HK be equal, and let AC, DF, GK be joined; I say that it is possible to construct a triangle out of straight lines equal to AC, DF, GK, that is, that any two of the straight lines AC, DF, GK are greater than the remaining one.

Now, if the angles ABC, DEF, GHK are equal to one another, it is manifest that, AC, DF, GK being equal also, it is possible to construct a triangle out of straight lines equal to AC, DF, GK.

But, if not, let them be unequal, and on the straight line HK, and at the point H on it, let the angle KHL be constructed equal to the angle ABC; let HL be made equal to one of the straight lines AB, BC, DE, EF, GH, HK, and let KL, GL be joined.

Now, since the two sides AB, BC are equal to the two sides KH, HL, and the angle at B is equal to the angle KHL, therefore the base AC is equal to the base KL. [I. 4]

-

And, since the angles ABC, GHK are greater than the angle DEF, while the angle ABC is equal to the angle KHL, therefore the angle GHL is greater than the angle DEF.

+

And, since the angles ABC, GHK are greater than the angle DEF, while the angle ABC is equal to the angle KHL, therefore the angle GHL is greater than the angle DEF.

And, since the two sides GH, HL are equal to the two sides DE, EF, and the angle GHL is greater than the angle DEF, therefore the base GL is greater than the base DF. [I. 24]

But GK, KL are greater than GL.

Therefore GK, KL are much greater than DF.

@@ -10753,7 +9179,7 @@ and, componendo, the squares on AB< PROPOSITION 23.

To construct a solid angle out of three plane angles two of which, taken together in any manner, are greater than the remaining one: thus the three angles must be less than four right angles. -

+

Let the angles ABC, DEF, GHK be the three given plane angles, and let two of these, taken together in any manner, be greater than the remaining one, while, further, the three are less than four right angles; thus it is required to construct a solid angle out of angles equal to the angles ABC, DEF, GHK.

Let AB, BC, DE, EF, GH, HK be cut off equal to one another, and let AC, DF, GK be joined; it is therefore possible to construct a triangle out of straight lines equal to AC, DF, GK. [XI. 22]

@@ -10762,7 +9188,7 @@ and, componendo, the squares on AB<

For, if not, AB is either equal to LO, or less.

First, let it be equal.

Then, since AB is equal to LO, while AB is equal to BC, and OL to OM, the two sides AB, BC are equal to the two sides LO, OM respectively; and, by hypothesis, the base AC is equal to the base LM; therefore the angle ABC is equal to the angle LOM. [I. 8]

-

For the same reason the angle DEF is also equal to the angle MON, and further the angle GHK to the angle NOL; therefore the three angles ABC, DEF, GHK are equal to the three angles LOM, MON, NOL.

+

For the same reason the angle DEF is also equal to the angle MON, and further the angle GHK to the angle NOL; therefore the three angles ABC, DEF, GHK are equal to the three angles LOM, MON, NOL.

But the three angles LOM, MON, NOL are equal to four right angles; therefore the angles ABC, DEF, GHK are equal to four right angles.

But they are also, by hypothesis, less than four right angles: which is absurd.

Therefore AB is not equal to LO.

@@ -10775,7 +9201,7 @@ and, componendo, the squares on AB<

Since, then, the two sides AB, BC are equal to the two sides PO, OQ, and the base AC is greater than the base PQ, therefore the angle ABC is greater than the angle POQ. [I. 25]

Similarly we can prove that the angle DEF is also greater than the angle MON, and the angle GHK greater than the angle NOL.

Therefore the three angles ABC, DEF, GHK are greater than the three angles LOM, MON, NOL.

-

But, by hypothesis, the angles ABC, DEF, GHK are less than four right angles; therefore the angles LOM, MON, NOL are much less than four right angles.

+

But, by hypothesis, the angles ABC, DEF, GHK are less than four right angles; therefore the angles LOM, MON, NOL are much less than four right angles.

But they are also equal to four right angles: which is absurd.

Therefore AB is not less than LO.

And it was proved that neither is it equal; therefore AB is greater than LO.

@@ -10785,7 +9211,7 @@ and, componendo, the squares on AB<

For the same reason RN is also equal to each of the straight lines RL, RM; therefore the three straight lines RL, RM, RN are equal to one another.

Next, since by hypothesis the square on OR is equal to that area by which the square on AB is greater than the square on LO, therefore the square on AB is equal to the squares on LO, OR.

But the square on LR is equal to the squares on LO, OR, for the angle LOR is right; [I. 47] therefore the square on AB is equal to the square on RL; therefore AB is equal to RL.

-

But each of the straight lines BC, DE, EF, GH, HK is equal to AB, while each of the straight lines RM, RN is equal to RL; therefore each of the straight lines AB, BC, DE, EF, GH, HK is equal to each of the straight lines RL, RM, RN.

+

But each of the straight lines BC, DE, EF, GH, HK is equal to AB, while each of the straight lines RM, RN is equal to RL; therefore each of the straight lines AB, BC, DE, EF, GH, HK is equal to each of the straight lines RL, RM, RN.

And, since the two sides LR, RM are equal to the two sides AB, BC, and the base LM is by hypothesis equal to the base AC, therefore the angle LRM is equal to the angle ABC. [I. 8]

For the same reason the angle MRN is also equal to the angle DEF, and the angle LRN to the angle GHK.

Therefore, out of the three plane angles LRM, MRN, LRN, which are equal to the three given angles ABC, DEF, GHK, the solid angle at R has been constructed, which is contained by the angles LRM, MRN, LRN. Q. E. F.

@@ -10808,7 +9234,7 @@ and, componendo, the squares on AB<

For let the solid CDHG be contained by the parallel planes AC, GF, AH, DF, BF, AE; I say that the opposite planes in it are equal and parallelogrammic.

-

For, since the two parallel planes BG, CE are cut by the plane AC, their common sections are parallel. [XI. 16]

+

For, since the two parallel planes BG, CE are cut by the plane AC, their common sections are parallel. [XI. 16]

Therefore AB is parallel to DC.

Again, since the two parallel planes BF, AE are cut by the plane AC, their common sections are parallel. [XI. 16]

Therefore BC is parallel to AD.

@@ -10821,7 +9247,7 @@ and, componendo, the squares on AB<

Similarly we can prove that AC is also equal to GF, and AE to BF.

Therefore etc. Q. E. D.

- + PROPOSITION 25.

If a parallelepipedal solid be cut by a plane which is parallel to the opposite planes, then, as the base is to the base, so will the solid be to the solid. @@ -10831,7 +9257,7 @@ and, componendo, the squares on AB<

For let AH be produced in each direction, let any number of straight lines whatever, AK, KL, be made equal to AE, and any number whatever, HM, MN, equal to EH; and let the parallelograms LP, KV, HW, MS and the solids LQ, KR, DM, MT be completed.

Then, since the straight lines LK, KA, AE are equal to one another, the parallelograms LP, KV, AF are also equal to one another, KO, KB, AG are equal to one another, and further LX, KQ, AR are equal to one another, for they are opposite. [XI. 24]

For the same reason the parallelograms EC, HW, MS are also equal to one another, HG, HI, IN are equal to one another, and further DH, MY, NT are equal to one another.

-

Therefore in the solids LQ, KR, AU three planes are equal to three planes.

+

Therefore in the solids LQ, KR, AU three planes are equal to three planes.

But the three planes are equal to the three opposite; therefore the three solids LQ, KR, AU are equal to one another.

For the same reason the three solids ED, DM, MT are also equal to one another.

Therefore, whatever multiple the base LF is of the base AF, the same multiple also is the solid LU of the solid AU.

@@ -10846,7 +9272,7 @@ and, componendo, the squares on AB<

Let AB be the given straight line, A the given point on it, and the angle at D, contained by the angles EDC, EDF, FDC, the given solid angle; thus it is required to construct on the straight line AB, and at the point A on it, a solid angle equal to the solid angle at D.

-

For let a point F be taken at random on DF, let FG be drawn from F perpendicular to the plane through ED, DC, and let it meet the plane at G, [XI. 11] let DG be joined, let there be constructed on the straight line AB and at the point A on it the angle BAL equal to the angle EDC, and the angle BAK equal to the angle EDG, [I. 23] let AK be made equal to DG, let KH be set up from the point K at right angles to the plane through BA, AL, [XI. 12] let KH be made equal to GF, and let HA be joined; I say that the solid angle at A, contained by the angles BAL, BAH, HAL is equal to the solid angle at D contained by the angles EDC, EDF, FDC.

+

For let a point F be taken at random on DF, let FG be drawn from F perpendicular to the plane through ED, DC, and let it meet the plane at G, [XI. 11] let DG be joined, let there be constructed on the straight line AB and at the point A on it the angle BAL equal to the angle EDC, and the angle BAK equal to the angle EDG, [I. 23] let AK be made equal to DG, let KH be set up from the point K at right angles to the plane through BA, AL, [XI. 12] let KH be made equal to GF, and let HA be joined; I say that the solid angle at A, contained by the angles BAL, BAH, HAL is equal to the solid angle at D contained by the angles EDC, EDF, FDC.

For let AB, DE be cut off equal to one another, and let HB, KB, FE, GE be joined.

Then, since FG is at right angles to the plane of reference, it will also make right angles with all the straight lines which meet it and are in the plane of reference; [XI. Def. 3] therefore each of the angles FGD, FGE is right.

For the same reason each of the angles HKA, HKB is also right.

@@ -10856,7 +9282,7 @@ and, componendo, the squares on AB<

But AB is also equal to DE; therefore the two sides HA, AB are equal to the two sides DF, DE.

And the base HB is equal to the base FE; therefore the angle BAH is equal to the angle EDF. [I. 8]

For the same reason the angle HAL is also equal to the angle FDC.

-

And the angle BAL is also equal to the angle EDC.

+

And the angle BAL is also equal to the angle EDC.

Therefore on the straight line AB, and at the point A on it, a solid angle has been constructed equal to the given solid angle at D. Q. E. F.

@@ -10869,7 +9295,7 @@ and, componendo, the squares on AB<

For on the straight line AB and at the point A on it let the solid angle, contained by the angles BAH, HAK, KAB, be constructed equal to the solid angle at C, so that the angle BAH is equal to the angle ECF, the angle BAK equal to the angle ECG, and the angle KAH to the angle GCF; and let it be contrived that, as EC is to CG, so is BA to AK, and, as GC is to CF, so is KA to AH. [VI. 12]

Therefore also, ex aequali, as EC is to CF, so is BA to AH. [V. 22]

Let the parallelogram HB and the solid AL be completed.

-

Now since, as EC is to CG, so is BA to AK, and the sides about the equal angles ECG, BAK are thus proportional, therefore the parallelogram GE is similar to the parallelogram KB.

+

Now since, as EC is to CG, so is BA to AK, and the sides about the equal angles ECG, BAK are thus proportional, therefore the parallelogram GE is similar to the parallelogram KB.

For the same reason the parallelogram KH is also similar to the parallelogram GF, and further FE to HB; therefore three parallelograms of the solid CD are similar to three parallelograms of the solid AL.

But the former three are both equal and similar to the three opposite parallelograms, and the latter three are both equal and similar to the three opposite parallelograms; therefore the whole solid CD is similar to the whole solid AL. [XI. Def. 9]

Therefore on the given straight line AB there has been described AL similar and similarly situated to the given parallelepipedal solid CD. Q. E. F.

@@ -10880,7 +9306,7 @@ and, componendo, the squares on AB<

For let the parallelepipedal solid AB be cut by the plane CDEF through the diagonals CF, DE of opposite planes; I say that the solid AB will be bisected by the plane CDEF.

-

For, since the triangle CGF is equal to the triangle CFB, [I. 34] and ADE to DEH, while the parallelogram CA is also equal to the parallelogram EB, for they are opposite, and GE to CH, therefore the prism contained by the two triangles CGF, ADE and the three parallelograms GE, AC, CE is also equal to the prism contained by the two triangles CFB, DEH and the three parallelograms CH, BE, CE; for they are contained by planes equal both in multitude and in magnitude. [XI. Def. 10]

+

For, since the triangle CGF is equal to the triangle CFB, [I. 34] and ADE to DEH, while the parallelogram CA is also equal to the parallelogram EB, for they are opposite, and GE to CH, therefore the prism contained by the two triangles CGF, ADE and the three parallelograms GE, AC, CE is also equal to the prism contained by the two triangles CFB, DEH and the three parallelograms CH, BE, CE; for they are contained by planes equal both in multitude and in magnitude. [XI. Def. 10]

Hence the whole solid AB is bisected by the plane CDEF. Q. E. D.

@@ -10891,7 +9317,7 @@ and, componendo, the squares on AB<

For, since each of the figures CH, CK is a parallelogram, CB is equal to each of the straight lines DH, EK, [I. 34] hence DH is also equal to EK.

Let EH be subtracted from each; therefore the remainder DE is equal to the remainder HK.

-

Hence the triangle DCE is also equal to the triangle HBK, [I. 8, 4] and the parallelogram DG to the parallelogram HN. [I. 36]

+

Hence the triangle DCE is also equal to the triangle HBK, [I. 8, 4] and the parallelogram DG to the parallelogram HN. [I. 36]

For the same reason the triangle AFG is also equal to the triangle MLN.

But the parallelogram CF is equal to the parallelogram BM, and CG to BN, for they are opposite; therefore the prism contained by the two triangles AFG, DCE and the three parallelograms AD, DG, CG is equal to the prism contained by the two triangles MLN, HBK and the three parallelograms BM, HN, BN.

Let there be added to each the solid of which the parallelogram AB is the base and GEHM its opposite; therefore the whole parallelepipedal solid CM is equal to the whole parallelepipedal solid CN.

@@ -10900,7 +9326,7 @@ and, componendo, the squares on AB< PROPOSITION 30.

Parallelepipedal solids which are on the same base and of the same height, and in which the extremities of the sides which stand up are not on the same straight lines, are equal to one another. -

+

Let CM, CN be parallelepipedal solids on the same base AB and of the same height, and let the extremities of their sides which stand up, namely AF, AG, LM, LN, CD, CE, BH, BK, not be on the same straight lines; I say that the solid CM is equal to the solid CN.

For let NK, DH be produced and meet one another at R, and further let FM, GE be produced to P, Q; let AO, LP, CQ, BR be joined.

@@ -10919,7 +9345,7 @@ and, componendo, the squares on AB<

First, let the sides which stand up, HK, BE, AG, LM, PQ, DF, CO, RS, be at right angles to the bases AB, CD; let the straight line RT be produced in a straight line with CR; on the straight line RT, and at the point R on it, let the angle TRU be constructed equal to the angle ALB, [I. 23] let RT be made equal to AL, and RU equal to LB, and let the base RW and the solid XU be completed.

Now, since the two sides TR, RU are equal to the two sides AL, LB, and they contain equal angles, therefore the parallelogram RW is equal and similar to the parallelogram HL.

-

Since again AL is equal to RT, and LM to RS, and they contain right angles, therefore the parallelogram RX is equal and similar to the parallelogram AM.

+

Since again AL is equal to RT, and LM to RS, and they contain right angles, therefore the parallelogram RX is equal and similar to the parallelogram AM.

For the same reason LE is also equal and similar to SU; therefore three parallelograms of the solid AE are equal and similar to three parallelograms of the solid XU.

But the former three are equal and similar to the three opposite, and the latter three to the three opposite; [XI. 24] therefore the whole parallelepipedal solid AE is equal to the whole parallelepipedal solid XU. [XI. Def. 10]

Let DR, WU be drawn through and meet one another at Y, let aTb be drawn through T parallel to DY, let PD be produced to a, and let the solids YX, RI be completed.

@@ -10927,7 +9353,7 @@ and, componendo, the squares on AB<

But the solid XU is equal to AE: therefore the solid XY is also equal to the solid AE.

And, since the parallelogram RUWT is equal to the parallelogram YT for they are on the same base RT and in the same parallels RT, YW, [I. 35] while RUWT is equal to CD, since it is also equal to AB, therefore the parallelogram YT is also equal to CD.

But DT is another parallelogram; therefore, as the base CD is to DT, so is YT to DT. [V. 7]

-

And, since the parallelepipedal solid CI has been cut by the plane RF which is parallel to opposite planes, as the base CD is to the base DT, so is the solid CF to the solid RI. [XI. 25]

+

And, since the parallelepipedal solid CI has been cut by the plane RF which is parallel to opposite planes, as the base CD is to the base DT, so is the solid CF to the solid RI. [XI. 25]

For the same reason, since the parallelepipedal solid YI has been cut by the plane RX which is parallel to opposite planes, as the base YT is to the base TD, so is the solid YX to the solid RI. [XI. 25]

But, as the base CD is to DT, so is YT to DT; therefore also, as the solid CF is to the solid RI, so is the solid YX to RI. [V. 11]

Therefore each of the solids CF, YX has to RI the same ratio; therefore the solid CF is equal to the solid YX. [V. 9]

@@ -10936,7 +9362,7 @@ and, componendo, the squares on AB<

For from the points K, E, G, M, Q, F, N, S let KO, ET, GU, MV, QW, FX, NY, SI be drawn perpendicular to the plane of reference, and let them meet the plane at the points O, T, U, V, W, X, Y, I, and let OT, OU, UV, TV, WX, WY, YI, IX be joined.

Then the solid KV is equal to the solid QI, for they are on the equal bases KM, QS and of the same height, and their sides which stand up are at right angles to their bases. [First part of this Prop.]

-

But the solid KV is equal to the solid AE, and QI to CF; for they are on the same base and of the same height, while the extremities of their sides which stand up are not on the same straight lines. [XI. 30]

+

But the solid KV is equal to the solid AE, and QI to CF; for they are on the same base and of the same height, while the extremities of their sides which stand up are not on the same straight lines. [XI. 30]

Therefore the solid AE is also equal to the solid CF.

Therefore etc. Q. E. D.

@@ -10961,7 +9387,7 @@ and, componendo, the squares on AB<

Let AB, CD be similar parallelepipedal solids, and let AE be the side corresponding to CF; I say that the solid AB has to the solid CD the ratio triplicate of that which AE has to CF.

For let EK, EL, EM be produced in a straight line with AE, GE, HE, let EK be made equal to CF, EL equal to FN, and further EM equal to FR, and let the parallelogram KL and the solid KP be completed.

-

Now, since the two sides KE, EL are equal to the two sides CF, FN, while the angle KEL is also equal to the angle CFN, inasmuch as the angle AEG is also equal to the angle CFN because of the similarity of the solids AB, CD, therefore the parallelogram KL is equal <and similar> to the parallelogram CN.

+

Now, since the two sides KE, EL are equal to the two sides CF, FN, while the angle KEL is also equal to the angle CFN, inasmuch as the angle AEG is also equal to the angle CFN because of the similarity of the solids AB, CD, therefore the parallelogram KL is equal <and similar> to the parallelogram CN.

For the same reason the parallelogram KM is also equal and similar to CR, and further EP to DF; therefore three parallelograms of the solid KP are equal and similar to three parallelograms of the solid CD.

But the former three parallelograms are equal and similar to their opposites, and the latter three to their opposites; [XI. 24] therefore the whole solid KP is equal and similar to the whole solid CD. [XI. Def. 10]

Let the parallelogram GK be completed, and on the parallelograms GK, KL as bases, and with the same height as that of AB, let the solids EO, LQ be completed.

@@ -10969,7 +9395,7 @@ and, componendo, the squares on AB<

But, as AE is to EK, so is AG to the parallelogram GK, as GE is to EL, so is GK to KL, and, as HE is to EM, so is QE to KM; [VI. 1] therefore also, as the parallelogram AG is to GK, so is GK to KL, and QE to KM.

But, as AG is to GK, so is the solid AB to the solid EO, as GK is to KL, so is the solid OE to the solid QL, and, as QE is to KM, so is the solid QL to the solid KP; [XI. 32] therefore also, as the solid AB is to EO, so is EO to QL, and QL to KP.

But, if four magnitudes be continuously proportional, the first has to the fourth the ratio triplicate of that which it has to the second; [V. Def. 10] therefore the solid AB has to KP the ratio triplicate of that which AB has to EO.

-

But, as AB is to EO, so is the parallelogram AG to GK, and the straight line AE to EK [VI. 1]; hence the solid AB has also to KP the ratio triplicate of that which AE has to EK.

+

But, as AB is to EO, so is the parallelogram AG to GK, and the straight line AE to EK [VI. 1]; hence the solid AB has also to KP the ratio triplicate of that which AE has to EK.

But the solid KP is equal to the solid CD, and the straight line EK to CF; therefore the solid AB has also to the solid CD the ratio triplicate of that which the corresponding side of it, AE, has to the corresponding side CF.

Therefore etc. Q. E. D.

@@ -10984,7 +9410,7 @@ and, componendo, the squares on AB<

Let AB, CD be equal parallelepipedal solids; I say that in the parallelepipedal solids AB, CD the bases are reciprocally proportional to the heights, that is, as the base EH is to the base NQ, so is the height of the solid CD to the height of the solid AB.

First, let the sides which stand up, namely AG, EF, LB, HK, CM, NO, PD, QR, be at right angles to their bases; I say that, as the base EH is to the base NQ, so is CM to AG.

-

If now the base EH is equal to the base NQ, while the solid AB is also equal to the solid CD, CM will also be equal to AG.

+

If now the base EH is equal to the base NQ, while the solid AB is also equal to the solid CD, CM will also be equal to AG.

For parallelepipedal solids of the same height are to one another as the bases; [XI. 32] and, as the base EH is to NQ, so will CM be to AG, and it is manifest that in the parallelepipedal solids AB, CD the bases are reciprocally proportional to the heights.

Next, let the base EH not be equal to the base NQ, but let EH be greater.

@@ -10992,7 +9418,7 @@ and, componendo, the squares on AB<

Let then CT be made equal to AG, and let the parallelepipedal solid VC be completed on NQ as base and with CT as height.

Now, since the solid AB is equal to the solid CD, and CV is outside them, while equals have to the same the same ratio, [V. 7] therefore, as the solid AB is to the solid CV, so is the solid CD to the solid CV.

But, as the solid AB is to the solid CV, so is the base EH to the base NQ, for the solids AB, CV are of equal height; [XI. 32] and, as the solid CD is to the solid CV, so is the base MQ to the base TQ [XI. 25] and CM to CT [VI. 1]; therefore also, as the base EH is to the base NQ, so is MC to CT.

-

But CT is equal to AG; therefore also, as the base EH is to the base NQ, so is MC to AG.

+

But CT is equal to AG; therefore also, as the base EH is to the base NQ, so is MC to AG.

Therefore in the parallelepipedal solids AB, CD the bases are reciprocally proportional to the heights.

Again, in the parallelepipedal solids AB, CD let the bases be reciprocally proportional to the heights, that is, as the base EH is to the base NQ, so let the height of the solid CD be to the height of the solid AB; I say that the solid AB is equal to the solid CD.

Let the sides which stand up be again at right angles to the bases.

@@ -11002,13 +9428,13 @@ and, componendo, the squares on AB<

Let CT be again made equal to AG, and let the solid CV be similarly completed.

Since, as the base EH is to the base NQ, so is MC to AG, while AG is equal to CT, therefore, as the base EH is to the base NQ, so is CM to CT.

But, as the base EH is to the base NQ, so is the solid AB to the solid CV, for the solids AB, CV are of equal height; [XI. 32] and, as CM is to CT, so is the base MQ to the base QT [VI. 1] and the solid CD to the solid CV. [XI. 25]

-

Therefore also, as the solid AB is to the solid CV, so is the solid CD to the solid CV; therefore each of the solids AB, CD has to CV the same ratio.

+

Therefore also, as the solid AB is to the solid CV, so is the solid CD to the solid CV; therefore each of the solids AB, CD has to CV the same ratio.

Therefore the solid AB is equal to the solid CD. [V. 9]

Now let the sides which stand up, FE, BL, GA, HK, ON, DP, MC, RQ, not be at right angles to their bases; let perpendiculars be drawn from the points F, G, B, K, O, M, D, R to the planes through EH, NQ, and let them meet the planes at S, T, U, V, W, X, Y, a, and let the solids FV, Oa be completed; I say that, in this case too, if the solids AB, CD are equal, the bases are reciprocally proportional to the heights, that is, as the base EH is to the base NQ, so is the height of the solid CD to the height of the solid AB.

Since the solid AB is equal to the solid CD, while AB is equal to BT, for they are on the same base FK and of the same height; [XI. 29, 30] and the solid CD is equal to DX, for they are again on the same base RO and of the same height; [id.] therefore the solid BT is also equal to the solid DX.

Therefore, as the base FK is to the base OR, so is the height of the solid DX to the height of the solid BT. [Part 1.]

-

But the base FK is equal to the base EH, and the base OR to the base NQ; therefore, as the base EH is to the base NQ, so is the height of the solid DX to the height of the solid BT.

+

But the base FK is equal to the base EH, and the base OR to the base NQ; therefore, as the base EH is to the base NQ, so is the height of the solid DX to the height of the solid BT.

But the solids DX, BT and the solids DC, BA have the same heights respectively; therefore, as the base EH is to the base NQ, so is the height of the solid DC to the height of the solid AB.

Therefore in the parallelepipedal solids AB, CD the bases are reciprocally proportional to the heights.

Again, in the parallelepipedal solids AB, CD let the bases be reciprocally proportional to the heights, that is, as the base EH is to the base NQ, so let the height of the solid CD be to the height of the solid AB; I say that the solid AB is equal to the solid CD.

@@ -11018,7 +9444,7 @@ and, componendo, the squares on AB<

But BT is equal to BA, for they are on the same base FK and of the same height; [XI. 29, 30] and the solid DX is equal to the solid DC. [id.]

Therefore the solid AB is also equal to the solid CD. Q. E. D.

- + PROPOSITION 35.

If there be two equal plane angles, and on their vertices there be set up elevated straight lines containing equal angles with the original straight lines respectively, if on the elevated straight lines points be taken at random and perpendiculars be drawn from them to the planes in which the original angles are, and if from the points so arising in the planes straight lines be joined to the vertices of the original angles, they will contain, with the elevated straight lines, equal angles.

@@ -11026,7 +9452,7 @@ and, componendo, the squares on AB<

Let AH be made equal to DM, and let HK be drawn through the point H parallel to GL.

But GL is perpendicular to the plane through BA, AC; therefore HK is also perpendicular to the plane through. BA, AC. [XI. 8]

-

From the points K, N let KC, NF, KB, NE be drawn perpendicular to the straight lines AC, DF, AB, DE, and let HC, CB, MF, FE be joined.

+

From the points K, N let KC, NF, KB, NE be drawn perpendicular to the straight lines AC, DF, AB, DE, and let HC, CB, MF, FE be joined.

Since the square on HA is equal to the squares on HK, KA, and the squares on KC, CA are equal to the square on KA, [I. 47] therefore the square on HA is also equal to the squares on HK, KC, CA.

But the square on HC is equal to the squares on HK, KC; [I. 47] therefore the square on HA is equal to the squares on HC, CA.

Therefore the angle HCA is right. [I. 48]

@@ -11039,7 +9465,7 @@ and, componendo, the squares on AB<

Since then AC is equal to DF, and AB to DE, the two sides CA, AB are equal to the two sides FD, DE.

But the angle CAB is also equal to the angle FDE; therefore the base BC is equal to the base EF, the triangle to the triangle, and the remaining angles to the remaining angles; [I. 4] therefore the angle ACB is equal to the angle DFE.

But the right angle ACK is also equal to the right angle DFN; therefore the remaining angle BCK is also equal to the remaining angle EFN.

-

For the same reason the angle CBK is also equal to the angle FEN.

+

For the same reason the angle CBK is also equal to the angle FEN.

Therefore BCK, EFN are two triangles which have two angles equal to two angles respectively, and one side equal to one side, namely, that adjacent to the equal angles, that is, BC equal to EF; therefore they will also have the remaining sides equal to the remaining sides. [I. 26]

Therefore CK is equal to FN.

But AC is also equal to DF; therefore the two sides AC, CK are equal to the two sides DF, FN; and they contain right angles.

@@ -11052,20 +9478,20 @@ and, componendo, the squares on AB< PORISM.

From this it is manifest that, if there be two equal plane angles, and if there be set up on them elevated straight lines which are equal and contain equal angles with the original straight lines respectively, the perpendiculars drawn from their extremities to the planes in which are the original angles are equal to one another. Q. E. D.

- + PROPOSITION 36.

If three straight lines be proportional, the parallelepipedal solid formed out of the three is equal to the parallelepipedal solid on the mean which is equilateral, but equiangular with the aforesaid solid.

Let A, B, C be three straight lines in proportion, so that, as A is to B, so is B to C; I say that the solid formed out of A, B, C is equal to the solid on B which is equilateral, but equiangular with the aforesaid solid.

-

Let there be set out the solid angle at E contained by the angles DEG, GEF, FED, let each of the straight lines DE, GE, EF be made equal to B, and let the parallelepipedal solid EK be completed, let LM be made equal to A, and on the straight line LM, and at the point L on it, let there be constructed a solid angle equal to the solid angle at E, namely that contained by NLO, OLM, MLN; let LO be made equal to B, and LN equal to C.

+

Let there be set out the solid angle at E contained by the angles DEG, GEF, FED, let each of the straight lines DE, GE, EF be made equal to B, and let the parallelepipedal solid EK be completed, let LM be made equal to A, and on the straight line LM, and at the point L on it, let there be constructed a solid angle equal to the solid angle at E, namely that contained by NLO, OLM, MLN; let LO be made equal to B, and LN equal to C.

Now, since, as A is to B, so is B to C, while A is equal to LM, B to each of the straight lines LO, ED, and C to LN, therefore, as LM is to EF, so is DE to LN.

Thus the sides about the equal angles NLM, DEF are reciprocally proportional; therefore the parallelogram MN is equal to the parallelogram DF. [VI. 14]

And, since the angles DEF, NLM are two plane rectilineal angles, and on them the elevated straight lines LO, EG are set up which are equal to one another and contain equal angles with the original straight lines respectively, therefore the perpendiculars drawn from the points G, O to the planes through NL, LM and DE, EF are equal to one another; [XI. 35, Por.] hence the solids LH, EK are of the same height.

But parallelepipedal solids on equal bases and of the same height are equal to one another; [XI. 31] therefore the solid HL is equal to the solid EK.

-

And LH is the solid formed out of A, B, C, and EK the solid on B; therefore the parallelepipedal solid formed out of A, B, C is equal to the solid on B which is equilateral, but equiangular with the aforesaid solid. Q. E. D.

+

And LH is the solid formed out of A, B, C, and EK the solid on B; therefore the parallelepipedal solid formed out of A, B, C is equal to the solid on B which is equilateral, but equiangular with the aforesaid solid. Q. E. D.

PROPOSITION 37. @@ -11073,7 +9499,7 @@ and, componendo, the squares on AB<

Let AB, CD, EF, GH be four straight lines in proportion, so that, as AB is to CD, so is EF to GH; and let there be described on AB, CD, EF, GH the similar and similarly situated parallelepipedal solids KA, LC, ME, NG; I say that, as KA is to LC, so is ME to NG.

-

For, since the parallelepipedal solid KA is similar to LC, therefore KA has to LC the ratio triplicate of that which AB has to CD. [XI. 33]

+

For, since the parallelepipedal solid KA is similar to LC, therefore KA has to LC the ratio triplicate of that which AB has to CD. [XI. 33]

For the same reason ME also has to NG the ratio triplicate of that which EF has to GH. [id.]

And, as AB is to CD, so is EF to GH.

Therefore also, as AK is to LC, so is ME to NG.

@@ -11085,7 +9511,7 @@ and, componendo, the squares on AB< PROPOSITION 38.

If the sides of the opposite planes of a cube be bisected, and planes be carried through the points of section, the common section of the planes and the diameter of the cube bisect one another.

-

For let the sides of the opposite planes CF, AH of the cube AF be bisected at the points K, L, M, N, O, Q, P, R, and through the points of section let the planes KN, OR be carried; let US be the common section of the planes, and DG the diameter of the cube AF.

+

For let the sides of the opposite planes CF, AH of the cube AF be bisected at the points K, L, M, N, O, Q, P, R, and through the points of section let the planes KN, OR be carried; let US be the common section of the planes, and DG the diameter of the cube AF.

I say that UT is equal to TS, and DT to TG.

For let DU, UE, BS, SG be joined.

Then, since DO is parallel to PE, the alternate angles DOU, UPE are equal to one another. [I. 29]

@@ -11093,7 +9519,7 @@ and, componendo, the squares on AB<

For this reason DUE is a straight line. [I. 14]

For the same reason, BSG is also a straight line, and BS is equal to SG.

-

Now, since CA is equal and parallel to DB, while CA is also equal and parallel to EG, therefore DB is also equal and parallel to EG. [XI. 9]

+

Now, since CA is equal and parallel to DB, while CA is also equal and parallel to EG, therefore DB is also equal and parallel to EG. [XI. 9]

And the straight lines DE, BG join their extremities; therefore DE is parallel to BG. [I. 33]

Therefore the angle EDT is equal to the angle BGT, for they are alternate; [I. 29] and the angle DTU is equal to the angle GTS. [I. 15]

Therefore DTU, GTS are two triangles which have two angles equal to two angles, and one side equal to one side, namely that subtending one of the equal angles, that is, DU equal to GS, for they are the halves of DE, BG; therefore they will also have the remaining sides equal to the remaining sides. [I. 26]

@@ -11110,12 +9536,12 @@ and, componendo, the squares on AB<

Since the parallelogram AF is double of the triangle GHK, while the parallelogram HK is also double of the triangle GHK, [I. 34] therefore the parallelogram AF is equal to the parallelogram HK.

But parallelepipedal solids which are on equal bases and of the same height are equal to one another; [XI. 31] therefore the solid AO is equal to the solid GP.

And the prism ABCDEF is half of the solid AO, and the prism GHKLMN is half of the solid GP; [XI. 28] therefore the prism ABCDEF is equal to the prism GHKLMN.

-

Therefore etc. Q. E. D.

+

Therefore etc. Q. E. D.

- - - +
+
+ BOOK XII. HISTORICAL NOTE.

The predominant feature of Book XII. is the use of the method of exhaustion, which is applied in Propositions 2, 3-5, 10, 11, 12, and (in a slightly different form) in Propositions 16-18. We conclude therefore that for the content of this Book Euclid was greatly indebted to Eudoxus, to whom the discovery of the method of exhaustion is attributed. The evidence for this attribution comes mainly from Archimedes. (1) In the preface to On the Sphere and Cylinder I., after stating the main results obtained by himself regarding the surface of a sphere or a segment thereof, and the volume and surface of a right cylinder with height equal to its diameter as compared with those of a sphere with the same diameter, Archimedes adds: Having now discovered that the properties mentioned are true of these figures, I cannot feel any hesitation in setting them side by side both with my former investigations and with those of the theorems of Eudoxus on solids which are held to be most irrefragably established, namely that any pyramid is one third part of the prism which has the same base with the pyramid and equal height [i.e. Eucl. XII. 7], and that any cone is one third part of the cylinder which has the same base with the cone and equal height [i.e. Eucl. XII. 10]. For, though these properties also were naturally inherent in the figures all along, yet they were in fact unknown to all the many able geometers who lived before Eudoxus and had not been observed by any one. @@ -11124,7 +9550,7 @@ and, componendo, the squares on AB< (see note on X. 1 above) and proceeds: Earlier geometers (oi( pro/teron gewme/trai) have also used this lemma; for it is by the use of this same lemma that they have shown that circles are to one another in the duplicate ratio of their diameters [Eucl. XII. 2], and that spheres are to one another in the triplicate ratio of their diameters [Eucl. XII. 18], and further that every pyramid is one third part of the prism which has the same base with the pyramid and equal height [Eucl. XII. 7]; also, that every cone is one third part of the cylinder which has the same base with the cone and equal height [Eucl. XII. 10] they proved by assuming a certain lemma similar to that aforesaid. Thus in the first passage two theorems of Eucl. XII. are definitely attributed to Eudoxus; and, when Archimedes says, in the second passage, that earlier geometers proved these two theorems by means of the lemma known as the Axiom of Archimedes - and of a lemma similar to it respectively, we can hardly suppose him to be alluding to any other proof than that given by Eudoxus. As a matter of fact, the lemma used by Euclid to prove both propositions (XII. 3-5 and 7, and XII. 10) is the theorem of Eucl. X. 1. As regards the connexion between the two lemmas + and of a lemma similar to it respectively, we can hardly suppose him to be alluding to any other proof than that given by Eudoxus. As a matter of fact, the lemma used by Euclid to prove both propositions (XII. 3-5 and 7, and XII. 10) is the theorem of Eucl. X. 1. As regards the connexion between the two lemmas see note on X. 1.

We are not, however, to suppose that none of the results obtained by the method of exhaustion had been discovered before the time of Eudoxus (fl. about 368-5 B.C.). Two at least are of earlier date, those of Eucl. XII. 2 and XII. 7.

(a) Simplicius (Comment. in Aristot. Phys. p. 61, ed. Diels) quotes Eudemus as saying, in his History of Geometry, that Hippocrates of Chios (fl. say 430 B.C.) first laid it down (e)/qeto) that similar segments of circles are in the ratio of the squares on their bases and that he proved this (e)dei/knuen) by proving (e)k tou= dei=cai) that the squares on the diameters have the same ratio as the (whole) circles. We know nothing of the method by which Hippocrates proved this proposition; but, having regard to the evidence from Archimedes quoted above, it is not permissible to suppose that the method was the fully developed method of exhaustion as we know it.

@@ -11132,7 +9558,7 @@ and, componendo, the squares on AB< (or Plan of attack ), addressed to Eratosthenes, is none other than the e)fo/dion on which, according to Suidas, Theodosius wrote a commentary, and which is several times cited by Heron in his Metrica; its discovery adds a new and important chapter to the history of the integral calculus. In the preface to this work Archimedes alludes to the theorems which he first discovered by means of mechanical considerations, but proved afterwards by geometry, because the investigation by means of mechanics did not constitute a rigid proof; he observes, however, that the mechanical method is of great use for the discovery of theorems, and it is much easier to provide the rigid proof when the fact to be proved has once been discovered than it would be if nothing were known to begin with. He goes on: Hence too, in the case of those theorems the proof of which was first discovered by Eudoxus, namely those relating to the cone and the pyramid, that the cone is one third part of the cylinder, and the pyramid one third part of the prism, having the same base and equal height, no small part of the credit will naturally be assigned to Democritus, who was the first to make the statement (of the fact) regarding the said figure [i.e. property], though without proving it. Hence the discovery of the two theorems must now be attributed to Democritus (fl. towards the end of 5th cent. B.C.). The words without proving it - (xwri\s a)podei/cews) do not mean that Democritus gave no sort of proof, but only that he did not give a proof on the rigorous lines required later; for the same words are used by Archimedes of his own investigations by means of mechanics, which, however, do constitute a reasoned argument. The character of Archimedes' mechanical arguments combined with a passage of Plutarch about a particular question in infinitesimals said to have been raised by Democritus may perhaps give a clue to the line of Democritus' argument as regards the pyramid. The essential feature of Archimedes' mechanical arguments in this tract is that he regards an area as the sum of an infinite number of siraight lines parallel to one another and terminated by the boundary or boundaries of the closed figure the area of which is to be found, and a volume as the sum of an infinite number of plane sections parallel to one another: which is of course the same thing as taking (as we do in the integral calculus) the sum of an infinite number of strips of breadth dx (say), when dx becomes indefinitely small, or the sum of an infinite number of parallel laminae of depth dz (say), when dz becomes indefinitely small. To give only one instance, we may take the case of the area of a segment of a parabola cut off by a chord.

+ (xwri\s a)podei/cews) do not mean that Democritus gave no sort of proof, but only that he did not give a proof on the rigorous lines required later; for the same words are used by Archimedes of his own investigations by means of mechanics, which, however, do constitute a reasoned argument. The character of Archimedes' mechanical arguments combined with a passage of Plutarch about a particular question in infinitesimals said to have been raised by Democritus may perhaps give a clue to the line of Democritus' argument as regards the pyramid. The essential feature of Archimedes' mechanical arguments in this tract is that he regards an area as the sum of an infinite number of siraight lines parallel to one another and terminated by the boundary or boundaries of the closed figure the area of which is to be found, and a volume as the sum of an infinite number of plane sections parallel to one another: which is of course the same thing as taking (as we do in the integral calculus) the sum of an infinite number of strips of breadth dx (say), when dx becomes indefinitely small, or the sum of an infinite number of parallel laminae of depth dz (say), when dz becomes indefinitely small. To give only one instance, we may take the case of the area of a segment of a parabola cut off by a chord.

Let CBA be the parabolic segment, CE the tangent at C meeting the

diameter EBD through the middle point of the chord CA in E, so that .

Draw AF parallel to ED meeting CE produced in F. Produce CB to H so that , where K is the point in which CH meets AF; and suppose CH to be a lever.

@@ -11141,14 +9567,14 @@ and, componendo, the squares on AB< ), whence , so that, if a straight line TG equal to PO be placed with its middle point at H, the straight line MO with centre of gravity at N, and the straight line TG with centre of gravity at H, will balance about K.

Taking all other parts of diameters like PO intercepted between the curve and CA, and placing equal straight lines with their centres of gravity at H, these straight lines collected at H will balance (about K) all the lines like MO parallel to FA intercepted within the triangle CFA in the positions in which they severally lie in the figure.

Hence Archimedes infers that an area equal to that of the parabolic segment hung at H will balance (about K) the triangle CFA hung at its centre of gravity, the point X (a point on CK such that ), and therefore that , - from which it follows that . + from which it follows that .

The same sort of argument is used for solids, plane sections taking the place of straight lines.

Archimedes is careful to state once more that this method of argument does not constitute a proof. Thus, at the end of the above proposition about the parabolic segment, he adds: This property is of course not proved by what has just been said; but it has furnished a sort of indication (e)/mfasi/n tina) that the conclusion is true.

Let us now turn to the passage of Plutarch (De Comm. Not. adv. Stoicos XXXIX 3) about Democritus above referred to. Plutarch speaks of Democritus as having raised the question in natural philosophy (fusikw=s): if a cone were cut by a plane parallel to the base [by which is clearly meant a plane indefinitely near to the base], what must we think of the surfaces of the sections, that they are equal or unequal? For, if they are unequal, they will make the cone irregular, as having many indentations, like steps, and unevennesses; but, if they are equal, the sections will be equal, and the cone will appear to have the property of the cylinder and to be made up of equal, not unequal circles, which is very absurd. The phrase made up of equal...circles - (e)c i)/swn sugkei/menos...ku/klwn) shows that Democritus already had the idea of a solid being the sum of an infinite number of parallel planes, or indefinitely thin laminae, indefinitely near together: a most important anticipation of the same thought which led to such fruitful results in Archimedes. If then one may hazard a conjecture as to Democritus' argument with regard to a pyramid, it seems probable that he would notice that, if two pyramids of the same height and equal triangular bases are respectively cut by planes parallel to the base and dividing the heights in the same ratio, the corresponding sections of the two pyramids are equal, whence he would infer that the pyramids are equal as being the sum of the same infinite number of equal plane sections or indefinitely thin laminae. (This would be a particular anticipation of Cavalieri's proposition that the areal or solid contents of two figures are equal if two sections of them taken at the same height, whatever the height may be, always give equal straight lines or equal surfaces respectively.) And Democritus would of course see that the three pyramids into which a prism on the same base and of equal height with the original pyramid is divided (as in Eucl. XII. 7) satisfy this test of equality, so that the pyramid would be one third part of the prism. The extension to a pyramid with a polygonal base would be easy. And Democritus may have stated the proposition for the cone (of course without an absolute proof) as a natural inference from the result of increasing indefinitely the number of sides in a regular polygon forming the base of a pyramid.

+ (e)c i)/swn sugkei/menos...ku/klwn) shows that Democritus already had the idea of a solid being the sum of an infinite number of parallel planes, or indefinitely thin laminae, indefinitely near together: a most important anticipation of the same thought which led to such fruitful results in Archimedes. If then one may hazard a conjecture as to Democritus' argument with regard to a pyramid, it seems probable that he would notice that, if two pyramids of the same height and equal triangular bases are respectively cut by planes parallel to the base and dividing the heights in the same ratio, the corresponding sections of the two pyramids are equal, whence he would infer that the pyramids are equal as being the sum of the same infinite number of equal plane sections or indefinitely thin laminae. (This would be a particular anticipation of Cavalieri's proposition that the areal or solid contents of two figures are equal if two sections of them taken at the same height, whatever the height may be, always give equal straight lines or equal surfaces respectively.) And Democritus would of course see that the three pyramids into which a prism on the same base and of equal height with the original pyramid is divided (as in Eucl. XII. 7) satisfy this test of equality, so that the pyramid would be one third part of the prism. The extension to a pyramid with a polygonal base would be easy. And Democritus may have stated the proposition for the cone (of course without an absolute proof) as a natural inference from the result of increasing indefinitely the number of sides in a regular polygon forming the base of a pyramid.

BOOK XII. PROPOSITIONS. @@ -11161,7 +9587,7 @@ and, componendo, the squares on AB<

For let BE, AM, GL, FN be joined.

Now, since the polygon ABCDE is similar to the polygon FGHKL, the angle BAE is equal to the angle GFL, and, as BA is to AE, so is GF to FL. [VI. Def. I]

Thus BAE, GFL are two triangles which have one angle equal to one angle, namely the angle BAE to the angle GFL, and the sides about the equal angles proportional; therefore the triangle ABE is equiangular with the triangle FGL. [VI. 6]

-

Therefore the angle AEB is equal to the angle FLG.

+

Therefore the angle AEB is equal to the angle FLG.

But the angle AEB is equal to the angle AMB, for they stand on the same circumference; [III. 27] and the angle FLG to the angle FNG; therefore the angle AMB is also equal to the angle FNG.

But the right angle BAM is also equal to the right angle GFN; [III. 31] therefore the remaining angle is equal to the remaining angle. [I. 32]

Therefore the triangle ABM is equiangular with the triangle FGN.

@@ -11169,7 +9595,7 @@ and, componendo, the squares on AB<

But the ratio of the square on BM to the square on GN is duplicate of the ratio of BM to GN, and the ratio of the polygon ABCDE to the polygon FGHKL is duplicate of the ratio of BA to GF; [VI. 20] therefore also, as the square on BM is to the square on GN, so is the polygon ABCDE to the polygon FGHKL.

Therefore etc. Q. E. D.

 - + PROPOSITION 2.

@@ -11180,14 +9606,14 @@ and, componendo, the squares on AB<

For, if the square on BD is not to the square on FH as the circle ABCD is to the circle EFGH, then, as the square on BD is to the square on FH, so will the circle ABCD be either to some less area than the circle EFGH, or to a greater.

First, let it be in that ratio to a less area S.

Let the square EFGH be inscribed in the circle EFGH; then the inscribed square is greater than the half of the circle EFGH, inasmuch as, if through the points E, F, G, H we draw tangents to the circle, the square EFGH is half the square circumscribed about the circle, and the circle is less than the circumscribed square; hence the inscribed square EFGH is greater than the half of the circle EFGH.

-

Let the circumferences EF, FG, GH, HE be bisected at the points K, L, M, N, and let EK, KF, FL, LG, GM, MH, HN, NE be joined; therefore each of the triangles EKF, FLG, GMH, HNE is also greater than the half of the segment of the circle about it, inasmuch as, if through the points K, L, M, N we draw tangents to the circle and complete the parallelograms on the straight lines EF, FG, GH, HE, each of the triangles EKF, FLG, GMH, HNE will be half of the parallelogram about it, while the segment about it is less than the parallelogram; hence each of the triangles EKF, FLG, GMH, HNE is greater than the half of the segment of the circle about it.

+

Let the circumferences EF, FG, GH, HE be bisected at the points K, L, M, N, and let EK, KF, FL, LG, GM, MH, HN, NE be joined; therefore each of the triangles EKF, FLG, GMH, HNE is also greater than the half of the segment of the circle about it, inasmuch as, if through the points K, L, M, N we draw tangents to the circle and complete the parallelograms on the straight lines EF, FG, GH, HE, each of the triangles EKF, FLG, GMH, HNE will be half of the parallelogram about it, while the segment about it is less than the parallelogram; hence each of the triangles EKF, FLG, GMH, HNE is greater than the half of the segment of the circle about it.

Thus, by bisecting the remaining circumferences and joining straight lines, and by doing this continually, we shall leave some segments of the circle which will be less than the excess by which the circle EFGH exceeds the area S.

For it was proved in the first theorem of the tenth book that, if two unequal magnitudes be set out, and if from the greater there be subtracted a magnitude greater than the half, and from that which is left a greater than the half, and if this be done continually, there will be left some magnitude which will be less than the lesser magnitude set out.

Let segments be left such as described, and let the segments of the circle EFGH on EK, KF, FL, LG, GM, MH, HN, NE be less than the excess by which the circle EFGH exceeds the area S.

Therefore the remainder, the polygon EKFLGMHN, is greater than the area S.

Let there be inscribed, also, in the circle ABCD the polygon AOBPCQDR similar to the polygon EKFLGMHN; therefore, as the square on BD is to the square on FH, so is the polygon AOBPCQDR to the polygon EKFLGMHN. [XII. 1]

But, as the square on BD is to the square on FH, so also is the circle ABCD to the area S; therefore also, as the circle ABCD is to the area S, so is the polygon AOBPCQDR to the polygon EKFLGMHN; [V. 11] therefore, alternately, as the circle ABCD is to the polygon inscribed in it, so is the area S to the polygon EKFLGMHN. [V. 16]

-

But the circle ABCD is greater than the polygon inscribed in it; therefore the area S is also greater than the polygon EKFLGMHN.

+

But the circle ABCD is greater than the polygon inscribed in it; therefore the area S is also greater than the polygon EKFLGMHN.

But it is also less: which is impossible.

Therefore, as the square on BD is to the square on FH, so is not the circle ABCD to any area less than the circle EFGH.

Similarly we can prove that neither is the circle EFGH to any area less than the circle ABCD as the square on FH is to the square on BD.

@@ -11203,7 +9629,7 @@ and, componendo, the squares on AB<

I say that, the area S being greater than the circle EFGH, as the area S is to the circle ABCD, so is the circle EFGH to some area less than the circle ABCD.

For let it be contrived that, as the area S is to the circle ABCD, so is the circle EFGH to the area T.

I say that the area T is less than the circle ABCD.

-

For since, as the area S is to the circle ABCD, so is the circle EFGH to the area T, therefore, alternately, as the area S is to the circle EFGH, so is the circle ABCD to the area T. [V. 16]

+

For since, as the area S is to the circle ABCD, so is the circle EFGH to the area T, therefore, alternately, as the area S is to the circle EFGH, so is the circle ABCD to the area T. [V. 16]

But the area S is greater than the circle EFGH; therefore the circle ABCD is also greater than the area T.

Hence, as the area S is to the circle ABCD, so is the circle EFGH to some area less than the circle ABCD. Q. E. D.

@@ -11215,7 +9641,7 @@ and, componendo, the squares on AB<

Let there be a pyramid of which the triangle ABC is the base and the point D the vertex; I say that the pyramid ABCD is divided into two pyramids equal to one another, having triangular bases and similar to the whole pyramid, and into two equal prisms; and the two prisms are greater than the half of the whole pyramid.

For let AB, BC, CA, AD, DB, DC be bisected at the points E, F, G, H, K, L, and let HE, EG, GH, HK, KL, LH, KF, FG be joined.

-

Since AE is equal to EB, and AH to DH, therefore EH is parallel to DB. [VI. 2]

+

Since AE is equal to EB, and AH to DH, therefore EH is parallel to DB. [VI. 2]

For the same reason HK is also parallel to AB.

Therefore HEBK is a parallelogram; therefore HK is equal to EB. [I. 34]

But EB is equal to EA; therefore AE is also equal to HK.

@@ -11227,7 +9653,7 @@ and, componendo, the squares on AB<

And, since the two straight lines EH, HG are equal to the two KD, DL respectively, and the angle EHG is equal to the angle KDL, therefore the base EG is equal to the base KL; [I. 4] therefore the triangle EHG is equal and similar to the triangle KDL.

For the same reason the triangle AEG is also equal and similar to the triangle HKL.

Therefore the pyramid of which the triangle AEG is the base and the point H the vertex is equal and similar to the pyramid of which the triangle HKL is the base and the point D the vertex. [XI. Def. 10]

-

And, since HK has been drawn parallel to AB, one of the sides of the triangle ADB, the triangle ADB is equiangular to the triangle DHK, [I. 29] and they have their sides proportional; therefore the triangle ADB is similar to the triangle DHK. [VI. Def. 1]

+

And, since HK has been drawn parallel to AB, one of the sides of the triangle ADB, the triangle ADB is equiangular to the triangle DHK, [I. 29] and they have their sides proportional; therefore the triangle ADB is similar to the triangle DHK. [VI. Def. 1]

For the same reason the triangle DBC is also similar to the triangle DKL, and the triangle ADC to the triangle DLH.

Now, since the two straight lines BA, AC meeting one another are parallel to the two straight lines KH, HL meeting one another, not in the same plane, they will contain equal angles. [XI. 10]

Therefore the angle BAC is equal to the angle KHL.

@@ -11237,7 +9663,7 @@ and, componendo, the squares on AB<

Therefore each of the pyramids AEGH, HKLD is similar to the whole pyramid ABCD.

Next, since BF is equal to FC, the parallelogram EBFG is double of the triangle GFC.

And since, if there be two prisms of equal height, and one have a parallelogram as base, and the other a triangle, and if the parallelogram be double of the triangle, the prisms are equal, [XI. 39] therefore the prism contained by the two triangles BKF, EHG, and the three parallelograms EBFG, EBKH, HKFG is equal to the prism contained by the two triangles GFC, HKL and the three parallelograms KFCL, LCGH, HKFG.

-

And it is manifest that each of the prisms, namely that in which the parallelogram EBFG is the base and the straight line HK is its opposite, and that in which the triangle GFC is the base and the triangle HKL its opposite, is greater than each of the pyramids of which the triangles AEG, HKL are the bases and the points H, D the vertices, inasmuch as, if we join the straight lines EF, EK, the prism in which the parallelogram EBFG is the base and the straight line HK its opposite is greater than the pyramid of which the triangle EBF is the base and the point K the vertex.

+

And it is manifest that each of the prisms, namely that in which the parallelogram EBFG is the base and the straight line HK is its opposite, and that in which the triangle GFC is the base and the triangle HKL its opposite, is greater than each of the pyramids of which the triangles AEG, HKL are the bases and the points H, D the vertices, inasmuch as, if we join the straight lines EF, EK, the prism in which the parallelogram EBFG is the base and the straight line HK its opposite is greater than the pyramid of which the triangle EBF is the base and the point K the vertex.

But the pyramid of which the triangle EBF is the base and the point K the vertex is equal to the pyramid of which the triangle AEG is the base and the point H the vertex; for they are contained by equal and similar planes.

Hence also the prism in which the parallelogram EBFG is the base and the straight line HK its opposite is greater than the pyramid of which the triangle AEG is the base and the point H the vertex.

But the prism in which the parallelogram EBFG is the base and the straight line HK its opposite is equal to the prism in which the triangle GFC is the base and the triangle HKL its opposite, and the pyramid of which the triangle AEG is the base and the point H the vertex is equal to the pyramid of which the triangle HKL is the base and the point D the vertex.

@@ -11248,14 +9674,14 @@ and, componendo, the squares on AB< PROPOSITION 4.

If there be two pyramids of the same height which have triangular bases, and each of them be divided into two pyramids equal to one another and similar to the whole, and into two equal prisms, then, as the base of the one pyramid is to the base of the other pyramid, so will all the prisms in the one pyramid be to all the prisms, being equal in multitude, in the other pyramid.

-

Let there be two pyramids of the same height which have the triangular bases ABC, DEF, and vertices the points G, H, and let each of them be divided into two pyramids equal to one another and similar to the whole and into two equal prisms; [XII. 3] I say that, as the base ABC is to the base DEF, so are all the prisms in the pyramid ABCG to all the prisms, being equal in multitude, in the pyramid DEFH,

+

Let there be two pyramids of the same height which have the triangular bases ABC, DEF, and vertices the points G, H, and let each of them be divided into two pyramids equal to one another and similar to the whole and into two equal prisms; [XII. 3] I say that, as the base ABC is to the base DEF, so are all the prisms in the pyramid ABCG to all the prisms, being equal in multitude, in the pyramid DEFH,

For, since BO is equal to OC, and AL to LC, therefore LO is parallel to AB, and the triangle ABC is similar to the triangle LOC.

For the same reason the triangle DEF is also similar to the triangle RVF.

And, since BC is double of CO, and EF of FV, therefore, as BC is to CO, so is EF to FV.

And on BC, CO are described the similar and similarly situated rectilineal figures ABC, LOC, and on EF, FV the similar and similarly situated figures DEF, RVF; therefore, as the triangle ABC is to the triangle LOC, so is the triangle DEF to the triangle RVF; [VI. 22] therefore, alternately, as the triangle ABC is to the triangle DEF, so is the triangle LOC to the triangle RVF. [V. 16]

But, as the triangle LOC is to the triangle RVF, so is the prism in which the triangle LOC is the base and PMN its opposite to the prism in which the triangle RVF is the base and STU its opposite; [Lemma following] therefore also, as the triangle ABC is to the triangle DEF, so is the prism in which the triangle LOC is the base and PMN its opposite to the prism in which the triangle RVF is the base and STU its opposite.

-

But, as the said prisms are to one another, so is the prism in which the parallelogram KBOL is the base and the straight line PM its opposite to the prism in which the parallelogram QEVR is the base and the straight line ST its opposite. [XI. 39; cf. XII. 3]

+

But, as the said prisms are to one another, so is the prism in which the parallelogram KBOL is the base and the straight line PM its opposite to the prism in which the parallelogram QEVR is the base and the straight line ST its opposite. [XI. 39; cf. XII. 3]

Therefore also the two prisms, that in which the parallelogram KBOL is the base and PM its opposite, and that in which the triangle LOC is the base and PMN its opposite, are to the prisms in which QEVR is the base and the straight line ST its opposite and in which the triangle RVF is the base and STU its opposite in the same ratio [V. 12]

Therefore also, as the base ABC is to the base DEF, so are the said two prisms to the said two prisms.

And similarly, if the pyramids PMNG, STUH be divided into two prisms and two pyramids, as the base PMN is to the base STU, so will the two prisms in the pyramid PMNG be to the two prisms in the pyramid STUH.

@@ -11266,7 +9692,7 @@ and, componendo, the squares on AB< LEMMA.

But that, as the triangle LOC is to the triangle RVF, so is the prism in which the triangle LOC is the base and PMN its opposite to the prism in which the triangle RVF is the base and STU its opposite, we must prove as follows.

For in the same figure let perpendiculars be conceived drawn from G, H to the planes ABC, DEF; these are of course equal because, by hypothesis, the pyramids are of equal height.

-

Now, since the two straight lines GC and the perpendicular from G are cut by the parallel planes ABC, PMN, they will be cut in the same ratios. [XI. 17]

+

Now, since the two straight lines GC and the perpendicular from G are cut by the parallel planes ABC, PMN, they will be cut in the same ratios. [XI. 17]

And GC is bisected by the plane PMN at N; therefore the perpendicular from G to the plane ABC will also be bisected by the plane PMN.

For the same reason the perpendicular from H to the plane DEF will also be bisected by the plane STU.

And the perpendiculars from G, H to the planes ABC, DEF are equal; therefore the perpendiculars from the triangles PMN, STU to the planes ABC, DEF are also equal.

@@ -11282,7 +9708,7 @@ and, componendo, the squares on AB<

Let there be pyramids of the same height, of which the triangles ABC, DEF are the bases and the points G, H the vertices; I say that, as the base ABC is to the base DEF, so is the pyramid ABCG to the pyramid DEFH.

For, if the pyramid ABCG is not to the pyramid DEFH as the base ABC is to the base DEF, then, as the base ABC is to the base DEF, so will the pyramid ABCG be either to some solid less than the pyramid DEFH or to a greater.

-

Let it, first, be in that ratio to a less solid W, and let the pyramid DEFH be divided into two pyramids equal to one another and similar to the whole and into two equal prisms; then the two prisms are greater than the half of the whole pyramid. [XII. 3]

+

Let it, first, be in that ratio to a less solid W, and let the pyramid DEFH be divided into two pyramids equal to one another and similar to the whole and into two equal prisms; then the two prisms are greater than the half of the whole pyramid. [XII. 3]

Again, let the pyramids arising from the division be similarly divided, and let this be done continually until there are left over from the pyramid DEFH some pyramids which are less than the excess by which the pyramid DEFH exceeds the solid W. [X. 1]

Let such be left, and let them be, for the sake of argument, DQRS, STUH; therefore the remainders, the prisms in the pyramid DEFH, are greater than the solid W.

Let the pyramid ABCG also be divided similarly, and a similar number of times, with the pyramid DEFH; therefore, as the base ABC is to the base DEF, so are the prisms in the pyramid ABCG to the prisms in the pyramid DEFH. [XII. 4]

@@ -11292,13 +9718,13 @@ and, componendo, the squares on AB<

Therefore the prism ABCG is not to any solid less than the pyramid DEFH as the base ABC is to the base DEF.

Similarly it can be proved that neither is the pyramid DEFH to any solid less than the pyramid ABCG as the base DEF is to the base ABC.

I say next that neither is the pyramid ABCG to any solid greater than the pyramid DEFH as the base ABC is to the base DEF.

-

For, if possible, let it be in that ratio to a greater solid W; therefore, inversely, as the base DEF is to the base ABC, so is the solid W to the pyramid ABCG.

+

For, if possible, let it be in that ratio to a greater solid W; therefore, inversely, as the base DEF is to the base ABC, so is the solid W to the pyramid ABCG.

But, as the solid W is to the solid ABCG, so is the pyramid DEFH to some solid less than the pyramid ABCG, as was before proved; [XII. 2, Lemma] therefore also, as the base DEF is to the base ABC, so is the pyramid DEFH to some solid less than the pyramid ABCG: [V. 11] which was proved absurd.

Therefore the pyramid ABCG is not to any solid greater than the pyramid DEFH as the base ABC is to the base DEF.

But it was proved that neither is it in that ratio to a less solid.

Therefore, as the base ABC is to the base DEF, so is the pyramid ABCG to the pyramid DEFH. Q. E. D.

- + PROPOSITION 6.

@@ -11312,14 +9738,14 @@ and, componendo, the squares on AB<

But also, as the base ACD is to the base ADE, so is the pyramid ACDM to the pyramid ADEM. [XII. 5]

Therefore, ex aequali, as the base ABCD is to the base ADE, so is the pyramid ABCDM to the pyramid ADEM. [V. 22]

And again componendo, as the base ABCDE is to the base ADE, so is the pyramid ABCDEM to the pyramid ADEM. [V. 18]

-

Similarly also it can be proved that, as the base FGHKL is to the base FGH, so is the pyramid FGHKLN to the pyramid FGHN.

+

Similarly also it can be proved that, as the base FGHKL is to the base FGH, so is the pyramid FGHKLN to the pyramid FGHN.

And, since ADEM, FGHN are two pyramids which have triangular bases and equal height, therefore, as the base ADE is to the base FGH, so is the pyramid ADEM to the pyramid FGHN. [XII. 5]

But, as the base ADE is to the base ABCDE, so was the pyramid ADEM to the pyramid ABCDEM.

Therefore also, ex aequali, as the base ABCDE is to the base FGH, so is the pyramid ABCDEM to the pyramid FGHN. [V. 22]

But further, as the base FGH is to the base FGHKL, so also was the pyramid FGHN to the pyramid FGHKLN.

Therefore also, ex aequali, as the base ABCDE is to the base FGHKL, so is the pyramid ABCDEM to the pyramid FGHKLN. [V. 22] Q. E. D.

 - + PROPOSITION 7.

@@ -11333,7 +9759,7 @@ and, componendo, the squares on AB<

Therefore the pyramid of which the triangle ABD is the base and the point C the vertex is also equal to the pyramid of which the triangle EBC is the base and the point D the vertex.

Again, since FCBE is a parallelogram, and CE is its diameter, the triangle CEF is equal to the triangle CBE. [I. 34]

Therefore also the pyramid of which the triangle BCE is the base and the point D the vertex is equal to the pyramid of which the triangle ECF is the base and the point D the vertex. [XII. 5]

-

But the pyramid of which the triangle BCE is the base and the point D the vertex was proved equal to the pyramid of which the triangle ABD is the base and the point C the vertex; therefore also the pyramid of which the triangle CEF is the base and the point D the vertex is equal to the pyramid of which the triangle ABD is the base and the point C the vertex; therefore the prism ABCDEF has been divided into three pyramids equal to one another which have triangular bases.

+

But the pyramid of which the triangle BCE is the base and the point D the vertex was proved equal to the pyramid of which the triangle ABD is the base and the point C the vertex; therefore also the pyramid of which the triangle CEF is the base and the point D the vertex is equal to the pyramid of which the triangle ABD is the base and the point C the vertex; therefore the prism ABCDEF has been divided into three pyramids equal to one another which have triangular bases.

And, since the pyramid of which the triangle ABD is the base and the point C the vertex is the same with the pyramid of which the triangle CAB is the base and the point D the vertex, for they are contained by the same planes, while the pyramid of which the triangle ABD is the base and the point C the vertex was proved to be a third of the prism in which the triangle ABC is the base and DEF its opposite, therefore also the pyramid of which the triangle ABC is the base and the point D the vertex is a third of the prism which has the same base, the triangle ABC, and DEF as its opposite.

PORISM. @@ -11345,7 +9771,7 @@ and, componendo, the squares on AB<

Similar pyramids which have triangular bases are in the triplicate ratio of their corresponding sides.

-

Let there be similar and similarly situated pyramids of which the triangles ABC, DEF, are the bases and the points G, H the vertices; I say that the pyramid ABCG has to the pyramid DEFH the ratio triplicate of that which BC has to EF.

+

Let there be similar and similarly situated pyramids of which the triangles ABC, DEF, are the bases and the points G, H the vertices; I say that the pyramid ABCG has to the pyramid DEFH the ratio triplicate of that which BC has to EF.

For let the parallelepipedal solids BGML, EHQP be completed.

Now, since the pyramid ABCG is similar to the pyramid DEFH, therefore the angle ABC is equal to the angle DEF, the angle GBC to the angle HEF, and the angle ABG to the angle DEH; and, as AB is to DE, so is BC to EF, and BG to EH.

@@ -11354,7 +9780,7 @@ and, componendo, the squares on AB<

But the three parallelograms MB, BK, BN are equal and similar to their three opposites, and the three EQ, EO, ER are equal and similar to their three opposites. [XI. 24]

Therefore the solids BGML, EHQP are contained by similar planes equal in multitude.

Therefore the solid BGML is similar to the solid EHQP.

-

But similar parallelepipedal solids are in the triplicate ratio of their corresponding sides. [XI. 33]

+

But similar parallelepipedal solids are in the triplicate ratio of their corresponding sides. [XI. 33]

Therefore the solid BGML has to the solid EHQP the ratio triplicate of that which the corresponding side BC has to the corresponding side EF.

But, as the solid BGML is to the solid EHQP, so is the pyramid ABCG to the pyramid DEFH, inasmuch as the pyramid is a sixth part of the solid, because the prism which is half of the parallelepipedal solid [XI. 28] is also triple of the pyramid. [XII. 7]

Therefore the pyramid ABCG also has to the pyramid DEFH the ratio triplicate of that which BC has to EF. Q. E. D.

@@ -11365,7 +9791,7 @@ and, componendo, the squares on AB<

But the pyramid which has a triangular base is to the pyramid which has a triangular base in the triplicate ratio of the corresponding sides; therefore also the pyramid which has a polygonal base has to the pyramid which has a similar base the ratio triplicate of that which the side has to the side.

- + PROPOSITION 9.

@@ -11377,14 +9803,14 @@ and, componendo, the squares on AB<

Now, since the pyramid ABCG is equal to the pyramid DEFH, and the solid BGML is six times the pyramid ABCG, and the solid EHQP six times the pyramid DEFH, therefore the solid BGML is equal to the solid EHQP.

But in equal parallelepipedal solids the bases are reciprocally proportional to the heights; [XI. 34] therefore, as the base BM is to the base EQ, so is the height of the solid EHQP to the height of the solid BGML.

But, as the base BM is to EQ, so is the triangle ABC to the triangle DEF. [I. 34]

-

Therefore also, as the triangle ABC is to the triangle DEF, so is the height of the solid EHQP to the height of the solid BGML. [V. 11]

+

Therefore also, as the triangle ABC is to the triangle DEF, so is the height of the solid EHQP to the height of the solid BGML. [V. 11]

But the height of the solid EHQP is the same with the height of the pyramid DEFH, and the height of the solid BGML is the same with the height of the pyramid ABCG, therefore, as the base ABC is to the base DEF, so is the height of the pyramid DEFH to the height of the pyramid ABCG.

Therefore in the pyramids ABCG, DEFH the bases are reciprocally proportional to the heights.

Next, in the pyramids ABCG, DEFH let the bases be reciprocally proportional to the heights; that is, as the base ABC is to the base DEF, so let the height of the pyramid DEFH be to the height of the pyramid ABCG; I say that the pyramid ABCG is equal to the pyramid DEFH.

For, with the same construction, since, as the base ABC is to the base DEF, so is the height of the pyramid DEFH to the height of the pyramid ABCG, while, as the base ABC is to the base DEF, so is the parallelogram BM to the parallelogram EQ, therefore also, as the parallelogram BM is to the parallelogram EQ, so is the height of the pyramid DEFH to the height of the pyramid ABCG. [V. 11]

But the height of the pyramid DEFH is the same with the height of the parallelepiped EHQP, and the height of the pyramid ABCG is the same with the height of the parallelepiped BGML; therefore, as the base BM is to the base EQ, so is the height of the parallelepiped EHQP to the height of the parallelepiped BGML.

But those parallelepipedal solids in which the bases are reciprocally proportional to the heights are equal; [XI. 34] therefore the parallelepipedal solid BGML is equal to the parallelepipedal solid EHQP.

-

And the pyramid ABCG is a sixth part of BGML, and the pyramid DEFH a sixth part of the parallelepiped EHQP; therefore the pyramid ABCG is equal to the pyramid DEFH.

+

And the pyramid ABCG is a sixth part of BGML, and the pyramid DEFH a sixth part of the parallelepiped EHQP; therefore the pyramid ABCG is equal to the pyramid DEFH.

Therefore etc Q. E. D.

 @@ -11397,10 +9823,10 @@ and, componendo, the squares on AB<

First let it be greater than triple, and let the square ABCD be inscribed in the circle ABCD; [IV. 6] then the square ABCD is greater than the half of the circle ABCD.

From the square ABCD let there be set up a prism of equal height with the cylinder.

-

Then the prism so set up is greater than the half of the cylinder, inasmuch as, if we also circumscribe a square about the circle ABCD [IV. 7], the square inscribed in the circle ABCD is half of that circumscribed about it, and the solids set up from them are parallelepipedal prisms of equal height, while parallelepipedal solids which are of the same height are to one another as their bases; [XI. 32] therefore also the prism set up on the square ABCD is half of the prism set up from the square circumscribed about the circle ABCD; [cf. XI. 28, or XII. 6 and 7, Por.] and the cylinder is less than the prism set up from the square circumscribed about the circle ABCD; therefore the prism set up from the square ABCD and of equal height with the cylinder is greater than the half of the cylinder.

+

Then the prism so set up is greater than the half of the cylinder, inasmuch as, if we also circumscribe a square about the circle ABCD [IV. 7], the square inscribed in the circle ABCD is half of that circumscribed about it, and the solids set up from them are parallelepipedal prisms of equal height, while parallelepipedal solids which are of the same height are to one another as their bases; [XI. 32] therefore also the prism set up on the square ABCD is half of the prism set up from the square circumscribed about the circle ABCD; [cf. XI. 28, or XII. 6 and 7, Por.] and the cylinder is less than the prism set up from the square circumscribed about the circle ABCD; therefore the prism set up from the square ABCD and of equal height with the cylinder is greater than the half of the cylinder.

Let the circumferences AB, BC, CD, DA be bisected at the points E, F, G, H, and let AE, EB, BF, FC, CG, GD, DH, HA be joined; then each of the triangles AEB, BFC, CGD, DHA is greater than the half of that segment of the circle ABCD which is about it, as we proved before. [XII. 2]

On each of the triangles AEB, BFC, CGD, DHA let prisms be set up of equal height with the cylinder; then each of the prisms so set up is greater than the half part of that segment of the cylinder which is about it, inasmuch as, if we draw through the points E, F, G, H parallels to AB, BC, CD, DA, complete the parallelograms on AB, BC, CD, DA, and set up from them parallelepipedal solids of equal height with the cylinder, the prisms on the triangles AEB, BFC, CGD, DHA are halves of the several solids set up; and the segments of the cylinder are less than the parallelepipedal solids set up; hence also the prisms on the triangles AEB, BFC, CGD, DHA are greater than the half of the segments of the cylinder about them.

-

Thus, bisecting the circumferences that are left, joining straight lines, setting up on each of the triangles prisms of equal height with the cylinder, and doing this continually, we shall leave some segments of the cylinder which will be less than the excess by which the cylinder exceeds the triple of the cone. [X. 1]

+

Thus, bisecting the circumferences that are left, joining straight lines, setting up on each of the triangles prisms of equal height with the cylinder, and doing this continually, we shall leave some segments of the cylinder which will be less than the excess by which the cylinder exceeds the triple of the cone. [X. 1]

Let such segments be left, and let them be AE, EB, BF, FC, CG, GD, DH, HA; therefore the remainder, the prism of which the polygon AEBFCGDH is the base and the height is the same as that of the cylinder, is greater than triple of the cone.

But the prism of which the polygon AEBFCGDH is the base and the height the same as that of the cylinder is triple of the pyramid of which the polygon AEBFCGDH is the base and the vertex is the same as that of the cone; [XII. 7, Por.] therefore also the pyramid of which the polygon AEBFCGDH is the base and the vertex is the same as that of the cone is greater than the cone which has the circle ABCD as base.

But it is also less, for it is enclosed by it: which is impossible.

@@ -11408,21 +9834,21 @@ and, componendo, the squares on AB<

I say next that neither is the cylinder less than triple of the cone,

For, if possible, let the cylinder be less than triple of the cone, therefore, inversely, the cone is greater than a third part of the cylinder.

Let the square ABCD be inscribed in the circle ABCD; therefore the square ABCD is greater than the half of the circle ABCD.

-

Now let there be set up from the square ABCD a pyramid having the same vertex with the cone; therefore the pyramid so set up is greater than the half part of the cone, seeing that, as we proved before, if we circumscribe a square about the circle, the square ABCD will be half of the square circumscribed about the circle, and if we set up from the squares parallelepipedal solids of equal height with the cone, which are also called prisms, the solid set up from the square ABCD will be half of that set up from the square circumscribed about the circle; for they are to one another as their bases. [XI. 32]

+

Now let there be set up from the square ABCD a pyramid having the same vertex with the cone; therefore the pyramid so set up is greater than the half part of the cone, seeing that, as we proved before, if we circumscribe a square about the circle, the square ABCD will be half of the square circumscribed about the circle, and if we set up from the squares parallelepipedal solids of equal height with the cone, which are also called prisms, the solid set up from the square ABCD will be half of that set up from the square circumscribed about the circle; for they are to one another as their bases. [XI. 32]

Hence also the thirds of them are in that ratio; therefore also the pyramid of which the square ABCD is the base is half of the pyramid set up from the square circumscribed about the circle.

And the pyramid set up from the square about the circle is greater than the cone, for it encloses it.

Therefore the pyramid of which the square ABCD is the base and the vertex is the same with that of the cone is greater than the half of the cone.

Let the circumferences AB, BC, CD, DA be bisected at the points E, F, G, H, and let AE, EB, BF, FC, CG, GD, DH, HA be joined; therefore also each of the triangles AEB, BFC, CGD, DHA is greater than the half part of that segment of the circle ABCD which is about it.

Now, on each of the triangles AEB, BFC, CGD, DHA let pyramids be set up which have the same vertex as the cone; therefore also each of the pyramids so set up is, in the same manner, greater than the half part of that segment of the cone which is about it.

Thus, by bisecting the circumferences that are left, joining straight lines, setting up on each of the triangles a pyramid which has the same vertex as the cone, and doing this continually, we shall leave some segments of the cone which will be less than the excess by which the cone exceeds the third part of the cylinder. [X. 1]

-

Let such be left, and let them be the segments on AE, EB, BF, FC, CG, GD, DH, HA; therefore the remainder, the pyramid of which the polygon AEBFCGDH is the base and the vertex the same with that of the cone, is greater than a third part of the cylinder.

+

Let such be left, and let them be the segments on AE, EB, BF, FC, CG, GD, DH, HA; therefore the remainder, the pyramid of which the polygon AEBFCGDH is the base and the vertex the same with that of the cone, is greater than a third part of the cylinder.

But the pyramid of which the polygon AEBFCGDH is the base and the vertex the same with that of the cone is a third part of the prism of which the polygon AEBFCGDH is the base and the height is the same with that of the cylinder; therefore the prism of which the polygon AEBFCGDH is the base and the height is the same with that of the cylinder is greater than the cylinder of which the circle ABCD is the base.

But it is also less, for it is enclosed by it: which is impossible.

Therefore the cylinder is not less than triple of the cone.

But it was proved that neither is it greater than triple; therefore the cylinder is triple of the cone; hence the cone is a third part of the cylinder.

Therefore etc. Q. E. D.

 - + PROPOSITION 11.

@@ -11433,7 +9859,7 @@ and, componendo, the squares on AB<

For, if not, then, as the circle ABCD is to the circle EFGH, so will the cone AL be either to some solid less than the cone EN or to a greater.

First, let it be in that ratio to a less solid O, and let the solid X be equal to that by which the solid O is less than the cone EN; therefore the cone EN is equal to the solids O, X.

Let the square EFGH be inscribed in the circle EFGH; therefore the square is greater than the half of the circle.

-

Let there be set up from the square EFGH a pyramid of equal height with the cone; therefore the pyramid so set up is greater than the half of the cone, inasmuch as, if we circumscribe a square about the circle, and set up from it a pyramid of equal height with the cone, the inscribed pyramid is half of the circumscribed pyramid, for they are to one another as their bases, [XII. 6] while the cone is less than the circumscribed pyramid.

+

Let there be set up from the square EFGH a pyramid of equal height with the cone; therefore the pyramid so set up is greater than the half of the cone, inasmuch as, if we circumscribe a square about the circle, and set up from it a pyramid of equal height with the cone, the inscribed pyramid is half of the circumscribed pyramid, for they are to one another as their bases, [XII. 6] while the cone is less than the circumscribed pyramid.

Let the circumferences EF, FG, GH, HE be bisected at the points P, Q, R, S, and let HP, PE, EQ, QF, FR, RG, GS, SH be joined.

Therefore each of the triangles HPE, EQF, FRG, GSH is greater than the half of that segment of the circle which is about it.

On each of the triangles HPE, EQF, FRG, GSH let there be set up a pyramid of equal height with the cone; therefore, also, each of the pyramids so set up is greater than the half of that segment of the cone which is about it.

@@ -11441,7 +9867,7 @@ and, componendo, the squares on AB<

Let such be left, and let them be the segments on HP, PE, EQ, QF, FR, RG, GS, SH; therefore the remainder, the pyramid of which the polygon HPEQFRGS is the base and the height the same with that of the cone, is greater than the solid O.

Let there also be inscribed in the circle ABCD the polygon DTAUBVCW similar and similarly situated to the polygon HPEQFRGS, and on it let a pyramid be set up of equal height with the cone AL.

Since then, as the square on AC is to the square on EG, so is the polygon DTAUBVCW to the polygon HPEQFRGS, [XII. 1] while, as the square on AC is to the square on EG, so is the circle ABCD to the circle EFGH, [XII. 2] therefore also, as the circle ABCD is to the circle EFGH, so is the polygon DTAUBVCW to the polygon HPEQFRGS.

-

But, as the circle ABCD is to the circle EFGH, so is the cone AL to the solid O, and, as the polygon DTAUBVCW is to the polygon HPEQFRGS, so is the pyramid of which the polygon DTAUBVCW is the base and the point L the vertex to the pyramid of which the polygon HPEQFRGS is the base and the point N the vertex. [XII. 6]

+

But, as the circle ABCD is to the circle EFGH, so is the cone AL to the solid O, and, as the polygon DTAUBVCW is to the polygon HPEQFRGS, so is the pyramid of which the polygon DTAUBVCW is the base and the point L the vertex to the pyramid of which the polygon HPEQFRGS is the base and the point N the vertex. [XII. 6]

Therefore also, as the cone AL is to the solid O, so is the pyramid of which the polygon DTAUBVCW is the base and the point L the vertex to the pyramid of which the polygon HPEQFRGS is the base and the point N the vertex; [V. 11] therefore, alternately, as the cone AL is to the pyramid in it, so is the solid O to the pyramid in the cone EN. [V. 16]

But the cone AL is greater than the pyramid in it; therefore the solid O is also greater than the pyramid in the cone EN.

But it is also less: which is absurd.

@@ -11452,11 +9878,11 @@ and, componendo, the squares on AB<

But, as the solid O is to the cone AL, so is the cone EN to some solid less than the cone AL; therefore also, as the circle EFGH is to the circle ABCD, so is the cone EN to some solid less than the cone AL: which was proved impossible.

Therefore the cone AL is not to any solid greater than the cone EN as the circle ABCD is to the circle EFGH.

But it was proved that neither is it in this ratio to a less solid; therefore, as the circle ABCD is to the circle EFGH, so is the cone AL to the cone EN.

-

But, as the cone is to the cone, so is the cylinder to the cylinder, for each is triple of each; [XII. 10]

+

But, as the cone is to the cone, so is the cylinder to the cylinder, for each is triple of each; [XII. 10]

Therefore also, as the circle ABCD is to the circle EFGH, so are the cylinders on them which are of equal height.

Therefore etc. Q. E. D.

 - + PROPOSITION 12.

@@ -11467,21 +9893,21 @@ and, componendo, the squares on AB<

For, if the cone ABCDL has not to the cone EFGHN the ratio triplicate of that which BD has to FH, the cone ABCDL will have that triplicate ratio either to some solid less than the cone EFGHN or to a greater.

First, let it have that triplicate ratio to a less solid O.

Let the square EFGH be inscribed in the circle EFGH; [IV. 6] therefore the square EFGH is greater than the half of the circle EFGH.

-

Now let there be set up on the square EFGH a pyramid having the same vertex with the cone; therefore the pyramid so set up is greater than the half part of the cone.

+

Now let there be set up on the square EFGH a pyramid having the same vertex with the cone; therefore the pyramid so set up is greater than the half part of the cone.

Let the circumferences EF, FG, GH, HE be bisected at the points P, Q, R, S, and let EP, PF, FQ, QG, GR, RH, HS, SE be joined.

Therefore each of the triangles EPF, FQG, GRH, HSE is also greater than the half part of that segment of the circle EFGH which is about it.

Now on each of the triangles EPF, FQG, GRH, HSE let a pyramid be set up having the same vertex with the cone; therefore each of the pyramids so set up is also greater than the half part of that segment of the cone which is about it.

Thus, bisecting the circumferences so left, joining straight lines, setting up on each of the triangles pyramids having the same vertex with the cone, and doing this continually, we shall leave some segments of the cone which will be less than the excess by which the cone EFGHN exceeds the solid O. [X. 1]

Let such be left, and let them be the segments on EP, PF, FQ, QG, GR, RH, HS, SE; therefore the remainder, the pyramid of which the polygon EPFQGRHS is the base and the point N the vertex, is greater than the solid O.

Let there be also inscribed in the circle ABCD the polygon ATBUCVDW similar and similarly situated to the polygon EPFQGRHS, and let there be set up on the polygon ATBUCVDW a pyramid having the same vertex with the cone; of the triangles containing the pyramid of which the polygon ATBUCVDW is the base and the point L the vertex let LBT be one, and of the triangles containing the pyramid of which the polygon EPFQGRHS is the base and the point N the vertex let NFP be one; and let KT, MP be joined.

-

Now, since the cone ABCDL is similar to the cone EFGHN, therefore, as BD is to FH, so is the axis KL to the axis MN. [XI. Def. 24]

+

Now, since the cone ABCDL is similar to the cone EFGHN, therefore, as BD is to FH, so is the axis KL to the axis MN. [XI. Def. 24]

But, as BD is to FH, so is BK to FM; therefore also, as BK is to FM, so is KL to MN.

And, alternately, as BK is to KL, so is FM to MN. [V. 16]

And the sides are proportional about equal angles, namely the angles BKL, FMN; therefore the triangle BKL is similar to the triangle FMN. [VI. 6]

Again, since, as BK is to KT, so is FM to MP, and they are about equal angles, namely the angles BKT, FMP, inasmuch as, whatever part the angle BKT is of the four right angles at the centre K, the same part also is the angle FMP of the four right angles at the centre M; since then the sides are proportional about equal angles, therefore the triangle BKT is similar to the triangle FMP. [VI. 6]

Again, since it was proved that, as BK is to KL, so is FM to MN, while BK is equal to KT, and FM to PM, therefore, as TK is to KL, so is PM to MN; and the sides are proportional about equal angles, namely the angles TKL, PMN, for they are right; therefore the triangle LKT is similar to the triangle NMP. [VI. 6]

And since, owing to the similarity of the triangles LKB, NMF, as LB is to BK, so is NF to FM, and, owing to the similarity of the triangles BKT, FMP, as KB is to BT, so is MF to FP, therefore, ex aequali, as LB is to BT, so is NF to FP. [V. 22]

-

Again since, owing to the similarity of the triangles LTK, NPM, as LT is to TK, so is NP to PM, and, owing to the similarity of the triangles TKB, PMF, as KT is to TB, so is MP to PF; therefore, ex aequali, as LT is to TB, so is NP to PF. [V. 22]

+

Again since, owing to the similarity of the triangles LTK, NPM, as LT is to TK, so is NP to PM, and, owing to the similarity of the triangles TKB, PMF, as KT is to TB, so is MP to PF; therefore, ex aequali, as LT is to TB, so is NP to PF. [V. 22]

But it was also proved that, as TB is to BL, so is PF to FN.

Therefore, ex aequali, as TL is to LB, so is PN to NF. [V. 22]

Therefore in the triangles LTB, NPF the sides are proportional; therefore the triangles LTB, NPF are equiangular; [VI. 5] hence they are also similar. [VI. Def. I]

@@ -11491,7 +9917,7 @@ and, componendo, the squares on AB<

Similarly, by joining straight lines from A, W, D, V, C, U to K, and from E, S, H, R, G, Q to M, and setting up on each of the triangles pyramids which have the same vertex with the cones, we can prove that each of the similarly arranged pyramids will also have to each similarly arranged pyramid the ratio triplicate of that which the corresponding side BK has to the corresponding side FM, that is, which BD has to FH.

And, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents; [V. 12] therefore also, as the pyramid BKTL is to the pyramid FMPN, so is the whole pyramid of which the polygon ATBUCVDW is the base and the point L the vertex to the whole pyramid of which the polygon EPFQGRHS is the base and the point N the vertex; hence also the pyramid of which ATBUCVDW is the base and the point L the vertex has to the pyramid of which the polygon EPFQGRHS is the base and the point N the vertex the ratio triplicate of that which BD has to FH.

But, by hypothesis, the cone of which the circle ABCD - is the base and the point L the vertex has also to the solid O the ratio triplicate of that which BD has to FH; therefore, as the cone of which the circle ABCD is the base and the point L the vertex is to the solid O, so is the pyramid of which the polygon ATBUCVDW is the base and L the vertex to the pyramid of which the polygon EPFQGRHS is the base and the point N the vertex; therefore, alternately, as the cone of which the circle ABCD is the base and L the vertex is to the pyramid contained in it of which the polygon ATBUCVDW is the base and L the vertex, so is the solid O to the pyramid of which the polygon EPFQGRHS is the base and N the vertex. [V. 16]

+ is the base and the point L the vertex has also to the solid O the ratio triplicate of that which BD has to FH; therefore, as the cone of which the circle ABCD is the base and the point L the vertex is to the solid O, so is the pyramid of which the polygon ATBUCVDW is the base and L the vertex to the pyramid of which the polygon EPFQGRHS is the base and the point N the vertex; therefore, alternately, as the cone of which the circle ABCD is the base and L the vertex is to the pyramid contained in it of which the polygon ATBUCVDW is the base and L the vertex, so is the solid O to the pyramid of which the polygon EPFQGRHS is the base and N the vertex. [V. 16]

But the said cone is greater than the pyramid in it; for it encloses it.

Therefore the solid O is also greater than the pyramid of which the polygon EPFQGRHS is the base and N the vertex.

But it is also less: which is impossible.

@@ -11501,7 +9927,7 @@ and, componendo, the squares on AB<

For, if possible, let it have that ratio to a greater solid O.

Therefore, inversely, the solid O has to the cone ABCDL the ratio triplicate of that which FH has to BD.

But, as the solid O is to the cone ABCDL, so is the cone EFGHN to some solid less than the cone ABCDL.

-

Therefore the cone EFGHN also has to some solid less than the cone ABCDL the ratio triplicate of that which FH has to BD: which was proved impossible.

+

Therefore the cone EFGHN also has to some solid less than the cone ABCDL the ratio triplicate of that which FH has to BD: which was proved impossible.

Therefore the cone ABCDL has not to any solid greater than the cone EFGHN the ratio triplicate of that which BD has to FH.

But it was proved that neither has it this ratio to a less solid than the cone EFGHN.

Therefore the cone ABCDL has to the cone EFGHN the ratio triplicate of that which BD has to FH.

@@ -11517,7 +9943,7 @@ and, componendo, the squares on AB<

For let the axis EF be produced in both directions to the points L, M, and let there be set out any number whatever of axes EN, NL equal to the axis EK, and any number whatever FO, OM equal to FK; and let the cylinder PW on the axis LM be conceived of which the circles PQ, VW are the bases.

Let planes be carried through the points N, O parallel to AB, CD and to the bases of the cylinder PW, and let them produce the circles RS, TU about the centres N, O.

-

Then, since the axes LN, NE, EK are equal to one another, therefore the cylinders QR, RB, BG are to one another as their bases. [XII. 11]

+

Then, since the axes LN, NE, EK are equal to one another, therefore the cylinders QR, RB, BG are to one another as their bases. [XII. 11]

But the bases are equal; therefore the cylinders QR, RB, BG are also equal to one another.

Since then the axes LN, NE, EK are equal to one another, and the cylinders QR, RB, BG are also equal to one another, and the multitude of the former is equal to the multitude of the latter, therefore, whatever multiple the axis KL is of the axis EK, the same multiple also will the cylinder QG be of the cylinder GB.

For the same reason, whatever multiple the axis MK is of the axis KF, the same multiple also is the cylinder WG of the cylinder GD.

@@ -11540,7 +9966,7 @@ and, componendo, the squares on AB<

But, as the cylinder EB is to the cylinder FD, so is the cone ABG to the cone CDK. [XII. 10]

Therefore also, as the axis GH is to the axis KL, so is the cone ABG to the cone CDK and the cylinder EB to the cylinder FD. Q. E. D.

 - + PROPOSITION 15.

@@ -11554,7 +9980,7 @@ and, componendo, the squares on AB<

Now the cylinder AO is also equal to the cylinder EP.

But cones and cylinders which are of the same height are to one another as their bases; [XII. 11] therefore the base ABCD is also equal to the base EFGH.

Hence also, reciprocally, as the base ABCD is to the base EFGH, so is the height MN to the height KL.

-

Next, let the height LK not be equal to MN, but let MN be greater; from the height MN let QN be cut off equal to KL, through the point Q let the cylinder EP be cut by the plane TUS parallel to the planes of the circles EFGH, RP, and let the cylinder ES be conceived erected from the circle EFGH as base and with height NQ.

+

Next, let the height LK not be equal to MN, but let MN be greater; from the height MN let QN be cut off equal to KL, through the point Q let the cylinder EP be cut by the plane TUS parallel to the planes of the circles EFGH, RP, and let the cylinder ES be conceived erected from the circle EFGH as base and with height NQ.

Now, since the cylinder AO is equal to the cylinder EP, therefore, as the cylinder AO is to the cylinder ES, so is the cylinder EP to the cylinder ES. [V. 7]

But, as the cylinder AO is to the cylinder ES, so is the base ABCD to the base EFGH, for the cylinders AO, ES are of the same height; [XII. 11] and, as the cylinder EP is to the cylinder ES, so is the height MN to the height QN, for the cylinder EP has been cut by a plane which is parallel to its opposite planes. [XII. 13]

Therefore also, as the base ABCD is to the base EFGH, so is the height MN to the height QN. [V. 11]

@@ -11563,14 +9989,14 @@ and, componendo, the squares on AB<

Next, in the cylinders AO, EP let the bases be reciprocally proportional to the heights, that is, as the base ABCD is to the base EFGH, so let the height MN be to the height KL; I say that the cylinder AO is equal to the cylinder EP.

For, with the same construction, since, as the base ABCD is to the base EFGH, so is the height MN to the height KL, while the height KL is equal to the height QN, therefore, as the base ABCD is to the base EFGH, so is the height MN to the height QN

-

But, as the base ABCD is to the base EFGH, so is the cylinder AO to the cylinder ES, for they are of the same height; [XII. 11] and, as the height MN is to QN, so is the cylinder EP to the cylinder ES; [XII. 13] therefore, as the cylinder AO is to the cylinder ES, so is the cylinder EP to the cylinder ES. [V. 11]

+

But, as the base ABCD is to the base EFGH, so is the cylinder AO to the cylinder ES, for they are of the same height; [XII. 11] and, as the height MN is to QN, so is the cylinder EP to the cylinder ES; [XII. 13] therefore, as the cylinder AO is to the cylinder ES, so is the cylinder EP to the cylinder ES. [V. 11]

Therefore the cylinder AO is equal to the cylinder EP. [V. 9]

And the same is true for the cones also. Q. E. D.

 PROPOSITION 16.

Given two circles about the same centre, to inscribe in the greater circle an equilateral polygon with an even number of sides which does not touch the lesser circle. -

+

Let ABCD, EFGH be the two given circles about the same centre K; thus it is required to inscribe in the greater circle ABCD an equilateral polygon with an even number of sides which does not touch the circle EFGH.

For let the straight line BKD be drawn through the centre K, and from the point G let GA be drawn at right angles to the straight line BD and carried through to C; therefore AC touches the circle EFGH. [III. 16, Por.]

@@ -11579,7 +10005,7 @@ and, componendo, the squares on AB<

Now, since LN is parallel to AC, and AC touches the circle EFGH, therefore LN does not touch the circle EFGH; therefore LD, DN are far from touching the circle EFGH.

If then we fit into the circle ABCD straight lines equal to the straight line LD and placed continuously, there will be inscribed in the circle ABCD an equilateral polygon with an even number of sides which does not touch the lesser circle EFGH. Q. E. F.

- + PROPOSITION 17.

Given two spheres about the same centre, to inscribe in the greater sphere a polyhedral solid which does not touch the lesser sphere at its surface. @@ -11587,14 +10013,14 @@ and, componendo, the squares on AB<

Let two spheres be conceived about the same centre A; thus it is required to inscribe in the greater sphere a polyhedral solid which does not touch the lesser sphere at its surface.

Let the spheres be cut by any plane through the centre; then the sections will be circles, inasmuch as the sphere was produced by the diameter remaining fixed and the semicircle being carried round it; [XI. Def. 14] hence, in whatever position we conceive the semicircle to be, the plane carried through it will produce a circle on the circumference of the sphere.

-

And it is manifest that this circle is the greatest possible, inasmuch as the diameter of the sphere, which is of course the diameter both of the semicircle and of the circle, is greater than all the straight lines drawn across in the circle or the sphere.

+

And it is manifest that this circle is the greatest possible, inasmuch as the diameter of the sphere, which is of course the diameter both of the semicircle and of the circle, is greater than all the straight lines drawn across in the circle or the sphere.

Let then BCDE be the circle in the greater sphere, and FGH the circle in the lesser sphere; let two diameters in them, BD, CE, be drawn at right angles to one another; then, given the two circles BCDE, FGH about the same centre, let there be inscribed in the greater circle BCDE an equilateral polygon with an even number of sides which does not touch the lesser circle FGH, let BK, KL, LM ME be its sides in the quadrant BE. let KA be joined and carried through to N, let AO be set up from the point A at right angles to the plane of the circle BCDE, and let it meet the surface of the sphere at O, and through AO and each of the straight lines BD, KN let planes be carried; they will then make greatest circles on the surface of the sphere, for the reason stated.

Let them make such, and in them let BOD, KON be the semicircles on BD, KN.

Now, since OA is at right angles to the plane of the circle BCDE, therefore all the planes through OA are also at right angles to the plane of the circle BCDE; [XI. 18] hence the semicircles BOD, KON are also at right angles to the plane of the circle BCDE.

And, since the semicircles BED, BOD, KON are equal, for they are on the equal diameters BD, KN, therefore the quadrants BE, BO, KO are also equal to one another.

Therefore there are as many straight lines in the quadrants BO, KO equal to the straight lines BK, KL, LM, ME as there are sides of the polygon in the quadrant BE.

-

Let them be inscribed, and let them be BP, PQ, QR, RO and KS, ST, TU, UO, let SP, TQ, UR be joined, and from P, S let perpendiculars be drawn to the plane of the circle BCDE; [XI. 11] these will fall on BD, KN, the common sections of the planes, inasmuch as the planes of BOD, KON are also at right angles to the plane of the circle BCDE. [cf. XI. Def. 4]

+

Let them be inscribed, and let them be BP, PQ, QR, RO and KS, ST, TU, UO, let SP, TQ, UR be joined, and from P, S let perpendiculars be drawn to the plane of the circle BCDE; [XI. 11] these will fall on BD, KN, the common sections of the planes, inasmuch as the planes of BOD, KON are also at right angles to the plane of the circle BCDE. [cf. XI. Def. 4]

Let them so fall, and let them be PV, SW, and let WV be joined.

Now since, in the equal semicircles BOD, KON, equal straight lines BP, KS have been cut off, and the perpendiculars PV, SW have been drawn, therefore PV is equal to SW, and BV to KW. [III. 27, I. 26]

But the whole BA is also equal to the whole KA; therefore the remainder VA is also equal to the remainder WA; therefore, as BV is to VA, so is KW to WA; therefore WV is parallel to KB. [VI. 2]

@@ -11604,7 +10030,7 @@ and, componendo, the squares on AB<

And BP, KS join their extremities; therefore the quadrilateral KBPS is in one plane, inasmuch as, if two straight lines be parallel, and points be taken at random on each of them, the straight line joining the points is in the same plane with the parallels. [XI. 7]

For the same reason each of the quadrilaterals SPQT, TQRU is also in one plane.

But the triangle URO is also in one plane. [XI. 2]

-

If then we conceive straight lines joined from the points P, S, Q, T, R, U to A, there will be constructed a certain polyhedral solid figure between the circumferences BO, KO, consisting of pyramids of which the quadrilaterals KBPS, SPQT, TQRU and the triangle URO are the bases and the point A the vertex.

+

If then we conceive straight lines joined from the points P, S, Q, T, R, U to A, there will be constructed a certain polyhedral solid figure between the circumferences BO, KO, consisting of pyramids of which the quadrilaterals KBPS, SPQT, TQRU and the triangle URO are the bases and the point A the vertex.

And, if we make the same construction in the case of each of the sides KL, LM, ME as in the case of BK, and further in the case of the remaining three quadrants, there will be constructed a certain polyhedral figure inscribed in the sphere and contained by pyramids, of which the said quadrilaterals and the triangle URO, and the others corresponding to them, are the bases and the point A the vertex.

I say that the said polyhedron will not touch the lesser sphere at the surface on which the circle FGH is.

Let AX be drawn from the point A perpendicular to the plane of the quadrilateral KBPS, and let it meet the plane at the point X; [XI. 11] let XB, XK be joined.

@@ -11614,7 +10040,7 @@ and, componendo, the squares on AB<

And the squares on AX, XB are equal to the square on AB, for the angle at X is right; [I. 47] and the squares on AX, XK are equal to the square on AK. [id.]

Therefore the squares on AX, XB are equal to the squares on AX, XK.

Let the square on AX be subtracted from each; therefore the remainder, the square on BX, is equal to the remainder, the square on XK; therefore BX is equal to XK.

-

Similarly we can prove that the straight lines joined from X to P, S are equal to each of the straight lines BX, XK.

+

Similarly we can prove that the straight lines joined from X to P, S are equal to each of the straight lines BX, XK.

Therefore the circle described with centre X and distance one of the straight lines XB, XK will pass through P, S also, and KBPS will be a quadrilateral in a circle.

Now, since KB is greater than WV, while WV is equal to SP, therefore KB is greater than SP.

But KB is equal to each of the straight lines KS, BP; therefore each of the straight lines KS, BP is greater than SP.

@@ -11624,7 +10050,7 @@ and, componendo, the squares on AB<

And, if KD be joined, the rectangle DB, BZ is equal to the square on BK, and the rectangle DZ, ZB equal to the square on KZ; [III. 31, VI. 8 and Por.] therefore the square on KB is less than double of the square on KZ.

But the square on KB is greater than double of the square on BX; therefore the square on KZ is greater than the square on BX.

And, since BA is equal to KA, the square on BA is equal to the square on AK.

-

And the squares on BX, XA are equal to the square on BA, and the squares on KZ, ZA equal to the square on KA; [I. 47] therefore the squares on BX, XA are equal to the squares on KZ, ZA, and of these the square on KZ is greater than the square on BX; therefore the remainder, the square on ZA, is less than the square on XA.

+

And the squares on BX, XA are equal to the square on BA, and the squares on KZ, ZA equal to the square on KA; [I. 47] therefore the squares on BX, XA are equal to the squares on KZ, ZA, and of these the square on KZ is greater than the square on BX; therefore the remainder, the square on ZA, is less than the square on XA.

Therefore AX is greater than AZ; therefore AX is much greater than AG.

And AX is the perpendicular on one base of the polyhedron, and AG on the surface of the lesser sphere; hence the polyhedron will not touch the lesser sphere on its surface.

Therefore, given two spheres about the same centre, a polyhedral solid has been inscribed in the greater sphere which does not touch the lesser sphere at its surface. Q. E. F.

@@ -11634,7 +10060,7 @@ and, componendo, the squares on AB<

For, the solids being divided into their pyramids similar in multitude and arrangement, the pyramids will be similar.

But similar pyramids are to one another in the triplicate ratio of their corresponding sides; [XII. 8, Por.] therefore the pyramid of which the quadrilateral KBPS is the base, and the point A the vertex, has to the similarly arranged pyramid in the other sphere the ratio triplicate of that which the corresponding side has to the corresponding side, that is, of that which the radius AB of the sphere about A as centre has to the radius of the other sphere.

Similarly also each pyramid of those in the sphere about A as centre has to each similarly arranged pyramid of those in the other sphere the ratio triplicate of that which AB has to the radius of the other sphere.

-

And, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents; [V. 12] hence the whole polyhedral solid in the sphere about A as centre has to the whole polyhedral solid in the other sphere the ratio triplicate of that which AB has to the radius of the other sphere, that is, of that which the diameter BD has to the diameter of the other sphere. Q. E. D.

+

And, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents; [V. 12] hence the whole polyhedral solid in the sphere about A as centre has to the whole polyhedral solid in the other sphere the ratio triplicate of that which AB has to the radius of the other sphere, that is, of that which the diameter BD has to the diameter of the other sphere. Q. E. D.

@@ -11644,12 +10070,12 @@ and, componendo, the squares on AB<

Let the spheres ABC, DEF be conceived, and let BC, EF be their diameters; I say that the sphere ABC has to the sphere DEF the ratio triplicate of that which BC has to EF.

For, if the sphere ABC has not to the sphere DEF the ratio triplicate of that which BC has to EF, then the sphere ABC will have either to some less sphere than the sphere DEF, or to a greater, the ratio triplicate of that which BC has to EF.

-

First, let it have that ratio to a less sphere GHK, let DEF be conceived about the same centre with GHK, let there be inscribed in the greater sphere DEF a polyhedral solid which does not touch the lesser sphere GHK at its surface, [XII. 17] and let there also be inscribed in the sphere ABC a polyhedral solid similar to the polyhedral solid in the sphere DEF; therefore the polyhedral solid in ABC has to the polyhedral solid in DEF the ratio triplicate of that which BC has to EF. [XII. 17, Por.]

+

First, let it have that ratio to a less sphere GHK, let DEF be conceived about the same centre with GHK, let there be inscribed in the greater sphere DEF a polyhedral solid which does not touch the lesser sphere GHK at its surface, [XII. 17] and let there also be inscribed in the sphere ABC a polyhedral solid similar to the polyhedral solid in the sphere DEF; therefore the polyhedral solid in ABC has to the polyhedral solid in DEF the ratio triplicate of that which BC has to EF. [XII. 17, Por.]

But the sphere ABC also has to the sphere GHK the ratio triplicate of that which BC has to EF; therefore, as the sphere ABC is to the sphere GHK, so is the polyhedral solid in the sphere ABC to the polyhedral solid in the sphere DEF; and, alternately, as the sphere ABC is to the polyhedron in it, so is the sphere GHK to the polyhedral solid in the sphere DEF. [V. 16]

But the sphere ABC is greater than the polyhedron in it; therefore the sphere GHK is also greater than the polyhedron in the sphere DEF.

But it is also less, for it is enclosed by it.

-

Therefore the sphere ABC has not to a less sphere than the sphere DEF the ratio triplicate of that which the diameter BC has to EF.

+

Therefore the sphere ABC has not to a less sphere than the sphere DEF the ratio triplicate of that which the diameter BC has to EF.

Similarly we can prove that neither has the sphere DEF to a less sphere than the sphere ABC the ratio triplicate of that which EF has to BC.

I say next that neither has the sphere ABC to any greater sphere than the sphere DEF the ratio triplicate of that which BC has to EF.

For, if possible, let it have that ratio to a greater, LMN; therefore, inversely, the sphere LMN has to the sphere ABC the ratio triplicate of that which the diameter EF has to the diameter BC.

@@ -11660,9 +10086,9 @@ and, componendo, the squares on AB<

Therefore the sphere ABC has to the sphere DEF the ratio triplicate of that which BC has to EF. Q. E. D.

- - - +
+
+ BOOK XIII. HISTORICAL NOTE.

I have already given, in the note to IV. 10, the evidence upon which the construction of the five regular solids is attributed to the Pythagoreans. Some of them, the cube, the tetrahedron (which is nothing but a pyramid), and the octahedron (which is only a double pyramid with a square base), cannot but have been known to the Egyptians. And it appears that dodecahedra have been found, of bronze or other material, which may belong to periods earlier than Pythagoras' time by some centuries (for references see Cantor's Geschichte der Mathematik I_{3}, pp. 175-6).

@@ -11672,10 +10098,10 @@ and, componendo, the squares on AB<

Theaetetus flourished about 380 B.C., and his work on the regular solids was soon followed by another, that of Aristaeus, an elder contemporary of Euclid, who also wrote an important book on Solid Loci, i.e. on conics treated as loci. This Aristaeus (known as the elder ) wrote in the period about 320 B.C. We hear of his Comparison of the five regular solids from Hypsicles (2nd cent. B.C.), the writer of the short book commonly included in the editions of the Elements as Book XIV. Hypsicles gives in this Book some six propositions supplementing Eucl. XIII.; and he introduces the second of the propositions (Heiberg's Euclid, Vol. v. p. 6) as follows:

The same circle circumscribes both the pentagon of the dodecahedron and the triangle of the icosahedron when both are inscribed in the same sphere. This is proved by Aristaeus in the book entitled Comparison of the five figures. -

+

Hypsicles proceeds (pp. 7 sqq.) to give a proof of this theorem. Allman pointed out (Greek Geometry from Thales to Euclid, 1889, pp. 201-2) that this proof depends on eight theorems, six of which appear in Euclid's Book XIII. (in Propositions 8, 10, 12, 15, 16 with Por., 17); two other propositions not mentioned by Allman are also used, namely XIII. 4 and 9. This seems, as Allman says, to confirm the inference of Bretschneider (p. 171) that, as Aristaeus' work was the newest and latest in which, before Euclid's time, this subject was treated, we have in Eucl. XIII. at least a partial recapitulation of the contents of the treatise of Aristaeus.

After Euclid, Apollonius wrote on the comparison of the dodecahedron and the icosahedron inscribed in one and the same sphere. This we also learn from Hypsicles, who says in the next words following those about Aristaeus above quoted: But it is proved by Apollonius in the second edition of his Comparison of the dodecahedron with the icosahedron that, as the surface of the dodecahedron is to the surface of the icosahedron [inscribed in the same sphere], so is the dodecahedron itself [i.e. its volume] to the icosahedron, because the perpendicular is the same from the centre of the sphere to the pentagon of the dodecahedron and to the triangle of the icosahedron. -

+

BOOK XIII. PROPOSITIONS. @@ -11690,7 +10116,7 @@ and, componendo, the squares on AB<

And, since BA is double of AD, while BA is equal to KA, and AD to AH, therefore KA is also double of AH.

But, as KA is to AH, so is CK to CH; [VI. 1] therefore CK is double of CH.

But LH, HC are also double of CH.

-

Therefore KC is equal to LH, HC.

+

Therefore KC is equal to LH, HC.

But CE was also proved equal to HF; therefore the whole square AE is equal to the gnomon MNO.

And, since BA is double of AD, the square on BA is quadruple of the square on AD, that is, AE is quadruple of DH.

But AE is equal to the gnomon MNO; therefore the gnomon MNO is also quadruple of AP; therefore the whole DF is five times AP.

@@ -11708,7 +10134,7 @@ and, componendo, the squares on AB<

Therefore the gnomon MNO is quadruple of AH.

And, since DC is double of CA, therefore the square on DC is quadruple of the square on CA, that is, CG is quadruple of AH.

But the gnomon MNO was also proved quadruple of AH; therefore the gnomon MNO is equal to CG.

-

And, since DC is double of CA, while DC is equal to CK, and AC to CH, therefore KB is also double of BH. [VI. 1]

+

And, since DC is double of CA, while DC is equal to CK, and AC to CH, therefore KB is also double of BH. [VI. 1]

But LH, HB are also double of HB; therefore KB is equal to LH, HB.

But the whole gnomon MNO was also proved equal to the whole CG; therefore the remainder HF is equal to BG.

And BG is the rectangle CD, DB, for CD is equal to DG; and HF is the square on CB; therefore the rectangle CD, DB is equal to the square on CB.

@@ -11730,7 +10156,7 @@ and, componendo, the squares on AB< PROPOSITION 3.

If a straight line be cut in extreme and mean ratio, the square on the lesser segment added to the half of the greater segment is five times the square on the half of the greater segment. -

+

For let any straight line AB be cut in extreme and mean ratio at the point C, let AC be the greater segment, and let AC be bisected at D; I say that the square on BD is five times the square on DC.

For let the square AE be described on AB, and let the figure be drawn double.

@@ -11746,7 +10172,7 @@ and, componendo, the squares on AB<

Therefore the gnomon OPQ and the square FG are five times FG.

But the gnomon OPQ and the square FG are the square DN.

And DN is the square on DB, and GF the square on DC.

-

Therefore the square on DB is five times the square on DC. Q. E. D.

+

Therefore the square on DB is five times the square on DC. Q. E. D.

PROPOSITION 4. @@ -11756,7 +10182,7 @@ and, componendo, the squares on AB<

For let the square ADEB be described on AB, and let the figure be drawn.

Since then AB has been cut in extreme and mean ratio at C, and AC is the greater segment, therefore the rectangle AB, BC is equal to the square on AC. [VI. Def. 3, VI. 17]

-

And AK is the rectangle AB, BC, and HG the square on AC; therefore AK is equal to HG.

+

And AK is the rectangle AB, BC, and HG the square on AC; therefore AK is equal to HG.

And, since AF is equal to FE, let CK be added to each; therefore the whole AK is equal to the whole CE; therefore AK, CE are double of AK.

But AK, CE are the gnomon LMN and the square CK; therefore the gnomon LMN and the square CK are double of AK.

But, further, AK was also proved equal to HG; therefore the gnomon LMN and the squares CK, HG are triple of the square HG.

@@ -11770,7 +10196,7 @@ and, componendo, the squares on AB<

For let the straight line AB be cut in extreme and mean ratio at the point C, let AC be the greater segment, and let AD be equal to AC.

I say that the straight line DB has been cut in extreme and mean ratio at A, and the original straight line AB is the greater segment.

-

For let the square AE be described on AB, and let the figure be drawn.

+

For let the square AE be described on AB, and let the figure be drawn.

Since AB has been cut in extreme and mean ratio at C, therefore the rectangle AB, BC is equal to the square on AC. [VI. Def. 3, VI. 17]

And CE is the rectangle AB, BC, and CH the square on AC; therefore CE is equal to HC.

But HE is equal to CE, and DH is equal to HC; therefore DH is also equal to HE.

@@ -11783,7 +10209,7 @@ and, componendo, the squares on AB< PROPOSITION 6.

If a rational straight line be cut in extreme and mean ratio, each of the segments is the irrational straight line called apotome. -

+

Let AB be a rational straight line, let it be cut in extreme and mean ratio at C, and let AC be the greater segment; I say that each of the straight lines AC, CB is the irrational straight line called apotome.

For let BA be produced, and let AD be made half of BA.

@@ -11793,7 +10219,7 @@ and, componendo, the squares on AB<

And, since the square on CD has not to the square on DA the ratio which a square number has to a square number, therefore CD is incommensurable in length with DA; [X. 9] therefore CD, DA are rational straight lines commensurable in square only; therefore AC is an apotome. [X. 73]

Again, since AB has been cut in extreme and mean ratio, and AC is the greater segment, therefore the rectangle AB, BC is equal to the square on AC. [VI. Def. 3, VI. 17]

Therefore the square on the apotome AC, if applied to the rational straight line AB, produces BC as breadth.

-

But the square on an apotome, if applied to a rational straight line, produces as breadth a first apotome; [X. 97] therefore CB is a first apotome.

+

But the square on an apotome, if applied to a rational straight line, produces as breadth a first apotome; [X. 97] therefore CB is a first apotome.

And CA was also proved to be an apotome.

Therefore etc. Q. E. D.

@@ -11804,7 +10230,7 @@ and, componendo, the squares on AB<

For in the equilateral pentagon ABCDE let, first, three angles taken in order, those at A, B, C, be equal to one another; I say that the pentagon ABCDE is equiangular.

For let AC, BE, FD be joined.

-

Now, since the two sides CB, BA are equal to the two sides BA, AE respectively, and the angle CBA is equal to the angle BAE, therefore the base AC is equal to the base BE, the triangle ABC is equal to the triangle ABE, and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend, [I. 4] that is, the angle BCA to the angle BEA, and the angle ABE to the angle CAB; hence the side AF is also equal to the side BF. [I. 6]

+

Now, since the two sides CB, BA are equal to the two sides BA, AE respectively, and the angle CBA is equal to the angle BAE, therefore the base AC is equal to the base BE, the triangle ABC is equal to the triangle ABE, and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend, [I. 4] that is, the angle BCA to the angle BEA, and the angle ABE to the angle CAB; hence the side AF is also equal to the side BF. [I. 6]

But the whole AC was also proved equal to the whole BE; therefore the remainder FC is also equal to the remainder FE.

But CD is also equal to DE.

Therefore the two sides FC, CD are equal to the two sides FE, ED; and the base FD is common to them; therefore the angle FCD is equal to the angle FED. [I. 8]

@@ -11816,7 +10242,7 @@ and, componendo, the squares on AB<

Then, since the two sides BA, AE are equal to the two sides BC, CD, and they contain equal angles, therefore the base BE is equal to the base BD, the triangle ABE is equal to the triangle BCD, and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend; [I. 4] therefore the angle AEB is equal to the angle CDB.

But the angle BED is also equal to the angle BDE, since the side BE is also equal to the side BD. [I. 5]

Therefore the whole angle AED is equal to the whole angle CDE.

-

But the angle CDE is, by hypothesis, equal to the angles at A, C; therefore the angle AED is also equal to the angles at A, C.

+

But the angle CDE is, by hypothesis, equal to the angles at A, C; therefore the angle AED is also equal to the angles at A, C.

For the same reason the angle ABC is also equal to the angles at A, C, D.

Therefore the pentagon ABCDE is equiangular. Q. E. D.

@@ -11826,7 +10252,7 @@ and, componendo, the squares on AB<

For in the equilateral and equiangular pentagon ABCDE let the straight lines AC, BE, cutting one another at the point H, subtend two angles taken in order, the angles at A, B; I say that each of them has been cut in extreme and mean ratio at the point H, and their greater segments are equal to the side of the pentagon.

-

For let the circle ABCDE be circumscribed about the pentagon ABCDE. [IV. 14]

+

For let the circle ABCDE be circumscribed about the pentagon ABCDE. [IV. 14]

Then, since the two straight lines EA, AB are equal to the two AB, BC, and they contain equal angles, therefore the base BE is equal to the base AC, the triangle ABE is equal to the triangle ABC, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend. [I. 4]

Therefore the angle BAC is equal to the angle ABE; therefore the angle AHE is double of the angle BAH. [I. 32]

But the angle EAC is also double of the angle BAC, inasmuch as the circumference EDC is also double of the circumference CB; [III. 28, VI. 33] therefore the angle HAE is equal to the angle AHE; hence the straight line HE is also equal to EA, that is, to AB. [I. 6]

@@ -11836,7 +10262,7 @@ and, componendo, the squares on AB<

But BA is equal to EH; therefore, as BE is to EH, so is EH to HB.

And BE is greater than EH; therefore EH is also greater than HB. [V. 14]

Therefore BE has been cut in extreme and mean ratio at H, and the greater segment HE is equal to the side of the pentagon.

-

Similarly we can prove that AC has also been cut in extreme and mean ratio at H, and its greater segment CH is equal to the side of the pentagon. Q. E. D.

+

Similarly we can prove that AC has also been cut in extreme and mean ratio at H, and its greater segment CH is equal to the side of the pentagon. Q. E. D.

PROPOSITION 9. @@ -11845,7 +10271,7 @@ and, componendo, the squares on AB<

Let ABC be a circle; of the figures inscribed in the circle ABC let BC be the side of a decagon, CD that of a hexagon, and let them be in a straight line; I say that the whole straight line BD has been cut in extreme and mean ratio, and CD is its greater segment.

For let the centre of the circle, the point E, be taken, let EB, EC, ED be joined, and let BE be carried through to A.

-

Since BC is the side of an equilateral decagon, therefore the circumference ACB is five times the circumference BC; therefore the circumference AC is quadruple of CB.

+

Since BC is the side of an equilateral decagon, therefore the circumference ACB is five times the circumference BC; therefore the circumference AC is quadruple of CB.

But, as the circumference AC is to CB, so is the angle AEC to the angle CEB; [VI. 33] therefore the angle AEC is quadruple of the angle CEB.

And, since the angle EBC is equal to the angle ECB, [I. 5] therefore the angle AEC is double of the angle ECB. [I. 32]

And, since the straight line EC is equal to CD, for each of them is equal to the side of the hexagon inscribed in the circle ABC, [IV. 15, Por.] the angle CED is also equal to the angle CDE; [I. 5] therefore the angle ECB is double of the angle EDC. [I. 32]

@@ -11864,7 +10290,7 @@ and, componendo, the squares on AB<

Let ABCDE be a circle, and let the equilateral pentagon ABCDE be inscribed in the circle ABCDE.

I say that the square on the side of the pentagon ABCDE is equal to the squares on the side of the hexagon and on that of the decagon inscribed in the circle ABCDE.

-

For let the centre of the circle, the point F, be taken, let AF be joined and carried through to the point G, let FB be joined, let FH be drawn from F perpendicular to AB and be carried through to K, let AK, KB be joined, let FL be again drawn from F perpendicular to AK, and be carried through to M, and let KN be joined.

+

For let the centre of the circle, the point F, be taken, let AF be joined and carried through to the point G, let FB be joined, let FH be drawn from F perpendicular to AB and be carried through to K, let AK, KB be joined, let FL be again drawn from F perpendicular to AK, and be carried through to M, and let KN be joined.

Since the circumference ABCG is equal to the circumference AEDG, and in them ABC is equal to AED, therefore the remainder, the circumference CG, is equal to the remainder GD.

But CD belongs to a pentagon; therefore CG belongs to a decagon.

@@ -11873,7 +10299,7 @@ and, componendo, the squares on AB<

For the same reason AK is also double of KM.

Now, since the circumference AB is double of the circumference BK, while the circumference CD is equal to the circumference AB, therefore the circumference CD is also double of the circumference BK.

But the circumference CD is also double of CG; therefore the circumference CG is equal to the circumference BK.

-

But BK is double of KM, since KA is so also; therefore CG is also double of KM.

+

But BK is double of KM, since KA is so also; therefore CG is also double of KM.

But, further, the circumference CB is also double of the circumference BK, for the circumference CB is equal to BA.

Therefore the whole circumference GB is also double of BM; hence the angle GFB is also double of the angle BFM. [VI. 33]

But the angle GFB is also double of the angle FAB, for the angle FAB is equal to the angle ABF.

@@ -11885,7 +10311,7 @@ and, componendo, the squares on AB<

And the angle at A is common to the two triangles AKB and AKN.

Therefore the remaining angle AKB is equal to the remaining angle KNA; [I. 32] therefore the triangle KBA is equiangular with the triangle KNA.

Therefore, proportionally, as the straight line BA is to AK, so is KA to AN; [VI. 4] therefore the rectangle BA, AN is equal to the square on AK. [VI. 17]

-

But the rectangle AB, BN was also proved equal to the square on BF; therefore the rectangle AB, BN together with the rectangle BA, AN, that is, the square on BA [II. 2], is equal to the square on BF together with the square on AK.

+

But the rectangle AB, BN was also proved equal to the square on BF; therefore the rectangle AB, BN together with the rectangle BA, AN, that is, the square on BA [II. 2], is equal to the square on BF together with the square on AK.

And BA is a side of the pentagon, BF of the hexagon [IV. 15, Por.], and AK of the decagon.

Therefore etc. Q. E. D.

@@ -11897,14 +10323,14 @@ and, componendo, the squares on AB<

For let the centre of the circle, the point F, be taken, let AF, FB be joined and carried through to the points, G, H, let AC be joined, and let FK be made a fourth part of AF.

Now AF is rational; therefore FK is also rational.

But BF is also rational; therefore the whole BK is rational.

-

And, since the circumference ACG is equal to the circumference ADG, and in them ABC is equal to AED, therefore the remainder CG is equal to the remainder GD.

+

And, since the circumference ACG is equal to the circumference ADG, and in them ABC is equal to AED, therefore the remainder CG is equal to the remainder GD.

And, if we join AD, we conclude that the angles at L are right, and CD is double of CL.

For the same reason the angles at M are also right, and AC is double of CM.

Since then the angle ALC is equal to the angle AMF, and the angle LAC is common to the two triangles ACL and AMF, therefore the remaining angle ACL is equal to the remaining angle MFA; [I. 32] therefore the triangle ACL is equiangular with the triangle AMF; therefore, proportionally, as LC is to CA, so is MF to FA.

And the doubles of the antecedents may be taken; therefore, as the double of LC is to CA, so is the double of MF to FA.

But, as the double of MF is to FA, so is MF to the half of FA; therefore also, as the double of LC is to CA, so is MF to the half of FA.

-

And the halves of the consequents may be taken; therefore, as the double of LC is to the half of CA, so is MF to the fourth of FA.

+

And the halves of the consequents may be taken; therefore, as the double of LC is to the half of CA, so is MF to the fourth of FA.

And DC is double of LC, CM is half of CA, and FK a fourth part of FA; therefore, as DC is to CM, so is MF to FK.

Componendo also, as the sum of DC, CM is to CM, so is MK to KF; [V. 18] therefore also, as the square on the sum of DC, CM is to the square on CM, so is the square on MK to the square on KF.

And since, when the straight line subtending two sides of the pentagon, as AC, is cut in extreme and mean ratio, the greater segment is equal to the side of the pentagon, that is, to DC, [XIII. 8] while the square on the greater segment added to the half of the whole is five times the square on the half of the whole, [XIII. 1] and CM is half of the whole AC, therefore the square on DC, CM taken as one straight line is five times the square on CM.

@@ -11913,7 +10339,7 @@ and, componendo, the squares on AB<

And, since BF is quadruple of FK, therefore BK is five times KF; therefore the square on BK is twenty-five times the square on KF.

But the square on MK is five times the square on KF; therefore the square on BK is five times the square on KM; therefore the square on BK has not to the square on KM the ratio which a square number has to a square number; therefore BK is incommensurable in length with KM. [X. 9]

And each of them is rational.

-

Therefore BK, KM are rational straight lines commensurable in square only.

+

Therefore BK, KM are rational straight lines commensurable in square only.

But, if from a rational straight line there be subtracted a rational straight line which is commensurable with the whole in square only, the remainder is irrational, namely an apotome; therefore MB is an apotome and MK the annex to it. [X. 73]

I say next that MB is also a fourth apotome.

Let the square on N be equal to that by which the square on BK is greater than the square on KM; therefore the square on BK is greater than the square on KM by the square on N.

@@ -11923,13 +10349,13 @@ and, componendo, the squares on AB<

Therefore, convertendo, the square on BK has to the square on N the ratio which 5 has to 4 [V. 19, Por.], and this is not the ratio which a square number has to a square number; therefore BK is incommensurable with N; [X. 9] therefore the square on BK is greater than the square on KM by the square on a straight line incommensurable with BK.

Since then the square on the whole BK is greater than the square on the annex KM by the square on a straight line incommensurable with BK, and the whole BK is commensurable with the rational straight line, BH, set out, therefore MB is a fourth apotome. [X. Deff. III. 4]

But the rectangle contained by a rational straight line and a fourth apotome is irrational, and its square root is irrational, and is called minor. [X. 94]

-

But the square on AB is equal to the rectangle HB, BM, because, when AH is joined, the triangle ABH is equiangular with the triangle ABM, and, as HB is to BA, so is AB to BM.

+

But the square on AB is equal to the rectangle HB, BM, because, when AH is joined, the triangle ABH is equiangular with the triangle ABM, and, as HB is to BA, so is AB to BM.

Therefore the side AB of the pentagon is the irrational straight line called minor. Q. E. D.

PROPOSITION 12.

If an equilateral triangle be inscribed in a circle, the square on the side of the triangle is triple of the square on the radius of the circle. -

+

Let ABC be a circle, and let the equilateral triangle ABC be inscribed in it; I say that the square on one side of the triangle ABC is triple of the square on the radius of the circle.

For let the centre D of the circle ABC be taken, let AD be joined and carried through to E, and let BE be joined.

@@ -11944,12 +10370,12 @@ and, componendo, the squares on AB< PROPOSITION 13.

To construct a pyramid, to comprehend it in a given sphere, and to prove that the square on the diameter of the sphere is one and a half times the square on the side of the pyramid. -

+

Let the diameter AB of the given sphere be set out, and let it be cut at the point C so that AC is double of CB; let the semicircle ADB be described on AB, let CD be drawn from the point C at right angles to AB, and let DA be joined; let the circle EFG which has its radius equal to DC be set out, let the equilateral triangle EFG be inscribed in the circle EFG, [IV. 2] let the centre of the circle, the point H, be taken, [III. 1] let EH, HF, HG be joined; from the point H let HK be set up at right angles to the plane of the circle EFG, [XI. 12] let HK equal to the straight line AC be cut off from HK, and let KE, KF, KG be joined.

Now, since KH is at right angles to the plane of the circle EFG, therefore it will also make right angles with all the straight lines which meet it and are in the plane of the circle EFG. [XI. Def. 3]

But each of the straight lines HE, HF, HG meets it: therefore HK is at right angles to each of the straight lines HE, HF, HG.

-

And, since AC is equal to HK, and CD to HE, and they contain right angles, therefore the base DA is equal to the base KE. [I. 4]

+

And, since AC is equal to HK, and CD to HE, and they contain right angles, therefore the base DA is equal to the base KE. [I. 4]

For the same reason each of the straight lines KF, KG is also equal to DA; therefore the three straight lines KE, KF, KG are equal to one another.

And, since AC is double of CB, therefore AB is triple of BC.

But, as AB is to BC, so is the square on AD to the square on DC, as will be proved afterwards.

@@ -11960,7 +10386,7 @@ and, componendo, the squares on AB<

It is next required to comprehend it in the given sphere and to prove that the square on the diameter of the sphere is one and a half times the square on the side of the pyramid.

For let the straight line HL be produced in a straight line with KH, and let HL be made equal to CB.

Now, since, as AC is to CD, so is CD to CB, [VI. 8, Por.] while AC is equal to KH, CD to HE, and CB to HL, therefore, as KH is to HE, so is EH to HL; therefore the rectangle KH, HL is equal to the square on EH. [VI. 17]

-

And each of the angles KHE. EHL is right; therefore the semicircle described on KL will pass through E also. [cf. VI. 8, III. 31.]

+

And each of the angles KHE. EHL is right; therefore the semicircle described on KL will pass through E also. [cf. VI. 8, III. 31.]

If then, KL remaining fixed, the semicircle be carried round and restored to the same position from which it began to be moved, it will also pass through the points F, G, since, if FL, LG be joined, the angles at F, G similarly become right angles; and the pyramid will be comprehended in the given sphere.

For KL, the diameter of the sphere, is equal to the diameter AB of the given sphere, inasmuch as KH was made equal to AC, and HL to CB.

I say next that the square on the diameter of the sphere is one and a half times the square on the side of the pyramid

@@ -11974,13 +10400,13 @@ and, componendo, the squares on AB<

It is to be proved that, as AB is to BC, so is the square on AD to the square on DC.

For let the figure of the semicircle be set out, let DB be joined, let the square EC be described on AC, and let the parallelogram FB be completed.

-

Since then, because the triangle DAB is equiangular with the triangle DAC, as BA is to AD, so is DA to AC, [VI. 8, VI. 4] therefore the rectangle BA, AC is equal to the square on AD. [VI. 17]

+

Since then, because the triangle DAB is equiangular with the triangle DAC, as BA is to AD, so is DA to AC, [VI. 8, VI. 4] therefore the rectangle BA, AC is equal to the square on AD. [VI. 17]

And since, as AB is to BC, so is EB to BF, [VI. 1] and EB is the rectangle BA, AC, for EA is equal to AC, and BF is the rectangle AC, CB, therefore, as AB is to BC, so is the rectangle BA, AC to the rectangle AC, CB.

And the rectangle BA, AC is equal to the square on AD, and the rectangle AC, CB to the square on DC, for the perpendicular DC is a mean proportional between the segments AC, CB of the base, because the angle ADB is right. [VI. 8, Por.]

Therefore, as AB is to BC, so is the square on AD to the square on DC. Q. E. D.

- + PROPOSITION 14.

To construct an octahedron and comprehend it in a sphere, as in the preceding case; and to prove that the square on the diameter of the sphere is double of the square on the side of the octahedron. @@ -11988,7 +10414,7 @@ and, componendo, the squares on AB<

Let the diameter AB of the given sphere be set out, and let it be bisected at C; let the semicircle ADB be described on AB, let CD be drawn from C at right angles to AB, let DB be joined; let the square EFGH, having each of its sides equal to DB, be set out, let HF, EG be joined, from the point K let the straight line KL be set up at right angles to the plane of the square EFGH [XI. 12], and let it be carried through to the other side of the plane, as KM; from the straight lines KL, KM let KL, KM be respectively cut off equal to one of the straight lines EK, FK, GK, HK, and let LE, LF, LG, LH, ME, MF, MG, MH be joined.

Then, since KE is equal to KH, and the angle EKH is right, therefore the square on HE is double of the square on EK. [I. 47]

-

Again, since LK is equal to KE, and the angle LKE is right, therefore the square on EL is double of the square on EK. [id.]

+

Again, since LK is equal to KE, and the angle LKE is right, therefore the square on EL is double of the square on EK. [id.]

But the square on HE was also proved double of the square on EK; therefore the square on LE is equal to the square on EH; therefore LE is equal to EH.

For the same reason LH is also equal to HE; therefore the triangle LEH is equilateral.

Similarly we can prove that each of the remaining triangles of which the sides of the square EFGH are the bases, and the points L, M the vertices, is equilateral; therefore an octahedron has been constructed which is contained by eight equilateral triangles.

@@ -11998,7 +10424,7 @@ and, componendo, the squares on AB<

I say next that it is also comprehended in the given sphere.

For, since LK is equal to KM, while KE is common, and they contain right angles, therefore the base LE is equal to the base EM. [I. 4]

And, since the angle LEM is right, for it is in a semicircle, [III. 31] therefore the square on LM is double of the square on LE. [I. 47]

-

Again, since AC is equal to CB, AB is double of BC.

+

Again, since AC is equal to CB, AB is double of BC.

But, as AB is to BC, so is the square on AB to the square on BD; therefore the square on AB is double of the square on BD.

But the square on LM was also proved double of the square on LE.

And the square on DB is equal to the square on LE, for EH was made equal to DB.

@@ -12006,7 +10432,7 @@ and, componendo, the squares on AB<

And AB is the diameter of the given sphere; therefore LM is equal to the diameter of the given sphere.

Therefore the octahedron has been comprehended in the given sphere, and it has been demonstrated at the same time that the square on the diameter of the sphere is double of the square on the side of the octahedron. Q. E. D.

- + PROPOSITION 15.

To construct a cube and comprehend it in a sphere, like the pyramid; and to prove that the square on the diameter of the sphere is triple of the square on the side of the cube. @@ -12016,7 +10442,7 @@ and, componendo, the squares on AB<

It is then required to comprehend it in the given sphere, and to prove that the square on the diameter of the sphere is triple of the square on the side of the cube.

For let KG, EG be joined.

Then, since the angle KEG is right, because KE is also at right angles to the plane EG and of course to the straight line EG also, [XI. Def. 3] therefore the semicircle described on KG will also pass through the point E.

-

Again, since GF is at right angles to each of the straight lines FL, FE, GF is also at right angles to the plane FK; hence also, if we join FK, GF will be at right angles to FK; and for this reason again the semicircle described on GK will also pass through F.

+

Again, since GF is at right angles to each of the straight lines FL, FE, GF is also at right angles to the plane FK; hence also, if we join FK, GF will be at right angles to FK; and for this reason again the semicircle described on GK will also pass through F.

Similarly it will also pass through the remaining angular points of the cube.

If then, KG remaining fixed, the semicircle be carried round and restored to the same position from which it began to be moved, the cube will be comprehended in a sphere.

I say next that it is also comprehended in the given sphere.

@@ -12028,12 +10454,12 @@ and, componendo, the squares on AB<

And AB is the diameter of the given sphere; therefore KG is also equal to the diameter of the given sphere.

Therefore the cube has been comprehended in the given sphere; and it has been demonstrated at the same time that the square on the diameter of the sphere is triple of the square on the side of the cube. Q. E. D.

 - + PROPOSITION 16.

To construct an icosahedron and comprehend it in a sphere, like the aforesaid figures; and to prove that the side of the icosahedron is the irrational straight line called minor.

-

Let the diameter AB of the given sphere be set out, and let it be cut at C so that AC is quadruple of CB, let the semicircle ADB be described on AB, let the straight line CD be drawn from C at right angles to AB, and let DB be joined;

let the circle EFGHK be set out and let its radius be equal to DB, let the equilateral and equiangular pentagon EFGHK be inscribed in the circle EFGHK, let the circumferences EF, FG, GH, HK, KE be bisected at the points L, M, N, O, P, and let LM, MN, NO, OP, PL, EP be joined.

+

Let the diameter AB of the given sphere be set out, and let it be cut at C so that AC is quadruple of CB, let the semicircle ADB be described on AB, let the straight line CD be drawn from C at right angles to AB, and let DB be joined;

let the circle EFGHK be set out and let its radius be equal to DB, let the equilateral and equiangular pentagon EFGHK be inscribed in the circle EFGHK, let the circumferences EF, FG, GH, HK, KE be bisected at the points L, M, N, O, P, and let LM, MN, NO, OP, PL, EP be joined.

Therefore the pentagon LMNOP is also equilateral, and the straight line EP belongs to a decagon.

Now from the points E, F, G, H, K let the straight lines EQ, FR, GS, HT, KU be set up at right angles to the plane of the circle, and let them be equal to the radius of the circle EFGHK, let QR, RS, ST, TU, UQ, QL, LR, RM, MS, SN, NT, TO, OU, UP, PQ be joined.

Now, since each of the straight lines EQ, KU is at right angles to the same plane, therefore EQ is parallel to KU. [XI. 6]

@@ -12044,7 +10470,7 @@ and, componendo, the squares on AB<

And, since QE belongs to a hexagon, and EP to a decagon, and the angle QEP is right, therefore QP belongs to a pentagon; for the square on the side of the pentagon is equal to the square on the side of the hexagon and the square on the side of the decagon inscribed in the same circle. [XIII. 10]

For the same reason PU is also a side of a pentagon.

But QU also belongs to a pentagon; therefore the triangle QPU is equilateral.

-

For the same reason each of the triangles QLR, RMS, SNT, TOU is also equilateral.

+

For the same reason each of the triangles QLR, RMS, SNT, TOU is also equilateral.

And, since each of the straight lines QL, QP was proved to belong to a pentagon, and LP also belongs to a pentagon, therefore the triangle QLP is equilateral.

For the same reason each of the triangles LRM, MSN, NTO, OUP is also equilateral.

Let the centre of the circle EFGHK the point V, be taken; from V let VZ be set up at right angles to the plane of the circle, let it be produced in the other direction, as VX, let there be cut off VW, the side of a hexagon, and each of the straight lines VX, WZ, being sides of a decagon, and let QZ, QW, UZ, EV, LV, LX, XM be joined.

@@ -12054,7 +10480,7 @@ and, componendo, the squares on AB<

And, since QW belongs to a hexagon, and WZ to a decagon, and the angle QWZ is right, therefore QZ belongs to a pentagon. [XIII. 10]

For the same reason UZ also belongs to a pentagon, inasmuch as, if we join VK, WU, they will be equal and opposite, and VK, being a radius, belongs to a hexagon; [IV. 15, Por.] therefore WU also belongs to a hexagon.

But WZ belongs to a decagon, and the angle UWZ is right; therefore UZ belongs to a pentagon. [XIII. 10]

-

But QU also belongs to a pentagon; therefore the triangle QUZ is equilateral.

+

But QU also belongs to a pentagon; therefore the triangle QUZ is equilateral.

For the same reason each of the remaining triangles of which the straight lines QR, RS, ST, TU are the bases, and the point Z the vertex, is also equilateral.

Again, since VL belongs to a hexagon, and VX to a decagon, and the angle LVX is right, therefore LX belongs to a pentagon. [XIII. 10]

For the same reason, if we join MV, which belongs to a hexagon, MX is also inferred to belong to a pentagon.

@@ -12065,7 +10491,7 @@ and, componendo, the squares on AB<

For, since VW belongs to a hexagon, and WZ to a decagon, therefore VZ has been cut in extreme and mean ratio at W, and VW is its greater segment; [XIII. 9] therefore, as ZV is to VW, so is VW to WZ.

But VW is equal to VE, and WZ to VX; therefore, as ZV is to VE, so is EV to VX.

And the angles ZVE, EVX are right; therefore, if we join the straight line EZ, the angle XEZ will be right because of the similarity of the triangles XEZ, VEZ.

-

For the same reason, since, as ZV is to VW, so is VW to WZ, and ZV is equal to XW, and VW to WQ, therefore, as XW is to WQ, so is QW to WZ.

+

For the same reason, since, as ZV is to VW, so is VW to WZ, and ZV is equal to XW, and VW to WQ, therefore, as XW is to WQ, so is QW to WZ.

And for this reason again, if we join QX, the angle at Q will be right; [VI. 8] therefore the semicircle described on XZ will also pass through Q. [III. 31]

And if, XZ remaining fixed, the semicircle be carried round and restored to the same position from which it began to be moved, it will also pass through Q and the remaining angular points of the icosahedron, and the icosahedron will have been comprehended in a sphere.

I say next that it is also comprehended in the given sphere.

@@ -12077,7 +10503,7 @@ and, componendo, the squares on AB<

But the square on ZX was also proved to be five times the square on VW.

And DB is equal to VW, for each of them is equal to the radius of the circle EFGHK; therefore AB is also equal to XZ.

And AB is the diameter of the given sphere; therefore XZ is also equal to the diameter of the given sphere.

-

Therefore the icosahedron has been comprehended in the given sphere

+

Therefore the icosahedron has been comprehended in the given sphere

I say next that the side of the icosahedron is the irrational straight line called minor.

For, since the diameter of the sphere is rational, and the square on it is five times the square on the radius of the circle EFGHK, therefore the radius of the circle EFGHK is also rational; hence its diameter is also rational.

But, if an equilateral pentagon be inscribed in a circle which has its diameter rational, the side of the pentagon is the irrational straight line called minor. [XIII. 11]

@@ -12088,7 +10514,7 @@ and, componendo, the squares on AB<

From this it is manifest that the square on the diameter of the sphere is five times the square on the radius of the circle from which the icosahedron has been described, and that the diameter of the sphere is composed of the side of the hexagon and two of the sides of the decagon inscribed in the same circle. Q. E. D.

- + PROPOSITION 17.

To construct a dodecahedron and comprehend it in a sphere, like the aforesaid figures, and to prove that the side of the dodecahedron is the irrational straight line called apotome. @@ -12096,7 +10522,7 @@ and, componendo, the squares on AB<

Let ABCD, CBEF, two planes of the aforesaid cube at right angles to one another, be set out, let the sides AB, BC, CD, DA, EF, EB, FC be bisected at G, H, K, L, M, N, O respectively, let GK, HL, MH, NO be joined, let the straight lines NP, PO, HQ be cut in extreme and mean ratio at the points R, S, T respectively, and let RP, PS, TQ be their greater segments; from the points R, S, T let RU, SV, TW be set up at right angles to the planes of the cube towards the outside of the cube, let them be made equal to RP, PS, TQ, and let UB, BW, WC, CV, VU be joined.

I say that the pentagon UBWCV is equilateral, and in one plane, and is further equiangular.

-

For let RB, SB, VB be joined.

+

For let RB, SB, VB be joined.

Then, since the straight line NP has been cut in extreme and mean ratio at R, and RP is the greater segment, therefore the squares on PN, NR are triple of the square on RP. [XIII. 4]

But PN is equal to NB, and PR to RU; therefore the squares on BN, NR are triple of the square on RU.

But the square on BR is equal to the squares on BN, NR; [I. 47] therefore the square on BR is triple of the square on RU; hence the squares on BR, RU are quadruple of the square on RU.

@@ -12108,7 +10534,7 @@ and, componendo, the squares on AB<

For let PX be drawn from P parallel to each of the straight lines RU, SV and towards the outside of the cube, and let XH, HW be joined; I say that XHW is a straight line.

For, since HQ has been cut in extreme and mean ratio at T, and QT is its greater segment, therefore, as HQ is to QT, so is QT to TH.

But HQ is equal to HP, and QT to each of the straight lines TW, PX; therefore, as HP is to PX, so is WT to TH.

-

And HP is parallel to TW, for each of them is at right angles to the plane BD; [XI. 6] and TH is parallel to PX, for each of them is at right angles to the plane BF. [id.]

+

And HP is parallel to TW, for each of them is at right angles to the plane BD; [XI. 6] and TH is parallel to PX, for each of them is at right angles to the plane BF. [id.]

But if two triangles, as XPH, HTW, which have two sides proportional to two sides be placed together at one angle so that their corresponding sides are also parallel, the remaining straight lines will be in a straight line; [VI. 32] therefore XH is in a straight line with HW.

But every straight line is in one plane; [XI. 1] therefore the pentagon UBWCV is in one plane.

I say next that it is also equiangular.

@@ -12117,7 +10543,7 @@ and, componendo, the squares on AB<

But the square on SB is equal to the squares on SN, NB; therefore the squares on BS, SV, that is, the square on BV —for the angle VSB is right—is quadruple of the square on NB; therefore VB is double of BN.

But BC is also double of BN; therefore BV is equal to BC.

And, since the two sides BU, UV are equal to the two sides BW, WC, and the base BV is equal to the base BC, therefore the angle BUV is equal to the angle BWC. [I. 8]

-

Similarly we can prove that the angle UVC is also equal to the angle BWC; therefore the three angles BWC, BUV, UVC are equal to one another.

+

Similarly we can prove that the angle UVC is also equal to the angle BWC; therefore the three angles BWC, BUV, UVC are equal to one another.

But if in an equilateral pentagon three angles are equal to one another, the pentagon will be equiangular, [XIII. 7] therefore the pentagon BUVCW is equiangular.

And it was also proved equilateral; therefore the pentagon BUVCW is equilateral and equiangular, and it is on one side BC of the cube.

Therefore, if we make the same construction in the case of each of the twelve sides of the cube, a solid figure will have been constructed which is contained by twelve equilateral and equiangular pentagons, and which is called a dodecahedron.

@@ -12128,7 +10554,7 @@ and, componendo, the squares on AB<

Now, since the straight line NS has been cut in extreme and mean ratio at P, and NP is its greater segment, therefore the squares on NS, SP are triple of the square on NP. [XIII. 4]

But NS is equal to XZ, inasmuch as NP is also equal to PZ, and XP to PS.

But further PS is also equal to XU, since it is also equal to RP; therefore the squares on ZX, XU are triple of the square on NP.

-

But the square on UZ is equal to the squares on ZX, XU; therefore the square on UZ is triple of the square on NP.

+

But the square on UZ is equal to the squares on ZX, XU; therefore the square on UZ is triple of the square on NP.

But the square on the radius of the sphere which comprehends the cube is also triple of the square on the half of the side of the cube, for it has previously been shown how to construct a cube and comprehend it in a sphere, and to prove that the square on the diameter of the sphere is triple of the square on the side of the cube. [XIII. 15]

But, if whole is so related to whole, so is half to half also; and NP is half of the side of the cube; therefore UZ is equal to the radius of the sphere which comprehends the cube.

And Z is the centre of the sphere which comprehends the cube; therefore the point U is on the surface of the sphere.

@@ -12137,7 +10563,7 @@ and, componendo, the squares on AB<

For since, when NP has been cut in extreme and mean ratio, RP is the greater segment, and, when PO has been cut in extreme and mean ratio, PS is the greater segment, therefore, when the whole NO is cut in extreme and mean ratio, RS is the greater segment.

[Thus, since, as NP is to PR, so is PR to RN, the same is true of the doubles also, for parts have the same ratio as their equimultiples; [V. 15] therefore as NO is to RS, so is RS to the sum of NR, SO.

But NO is greater than RS; therefore RS is also greater than the sum of NR, SO; therefore NO has been cut in extreme and mean ratio, and RS is its greater segment.]

-

But RS is equal to UV; therefore, when NO is cut in extreme and mean ratio, UV is the greater segment.

+

But RS is equal to UV; therefore, when NO is cut in extreme and mean ratio, UV is the greater segment.

And, since the diameter of the sphere is rational, and the square on it is triple of the square on the side of the cube, therefore NO, being a side of the cube, is rational.

[But if a rational line be cut in extreme and mean ratio, each of the segments is an irrational apotome.]

Therefore UV, being a side of the dodecahedron, is an irrational apotome. [XIII. 6]

@@ -12154,7 +10580,7 @@ and, componendo, the squares on AB<

Let AB, the diameter of the given sphere, be set out, and let it be cut at C so that AC is equal to CB, and at D so that AD is double of DB; let the semicircle AEB be described on AB, from C, D let CE, DF be drawn at right angles to AB, and let AF, FB, EB be joined.

Then, since AD is double of DB, therefore AB is triple of BD.

-

Convertendo, therefore, BA is one and a half times AD.

+

Convertendo, therefore, BA is one and a half times AD.

But, as BA is to AD, so is the square on BA to the square on AF, [V. Def. 9, VI. 8] for the triangle AFB is equiangular with the triangle AFD; therefore the square on BA is one and a half times the square on AF.

But the square on the diameter of the sphere is also one and a half times the square on the side of the pyramid. [XIII. 13]

And AB is the diameter of the sphere; therefore AF is equal to the side of the pyramid.

@@ -12167,7 +10593,7 @@ and, componendo, the squares on AB<

But the square on the diameter of the sphere is also double of the square on the side of the octahedron. [XIII. 14]

And AB is the diameter of the given sphere; therefore BE is the side of the octahedron.

Next, let AG be drawn from the point A at right angles to the straight line AB, let AG be made equal to AB, let GC be joined, and from H let HK be drawn perpendicular to AB.

-

Then, since GA is double of AC, for GA is equal to AB, and, as GA is to AC, so is HK to KC, therefore HK is also double of KC.

+

Then, since GA is double of AC, for GA is equal to AB, and, as GA is to AC, so is HK to KC, therefore HK is also double of KC.

Therefore the square on HK is quadruple of the square on KC; therefore the squares on HK, KC, that is, the square on HC, is five times the square on KC.

But HC is equal to CB; therefore the square on BC is five times the square on CK.

And, since AB is double of CB, and, in them, AD is double of DB, therefore the remainder BD is double of the remainder DC.

@@ -12177,7 +10603,7 @@ and, componendo, the squares on AB<

Now, since the square on BC is five times the square on CK, and AB is double of BC, and KL double of CK, therefore the square on AB is five times the square on KL.

But the square on the diameter of the sphere is also five times the square on the radius of the circle from which the icosahedron has been described. [XIII. 16, Por.]

And AB is the diameter of the sphere; therefore KL is the radius of the circle from which the icosahedron has been described; therefore KL is a side of the hexagon in the said circle. [IV. 15, Por.]

-

And, since the diameter of the sphere is made up of the side of the hexagon and two of the sides of the decagon inscribed in the same circle, [XIII. 16, Por.] and AB is the diameter of the sphere, while KL is a side of the hexagon, and AK is equal to LB, therefore each of the straight lines AK, LB is a side of the decagon inscribed in the circle from which the icosahedron has been described.

+

And, since the diameter of the sphere is made up of the side of the hexagon and two of the sides of the decagon inscribed in the same circle, [XIII. 16, Por.] and AB is the diameter of the sphere, while KL is a side of the hexagon, and AK is equal to LB, therefore each of the straight lines AK, LB is a side of the decagon inscribed in the circle from which the icosahedron has been described.

And, since LB belongs to a decagon, and ML to a hexagon, for ML is equal to KL, since it is also equal to HK, being the same distance from the centre, and each of the straight lines HK, KL is double of KC, therefore MB belongs to a pentagon. [XIII. 10]

But the side of the pentagon is the side of the icosahedron; [XIII. 16] therefore MB belongs to the icosahedron.

Now, since FB is a side of the cube, let it be cut in extreme and mean ratio at N, and let NB be the greater segment; therefore NB is a side of the dodecahedron. [XIII. 17, Por.]

@@ -12186,7 +10612,7 @@ and, componendo, the squares on AB<

The said sides, therefore, of the three figures, I mean the pyramid, the octahedron and the cube, are to one another in rational ratios.

But the remaining two, I mean the side of the icosahedron and the side of the dodecahedron, are not in rational ratios either to one another or to the aforesaid sides; for they are irrational, the one being minor [XIII. 16] and the other an apotome [XIII. 17].

That the side MB of the icosahedron is greater than the side NB of the dodecahedron we can prove thus.

-

For, since the triangle FDB is equiangular with the triangle FAB, [VI. 8] proportionally, as DB is to BF, so is BF to BA. [VI. 4]

+

For, since the triangle FDB is equiangular with the triangle FAB, [VI. 8] proportionally, as DB is to BF, so is BF to BA. [VI. 4]

And, since the three straight lines are proportional, as the first is to the third, so is the square on the first to the square on the second; [V. Def. 9, VI. 20, Por.] therefore, as DB is to BA, so is the square on DB to the square on BF; therefore, inversely, as AB is to BD, so is the square on FB to the square on BD.

But AB is triple of BD; therefore the square on FB is triple of the square on BD.

But the square on AD is also quadruple of the square on DB, for AD is double of DB; therefore the square on AD is greater than the square on FB; therefore AD is greater than FB; therefore AL is by far greater than FB.

@@ -12196,7 +10622,7 @@ and, componendo, the squares on AB<

I say next that no other figure, besides the said five figures, can be constructed which is contained by equilateral and equiangular figures equal to one another.

For a solid angle cannot be constructed with two triangles, or indeed planes.

-

With three triangles the angle of the pyramid is constructed, with four the angle of the octahedron, and with five the angle of the icosahedron; but a solid angle cannot be formed by six equilateral and equiangular triangles placed together at one point, for, the angle of the equilateral triangle being two-thirds of a right angle, the six will be equal to four right angles: which is impossible, for any solid angle is contained by angles less than four right angles. [XI. 21]

+

With three triangles the angle of the pyramid is constructed, with four the angle of the octahedron, and with five the angle of the icosahedron; but a solid angle cannot be formed by six equilateral and equiangular triangles placed together at one point, for, the angle of the equilateral triangle being two-thirds of a right angle, the six will be equal to four right angles: which is impossible, for any solid angle is contained by angles less than four right angles. [XI. 21]

For the same reason, neither can a solid angle be constructed by more than six plane angles.

By three squares the angle of the cube is contained, but by four it is impossible for a solid angle to be contained, for they will again be four right angles.

By three equilateral and equiangular pentagons the angle of the dodecahedron is contained; but by four such it is impossible for any solid angle to be contained, for, the angle of the equilateral pentagon being a right angle and a fifth, the four angles will be greater than four right angles: which is impossible.

@@ -12208,15 +10634,13 @@ and, componendo, the squares on AB<

Let ABCDE be an equilateral and equiangular pentagon, let the circle ABCDE be circumscribed about it, let its centre F be taken, and let FA, FB, FC, FD, FE be joined.

Therefore they bisect the angles of the pentagon at A, B, C, D, E.

-

And, since the angles at F are equal to four right angles and are equal, therefore one of them, as the angle AFB, is one right angle less a fifth; therefore the remaining angles FAB, ABF consist of one right angle and a fifth.

+

And, since the angles at F are equal to four right angles and are equal, therefore one of them, as the angle AFB, is one right angle less a fifth; therefore the remaining angles FAB, ABF consist of one right angle and a fifth.

But the angle FAB is equal to the angle FBC; therefore the whole angle ABC of the pentagon consists of one right angle and a fifth. Q. E. D.

- +
+
- - - - + diff --git a/data/tlg1799/tlg001/tlg1799.tlg001.perseus-eng2.xml b/data/tlg1799/tlg001/tlg1799.tlg001.perseus-eng2.xml new file mode 100644 index 000000000..d4ef73f4e --- /dev/null +++ b/data/tlg1799/tlg001/tlg1799.tlg001.perseus-eng2.xml @@ -0,0 +1,10456 @@ + + + + + + + The Thirteen Books of Euclid's Elements + Euclid + Thomas Little Heath + Perseus Project, Tufts University + Gregory Crane + + Prepared under the supervision of + Lisa Cerrato + David Smith + + The National Science Foundation + + + Trustees of Tufts University + Medford, MA + Perseus Digital Library Project + Perseus 3.0 + tlg1799.tlg001.perseus-eng2.xml + + Available under a Creative Commons Attribution-ShareAlike 4.0 International License + + + + + + + + Euclid + The Thirteen Books of Euclid's Elements + Thomas Little Heath + + Cambridge, UK + Cambridge University Press + 1908 + + 1-3 + +HathiTrust +HathiTrust + HathiTrust + + + + + + + +

This pointer pattern extracts book and type and number.

+ + +

This pointer pattern extracts book and type.

+ + +

This pointer pattern extracts book.

+ + + + + + + + + English + Greek + Latin + + + + Epidoc and CTS conversion and header review for best practices + + + + + + + +
+
+ + BOOK I. +
+ DEFINITIONS. +
+ 1 +

A point is that which has no part.

+
+
+ 2 +

A line is breadthless length.

+
+
+ 3 +

The extremities of a line are points.

+
+
+ 4 +

A straight line is a line which lies evenly with the points on itself.

+
+
+ 5 +

A surface is that which has length and breadth only.

+
+
+ 6 +

The extremities of a surface are lines.

+
+
+ 7 +

A plane surface is a surface which lies evenly with the straight lines on itself.

+
+
+ 8 +

A plane angle is the inclination to one another of two lines in a plane which meet one another and do not lie in a straight line.

+
+
+ 9 +

And when the lines containing the angle are straight, the angle is called rectilineal.

+
+
+ 10 +

When a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right, and the straight line standing on the other is called a perpendicular to that on which it stands.

+
+
+ 11 +

An obtuse angle is an angle greater than a right angle.

+
+
+ 12 +

An acute angle is an angle less than a right angle.

+
+
+ 13 +

A boundary is that which is an extremity of anything.

+
+
+ 14 +

A figure is that which is contained by any boundary or boundaries.

+
+
+ 15 +

A circle is a plane figure contained by one line such that all the straight lines falling upon it from one point among those lying within the figure are equal to one another;

+
+
+ 16 + +

And the point is called the centre of the circle.

+
+
+ 17 +

A diameter of the circle is any straight line drawn through the centre and terminated in both directions by the circumference of the circle, and such a straight line also bisects the circle.

+
+
+ 18 +

A semicircle is the figure contained by the diameter and the circumference cut off by it. And the centre of the semicircle is the same as that of the circle.

+
+
+ 19 +

Rectilineal figures are those which are contained by straight lines, trilateral figures being those contained by three, quadrilateral those contained by four, and multilateral those contained by more than four straight lines.

+
+
+ 20 +

Of trilateral figures, an equilateral triangle is that which has its three sides equal, an isosceles triangle that which has two of its sides alone equal, and a scalene triangle that which has its three sides unequal.

+
+
+ 21 +

Further, of trilateral figures, a right-angled triangle is that which has a right angle, an obtuse-angled triangle that which has an obtuse angle, and an acuteangled triangle that which has its three angles acute.

+
+
+ 22 +

Of quadrilateral figures, a square is that which is both equilateral and right-angled; an oblong that which is right-angled but not equilateral; a rhombus that which is equilateral but not right-angled; and a rhomboid that which has its opposite sides and angles equal to one another but is neither equilateral nor right-angled. And let quadrilaterals other than these be called trapezia.

+
+
+ 23 +

Parallel straight lines are straight lines which, being in the same plane and being produced indefinitely in both directions, do not meet one another in either direction.

+
+
+
+ POSTULATES. +
+ 1 +

Let the following be postulated:

+

To draw a straight line from any point to any point.

+
+
+ 2 +

To produce a finite straight line continuously in a straight line.

+
+
+ 3 +

To describe a circle with any centre and distance.

+
+
+ 4 +

That all right angles are equal to one another.

+
+
+ 5 + +

That, if a straight line falling on two straight lines make the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles.

+
+
+
+ COMMON NOTIONS. +
+ 1 +

Things which are equal to the same thing are also equal to one another.

+
+
+ 2 +

If equals be added to equals, the wholes are equal.

+
+
+ 3 +

If equals be subtracted from equals, the remainders are equal.

+
+
+ 4 +

[7] Things which coincide with one another are equal to one another.

+
+
+ 5 +

[8] The whole is greater than the part.

+
+
+
+ + BOOK I. PROPOSITIONS. +
+ Proposition 1. +

On a given finite straight line to construct an equilateral triangle.

+ +

Let AB be the given finite straight line.

+

Thus it is required to construct an equilateral triangle on the straight line AB.

+

With centre A and distance AB let the circle BCD be described; [Post. 3] again, with centre B and distance BA let the circle ACE be described; [Post. 3] and from the point C, in which the circles cut one another, to the points A, B let the straight lines CA, CB be joined. [Post. 1]

+

Now, since the point A is the centre of the circle CDB, AC is equal to AB. [Def. 15]

+

Again, since the point B is the centre of the circle CAE, BC is equal to BA. [Def. 15]

+

But CA was also proved equal to AB; therefore each of the straight lines CA, CB is equal to AB.

+

And things which are equal to the same thing are also equal to one another; [C.N. 1] therefore CA is also equal to CB.

+

Therefore the three straight lines CA, AB, BC are equal to one another.

+

Therefore the triangle ABC is equilateral; and it has been constructed on the given finite straight line AB.

+ +

(Being) what it was required to do.

+ + +

The Greek usage differs from ours in that the definite article is employed in such a phrase as this where we have the indefinite. ἐπὶ τῆς δοθείσης εὐθείας πεπερασμένης, on the given finite straight line, + i.e. the finite straight line which we choose to take.

+ + +

To be strictly literal we should have to translate in the reverse order let the given finite straight line be the (straight line) AB +; but this order is inconvenient in other cases where there is more than one datum, e.g. in the setting-out of I. 2, let the given point be A, and the given straight line BC, + the awkwardness arising from the omission of the verb in the second clause. Hence I have, for clearness' sake, adopted the other order throughout the book.

+ + +

Two things are here to be noted, (1) the elegant and practically universal use of the perfect passive imperative in constructions, γεγράφθω meaning of course let it have been described + or suppose it described, + (2) the impossibility of expressing shortly in a translation the force of the words in their original order. κύκλος γεγράφθω ὸ ΒΓΔ means literally let a circle have been described, the (circle, namely, which I denote by) BCD. + Similarly we have lower down let straight lines, (namely) the (straight lines) CA, CB, be joined, + ἐπεζεύχθωσαν εὐθεῖαι αί ΓΑ, ΓΒ. There seems to be no practicable alternative, in English, but to translate as I have done in the text.

+ + +

Euclid is careful to adhere to the phraseology of Postulate 1 except that he speaks of joining + (ἐπεζεύχθωσαν) instead of drawing + (γράφειν). He does not allow himself to use the shortened expression let the straight line FC be joined + (without mention of the points F, C) until I. 5.

+ + +

, ἑκατέρα τῶν ΓΑ, ΓΒ and 24. the three straight lines CA, AB, BC, αἱ τρεῖς αἱ ΓΑ, ΑΒ, ΒΓ. I have, here and in all similar expressions, inserted the words straight lines + which are not in the Greek. The possession of the inflected definite article enables the Greek to omit the words, but this is not possible in English, and it would scarcely be English to write each of CA, CB + or the three CA, AB, BC. +

+ +
+
+ + Proposition 2. +

To place at a given point (as an extremity) a straight line equal to a given straight line.

+ +

Let A be the given point, and BC the given straight line.

+

Thus it is required to place at the point A (as an extremity) a straight line equal to the given straight line BC.

+

From the point A to the point B let the straight line AB be joined; [Post. 1] and on it let the equilateral triangle DAB be constructed. [I. 1]

+

Let the straight lines AE, BF be produced in a straight line with DA, DB; [Post. 2] with centre B and distance BC let the circle CGH be described; [Post. 3] and again, with centre D and distance DG let the circle GKL be described. [Post. 3]

+

Then, since the point B is the centre of the circle CGH, BC is equal to BG. +

+

Again, since the point D is the centre of the circle GKL, DL is equal to DG.

+

And in these DA is equal to DB; therefore the remainder AL is equal to the remainder BG. [C.N. 3] +

+

But BC was also proved equal to BG; therefore each of the straight lines AL, BC is equal to BG.

+

And things which are equal to the same thing are also equal to one another; [C.N. 1] therefore AL is also equal to BC.

+

Therefore at the given point A the straight line AL is placed equal to the given straight line BC.

+ +

(Being) what it was required to do.

+ + +

I have inserted these words because to place a straight line at a given point + (πρὸς τῷ δοθέντι σημείῳ) is not quite clear enough, at least in English.

+ + +

It will be observed that in this first application of Postulate 2, and again in I. 5, Euclid speaks of the continuation of the straight line as that which is produced in such cases, ἐκβεβλήσθωσαν and προσεκβεβλήσθωσαν meaning little more than drawing straight lines in a straight line with + the given straight lines. The first place in which Euclid uses phraseology exactly corresponding to ours when speaking of a straight line being produced is in I. 16: let one side of it, BC, be produced to D + (προσεκβεβλήσθω αὐτοῦ μία πλευρὰ ἡ ΒΓ ἐπὶ τὸ Δ).

+ + +

The Greek expressions are λοιπὴ ἡ ΑΛ and λοιπῇ τῇ BH, and the literal translation would be AL (or BG) remaining, + but the shade of meaning conveyed by the position of the definite article can hardly be expressed in English.

+ +
+
+ Proposition 3. +

Given two unequal straight lines, to cut off from the greater a straight line equal to the less.

+ +

Let AB, C be the-two given unequal straight lines, and let AB be the greater of them.

+

Thus it is required to cut off from AB the greater a straight line equal to C the less.

+

At the point A let AD be placed equal to the straight line C; [I. 2] and with centre A and distance AD let the circle DEF be described. [Post. 3] Now, since the point A is the centre of the circle DEF, AE is equal to AD. [Def. 15] But C is also equal to AD. Therefore each of the straight lines AE, C is equal to AD; so that AE is also equal to C. [C.N. 1]

+

Therefore, given the two straight lines AB, C, from AB the greater AE has been cut off equal to C the less.

+ +

(Being) what it was required to do.

+ +
+
+ Proposition 4. +

If two triangles have the two sides equal to two sides respectively, and have the angles contained by the equal straight lines equal, they will also have the base equal to the base, the triangle will be equal to the triangle, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend.

+ +

Let ABC, DEF be two triangles having the two sides AB, AC equal to the two sides DE, DF respectively, namely AB to DE and AC to DF, and the angle BAC equal to the angle EDF.

+

I say that the base BC is also equal to the base EF, the triangle ABC will be equal to the triangle DEF, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend, that is, the angle ABC to the angle DEF, and the angle ACB to the angle DFE.

+

For, if the triangle ABC be applied to the triangle DEF, and if the point A be placed on the point D and the straight line AB on DE, then the point B will also coincide with E, because AB is equal to DE.

+

+

Again, AB coinciding with DE, the straight line AC will also coincide with DF, because the angle BAC is equal to the angle EDF; hence the point C will also coincide with the point F, because AC is again equal to DF. +

+

But B also coincided with E; hence the base BC will coincide with the base EF.

+

[For if, when B coincides with E and C with F, the base BC does not coincide with the base EF, two straight lines will enclose a space: which is impossible. Therefore the base BC will coincide with EF] and will be equal to it. [C.N. 4]

+

Thus the whole triangle ABC will coincide with the whole triangle DEF, and will be equal to it. +

+

And the remaining angles will also coincide with the remaining angles and will be equal to them, the angle ABC to the angle DEF, and the angle ACB to the angle DFE.

+

Therefore etc.

+ +

(Being) what it was required to prove.

+ + +

It is a fact that Euclid's enunciations not infrequently leave something to be desired in point of clearness and precision. Here he speaks of the triangles having the angle equal to the angle, namely the angle contained by the equal straight lines + (τὴν γωνίαν τῇ γωνίᾳ ἴσην ἔχῃ τὴν ὑπὸ τῶν ἴσων εὐθειῶν περιεχομένην), only one of the two angles being described in the latter expression (in the accusative), and a similar expression in the dative being left to be understood of the other angle. It is curious too that, after mentioning two sides, + he speaks of the angles contained by the equal straight lines, + not sides. + It may be that he wished to adhere scrupulously, at the outset, to the phraseology of the definitions, where the angle is the inclination to one another of two lines or straight lines. Similarly in the enunciation of I. 5 he speaks of producing the equal straight lines + as if to keep strictly to the wording of Postulate 2.

+ + +

I agree with Mr H. M. Taylor (Euclid, p. ix) that it is best to abandon the traditional translation of each to each, + which would naturally seem to imply that all the four magnitudes are equal rather than (as the Greek ὲκατέρα ὲκατέρᾳ does) that one is equal to one and the other to the other.

+ + +

Here we have the word base used for the first time in the Elements. Proclus explains it (p. 236, 12-15) as meaning (1), when no side of a triangle has been mentioned before, the side which is on a level with the sight + (τὴν πρὸς τῇ ὄψει κειμένην), and (2), when two sides have already been mentioned, the third side. Proclus thus avoids the mistake made by some modern editors who explain the term exclusively with reference to the case where two sides have been mentioned before. That this is an error is proved (1) by the occurrence of the term in the enunciations of I. 37 etc. about triangles on the same base and equal bases, (2) by the application of the same term to the bases of parallelograms in I. 35 etc. The truth is that the use of the term must have been suggested by the practice of drawing the particular side horizontally, as it were, and the rest of the figure above it. The base of a figure was therefore spoken of, primarily, in the same sense as the base of anything else, e.g. of a pedestal or column; but when, as in I. 5, two triangles were compared occupying other than the normal positions which gave rise to the name, and when two sides had been previously mentioned, the base was, as Proclus says, necessarily the third side.

+ + +

ὑποτείνειν ὑπό, to stretch under, + with accusative.

+ + +

The full Greek expression would be ἡ ὑπὸ τῶν ΒΑ, ΑΓ περιεχομένη γωνία, the angle contained by the (straight lines) BA, AC. + But it was a common practice of Greek geometers, e.g. of Archimedes and Apollonius (and Euclid too in Books X.—XIII.), to use the abbreviation αἱ ΒΑΓ for αἱ ΒΑ, ΑΓ, the (straight lines) BA, AC. + Thus, on περιεχομένη being dropped, the expression would become first ἡ ὑπὸ τῶν ΒΑΓ γωνία, then ἡ ὑπὸ ΒΑΓ γωνία, and finally ἡ ὑπὸ ΒΑΓ, without γωνία, as we regularly find it in Euclid.

+ + +

The difference between the technical use of the passive ἐφαρμόζεσθαι to be applied (to), + and of the active ἐφαρμόζειν to coincide (with) + has been noticed above (note on Common Notion 4, pp. 224-5).

+ + +

Heiberg (Paralipomena su Euklid in Hermes, XXXVIII., 1903, p. 56) has pointed out, as a conclusive reason for regarding these words as an early interpolation, that the text of an-Nairīzī (Codex Leidensis 399, 1, ed. Besthorn-Heiberg, p. 55) does not give the words in this place but after the conclusion Q.E.D., which shows that they constitute a scholium only. They were doubtless added by some commentator who thought it necessary to explain the immediate inference that, since B coincides with E and C with F, the straight line BC coincides with the straight line EF, an inference which really follows from the definition of a straight line and Post. 1; and no doubt the Postulate that Two straight lines cannot enclose a space + (afterwards placed among the Common Notions) was interpolated at the same time.

+ + +

Where (as here) Euclid's conclusion merely repeats the enunciation word for word, I shall avoid the repetition and write Therefore etc. + simply.

+ +
+
+ + Proposition 5. +

In isosceles triangles the angles at the base are equal to one another, and, if the equal straight lines be produced further, the angles under the base will be equal to one another.

+ +

Let ABC be an isosceles triangle having the side AB + equal to the side AC; and let the straight lines BD, CE be produced further in a straight line with AB, AC. [Post. 2]

+

I say that the angle ABC is equal to the angle ACB, and the angle CBD to the angle BCE.

+

Let a point F be taken at random on BD;

from AE the greater let AG be cut off equal to AF the less; [I. 3] and let the straight lines FC, GB be joined. [Post. 1]

+

Then, since AF is equal to AG and AB to AC, the two sides FA, AC are equal to the two sides GA, AB, respectively; and they contain a common angle, the angle FAG. Therefore the base FC is equal to the base GB, and the triangle AFC is equal to the triangle AGB, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend, that is, the angle ACF to the angle ABG, and the angle AFC to the angle AGB. [I. 4]

+

And, since the whole AF is equal to the whole AG, and in these AB is equal to AC, the remainder BF is equal to the remainder CG.

+

But FC was also proved equal to GB; therefore the two sides BF, FC are equal to the two sides CG, GB respectively; and the angle BFC is equal to the angle CGB, while the base BC is common to them; therefore the triangle BFC is also equal to the triangle CGB, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend; therefore the angle FBC is equal to the angle GCB, and the angle BCF to the angle CBG. +

+

Accordingly, since the whole angle ABG was proved equal to the angle ACF, and in these the angle CBG is equal to the angle BCF, the remaining angle ABC is equal to the remaining angle ACB; and they are at the base of the triangle ABC. But the angle FBC was also proved equal to the angle GCB; and they are under the base.

+

Therefore etc.

+ +

Q. E. D.

+ + +

(meaning the equal sides). Cf. note on the similar expression in Prop. 4, lines 2, 3.

+ + +

, εἰλήφθω ἐπὶ τῆς ΒΔ τυχὸν σημεῖον τὸ Ζ, where τυχὸν σημεῖονmeans a chance point. +

+ + +

δύο αἱ ΖΑ, ΑΓ δυσὶ ταῖς ΗΑ, ΑΒ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ. Here, and in numberless later passages, I have inserted the word sides + for the reason given in the note on I. 1, line 20. It would have been permissible to supply either straight lines + or sides +; but on the whole sides + seems to be more in accordance with the phraseology of I. 4.

+ + +

i.e., apparently, common to the angles, as the αὐτῶν in βάσις αὐτῶν κοινὴ can only refer to γωνία and γωνίᾳ preceding. Simson wrote and the base BC is common to the two triangles BFC, CGB +; Todhunter left out these words as being of no use and tending to perplex a beginner. But Euclid evidently chose to quote the conclusion of I. 4 exactly; the first phrase of that conclusion is that the bases (of the two triangles) are equal, and, as the equal bases are here the same base, Euclid naturally substitutes the word common + for equal. +

+ + +

As (Being) what it was required to prove + (or do +) is somewhat long, I shall henceforth write the time-honoured Q. E. D. + and Q. E. F. + for ὅπερ ἔδει δεῖξαι and ὅπερ ἔδει ποιῆσαι.

+ +
+
+ Proposition 6. +

If in a triangle two angles be equal to one another, the sides which subtend the equal angles will also be equal to one another.

+ +

Let ABC be a triangle having the angle ABC equal to the angle ACB;

+

I say that the side AB is also equal to the side AC.

+

For, if AB is unequal to AC, one of them is greater.

+

Let AB be greater; and from AB the greater let DB be cut off equal to AC the less;

+

let DC be joined.

+

Then, since DB is equal to AC, and BC is common, the two sides DB, BC are equal to the two sides AC, CB respectively; + and the angle DBC is equal to the angle ACB; therefore the base DC is equal to the base AB, and the triangle DBC will be equal to the triangle ACB, the less to the greater: which is absurd. Therefore AB is not unequal to AC; it is therefore equal to it.

+

Therefore etc.

+ +

Q. E. D.

+ +
+
+ Proposition 7. +

Given two straight lines constructed on a straight line (from its extremities) and meeting in a point, there cannot be constructed on the same straight line (from its extremities), and on the same side of it, two other straight lines meeting in another point and equal to the former two respectively, namely each to that which has the same extremity with it.

+ +

For, if possible, given two straight lines AC, CB constructed on the straight line AB and meeting at the point C, let two other straight lines

+ AD, DB be constructed on the same straight line AB, on the same side of it, meeting in another point D and equal to the former two respectively, namely each to that which has the same extremity with it, so that CA is equal to DA which has the same extremity A with it, and CB to DB which has the same extremity B with it; and let CD be joined.

+

Then, since AC is equal to AD, the angle ACD is also equal to the angle ADC; [I. 5] therefore the angle ADC is greater than the angle DCB; therefore the angle CDB is much greater than the angle DCB.

+

Again, since CB is equal to DB, the angle CDB is also equal to the angle DCB. + But it was also proved much greater than it: which is impossible.

+

Therefore etc.

+ +

Q. E. D.

+ + +

In an English translation of the enunciation of this proposition it is absolutely necessary, in order to make it intelligible, to insert some words which are not in the Greek. The reason is partly that the Greek enunciation is itself very elliptical, and partly that some words used in it conveyed more meaning than the corresponding words in English do. Particularly is this the case with οὐ συσταθήσονται ἐπί there shall not be constructed upon, + since συνίστασθαι is the regular word for constructing a triangle in particular. Thus a Greek would easily understahd συσταθήσονται ἐπί as meaning the construction of two lines forming a triangle on a given straight line as base; whereas to construct two straight lines on a straight line + is not in English sufficiently definite unless we explain that they are drawn from the ends of the straight line to meet at a point. I have had the less hesitation in putting in the words from its extremities + because they are actually used by Euclid in the somewhat similar enunciation of I. 21.

+

How impossible a literal translation into English is, if it is to convey the meaning of the enunciation intelligibly, will be clear from the following attempt to render literally: On the same straight line there shall not be constructed two other straight lines equal, each to each, to the same two straight lines, (terminating) at different points on the same side, having the same extremities as the original straight lines + (ἐπὶ τῆς αὐτῆς εὐθείας δύο ταῖς αὐταῖς εὐθείαις ἄλλαι δύο εὐθεῖαι ἴσαι ἑκατὲρα ἑκατέρα οὐ συσταθήσονται πρὸς ἄλλῳ καἰ ἄλλῳ σημείῳ ἐπἰ τὰ αὐτὰ μέρη τὰ αὐτὰ πέρατα ἔχουσαι ταῖς ἐξ ἀρχῆς εὐθείαις).

+

The reason why Euclid allowed himself to use, in this enunciation, language apparently so obscure is no doubt that the phraseology was traditional and therefore, vague as it was, had a conventional meaning which the contemporary geometer well understood. This is proved, I think, by the occurrence in Aristotle (Meteorologica III. 5, 376 a 2 sqq.) of the very same, evidently technical, expressions. Aristotle is there alluding to the theorem given by Eutocius from Apollonius' Plane Loci to the effect that, if H, K be two fixed points and M such a variable point that the ratio of MH to MK is a given ratio (not one of equality), the locus of M is a circle. (For an account of this theorem see note on VI. 3 below.) Now Aristotle says The lines drawn up from H, K in this ratio cannot be constructed to two different points of the semicircle A + (αἰ οὖν ἀπὸ τῶν ΗΚ ἀναγόμεναι γραμμαὶ ἐν τούτῳ τῷ λόγῳ οὐ συσταθήσουται τοῦ ἐφ̓ ᾦ Α ἡμικυκλίου πρὸς ἄλλο καὶ ἄλλο σημεῖον).

+

If a paraphrase is allowed instead of a translation adhering as closely as possible to the original, Simson's is the best that could be found, since the fact that the straight lines form triangles on the same base is really conveyed in the Greek. Simson's enunciation is, Upon the same base, and on the same side of it, there cannot be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise those which are terminated at the other extremity. Th. Taylor (the translator of Proclus) attacks Simson's alteration as indiscreet + and as detracting from the beauty and accuracy of Euclid's enunciation which are enlarged upon by Proclus in his commentary. Yet, when Taylor says Whatever difficulty learners may find in conceiving this proposition abstractedly is easily removed by its exposition in the figure, + he really gives his case away. The fact is that Taylor, always enthusiastic over his author, was nettled by Simson's slighting remarks on Proclus' comments on the proposition. Simson had said, with reference to Proclus' explanation of the bearing of the second part of I. 5 on I. 7, that it was not worth while to relate his trifles at full length, + to which Taylor retorts But Mr Simson was no philosopher; and therefore the greatest part of these Commentaries must be considered by him as trifles, from the want of a philosophic genius to comprehend their meaning, and a taste superior to that of a mere mathematician, to discover their beauty and elegance. +

+ + +

It would be natural to insert here the step but the angle ACD is greater than the angle BCD. [C.N. 5]. +

+ + +

much greater, literally greater by much + (πολλῷ μεἰζων). Simson and those who follow him translate: much more then is the angle BDC greater than the angle BCD, + but the Greek for this would have to be πολλῷ (or πολὺ[ρπαρ ] μᾶλλόν ἐστι...μείζων. πολλῷ μᾶλλον, however, though used by Apollonius, is not, apparently, found in Euclid or Archimedes.

+ +
+
+ Proposition 8. +

If two triangles have the two sides equal to two sides respectively, and have also the base equal to the base, they will also have the angles equal which are contained by the equal straight lines.

+ +

Let ABC, DEF be two triangles having the two sides AB, AC equal to the two sides DE, DF respectively, namely AB to DE, and AC to DF; and let them have the base BC equal to the base EF;

+

I say that the angle BAC is also equal to the angle EDF.

+

For, if the triangle ABC be applied to the triangle DEF, and if the point B be placed on the point E and the straight line BC on EF, the point C will also coincide with F, because BC is equal to EF. +

+

Then, BC coinciding with EF, BA, AC will also coincide with ED, DF; + for, if the base BC coincides with the base EF, and the sides BA, AC do not coincide with ED, DF but fall beside them as EG, GF, then, given two straight lines constructed on a straight line (from its extremities) and meeting in a point, there will have been constructed on the same straight line (from its extremities), and on the same side of it, two other straight lines meeting in another point and equal to the former two respectively, namely each to that which has the same extremity with it. But they cannot be so constructed. [I. 7]

+

Therefore it is not possible that, if the base BC be applied to the base EF, the sides BA, AC should not coincide with ED, DF; they will therefore coincide, + so that the angle BAC will also coincide with the angle EDF, and will be equal to it.

+

If therefore etc.

+ +

Q. E. D.

+ + +

The text has here BA, CA. +

+ + +

fall beside them. The Greek has the future, παραλλάξουσι. παραλλάττω means to pass by without touching, + to miss + or to deviate. +

+ +
+
+ Proposition 9. +

To bisect a given rectilineal angle.

+ +

Let the angle BAC be the given rectilineal angle.

+

Thus it is required to bisect it.

+

Let a point D be taken at random on AB; let AE be cut off from AC equal to AD; [I. 3] let DE be joined, and on DE let the equilateral triangle DEF be constructed; let AF be joined.

+

I say that the angle BAC has been bisected by the straight line AF.

+

For, since AD is equal to AE, and AF is common, the two sides DA, AF are equal to the two sides EA, AF respectively.

+

And the base DF is equal to the base EF; therefore the angle DAF is equal to the angle EAF. [I. 8]

+

Therefore the given rectilineal angle BAC has been bisected by the straight line AF.

+ +

Q. E. F.

+ +
+
+ Proposition 10. +

To bisect a given finite straight line.

+ +

Let AB be the given finite straight line.

+

Thus it is required to bisect the finite straight line AB.

+

Let the equilateral triangle ABC be constructed on it, [I. 1] and let the angle ACB be bisected by the straight line CD; [I. 9]

+

I say that the straight line AB has been bisected at the point D.

+

For, since AC is equal to CB, and CD is common,

+ the two sides AC, CD are equal to the two sides BC, CD respectively; and the angle ACD is equal to the angle BCD; therefore the base AD is equal to the base BD. [I. 4]

+

Therefore the given finite straight line AB has been bisected at D.

+ +

Q. E. F.

+ +
+
+ + Proposition 11. +

To draw a straight line at right angles to a given straight line from a given point on it.

+ +

Let AB be the given straight line, and C the given point on it.

+

Thus it is required to draw from the point C a straight line at right angles to the straight line AB.

+

Let a point D be taken at random on AC; let CE be made equal to CD; [I. 3] on DE let the equilateral triangle FDE be constructed, [I. 1] and let FC be joined;

+

I say that the straight line FC has been drawn at right angles to the given straight line AB from C the given point on it.

+

For, since DC is equal to CE, and CF is common, the two sides DC, CF are equal to the two sides EC, CF respectively; and the base DF is equal to the base FE; therefore the angle DCF is equal to the angle ECF; [I. 8] and they are adjacent angles.

+

But, when a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right; [Def. 10] therefore each of the angles DCF, FCE is right.

+

Therefore the straight line CF has been drawn at right angles to the given straight line AB from the given point C on it.

+ +

Q. E. F.

+ + +

The verb is κείσθω which, as well as the other parts of κεῖμαι, is constantly used for the passive of τίθημι to place +; and the latter word is constantly used in the sense of making, e.g., one straight line equal to another straight line.

+ +
+
+ Proposition 12. +

To a given infinite straight line, from a given point which is not on it, to draw a perpendicular straight line.

+ +

Let AB be the given infinite straight line, and C the given point which is not on it; thus it is required to draw to the given infinite straight line AB, from the given point C which is not on it, a perpendicular straight line.

+

For let a point D be taken at random on the other side of the straight line AB, and with centre C and distance CD let the circle EFG be described; [Post. 3]

+ let the straight line EG + be bisected at H, [I. 10] and let the straight lines CG, CH, CE be joined. [Post. 1]

+

I say that CH has been drawn perpendicular to the given infinite straight line AB from the given point C which is not on it.

+

For, since GH is equal to HE, and HC is common, the two sides GH, HC are equal to the two sides EH, HC respectively; and the base CG is equal to the base CE; therefore the angle CHG is equal to the angle EHC. [I. 8] And they are adjacent angles.

+

But, when a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right, and the straight line standing on the other is called a perpendicular to that on which it stands. [Def. 10]

+

Therefore CH has been drawn perpendicular to the given infinite straight line AB from the given point C which is not on it.

+ +

Q. E. F.

+ + +

, κάθετον εὐθεῖαν γραμμἡν. This is the full expression for a perpendicular, κάθετος meaning let fall or let down, so that the expression corresponds to our plumb-line. ἡ κάθετος is however constantly used alone for a perpendicular, γραμμἡ being understood.

+ + +

, literally towards the other parts of the straight line AB, + ἐπὶ τὰ ἕτερα μέρη τῆς AB. Cf. on the same side + (ἐπὶ τὰ αὐτὰ μέρη) in Post. 5 and in both directions + (ἐφ̓ ἑκάτερα τὰ μἑρη) in Def. 23.

+ +
+
+ Proposition 13. +

If a straight line set up on a straight line make angles, it will make either two right angles or angles equal to two right angles.

+ +

For let any straight line AB set up on the straight line CD make the angles CBA, ABD;

+

I say that the angles CBA, ABD are either two right angles or equal to two right angles.

+

Now, if the angle CBA is equal to the angle ABD, they are two right angles. [Def. 10]

+

But, if not, let BE be drawn from the point B at right angles to CD; [I. 11] therefore the angles CBE, EBD are two right angles. +

+

Then, since the angle CBE is equal to the two angles CBA, ABE, let the angle EBD be added to each; therefore the angles CBE, EBD are equal to the three angles CBA, ABE, EBD. [C. N. 2] +

+

Again, since the angle DBA is equal to the two angles DBE, EBA, let the angle ABC be added to each; therefore the angles DBA. ABC are equal to the three angles DBE, EBA, ABC. [C. N. 2] +

+

But the angles CBE, EBD were also proved equal to the same three angles; and things which are equal to the same thing are also equal to one another; [C. N. 1] therefore the angles CBE, EBD are also equal to the angles DBA, ABC. + But the angles CBE, EBD are two right angles; therefore the angles DBA, ABC are also equal to two right angles.

+

Therefore etc.

+ +

Q. E. D.

+ + +

literally let the angle EBD be added (so as to be) common, + κοινὴ προσκείσθω ἡ ὑπὸ ΕΒΔ. Similarly κοινὴ ἀφηρήσθω is used of subtracting a straight line or angle from each of two others. Let the common angle EBD be added + is clearly an inaccurate translation, for the angle is not common before it is added, i.e. the κοινὴ is proleptic. Let the common angle be subtracted + as a translation of κοινὴ ἀφηρήσθω would be less unsatisfactory, it is true, but, as it is desirable to use corresponding words when translating the two expressions, it seems hopeless to attempt to keep the word common, + and I have therefore said to each + and from each + simply.

+ +
+
+ Proposition 14. +

If with any straight line, and at a point on it, two straight lines not lying on the same side make the adjacent angles equal to two right angles, the two straight lines will be in a straight line with one another.

+ +

For with any straight line AB, and at the point B on it, let the two straight lines BC, BD not lying on the same side make the adjacent angles ABC, ABD equal to two right angles;

+

I say that BD is in a straight line with CB.

+

For, if BD is not in a straight line with BC, let BE be in a straight line with CB.

+

Then, since the straight line AB stands on the straight line CBE,

+ the angles ABC, ABE are equal to two right angles. [I. 13] But the angles ABC, ABD are also equal to two right angles; therefore the angles CBA, ABE are equal to the angles CBA, ABD. [Post. 4 and C.N. 1]

+

Let the angle CBA be subtracted from each; therefore the remaining angle ABE is equal to the remaining angle ABD, [C.N. 3] the less to the greater: which is impossible. Therefore BE is not in a straight line with CB.

+

Similarly we can prove that neither is any other straight line except BD. Therefore CB is in a straight line with BD.

+

Therefore etc.

+ +

Q. E. D.

+ + +

There is no greater difficulty in translating the works of the Greek geometers than that of accurately giving the force of prepositions. πρός, for instance, is used in all sorts of expressions with various shades of meaning. The present enunciation begins ἐὰν πρός τινι εὐθείᾳ καὶ τῷ πρὸς αὐτῆ σημείῳ, and it is really necessary in this one sentence to translate πρός by three different words, with, at, and on. The first πρός must be translated by with because two straight lines make + an angle with one another. On the other hand, where the similar expression πρὸς τῇ δοθείση εὐθείᾳ occurs in I. 23, but it is a question of constructing + an angle (συστἡσασθαι), we have to say to construct on a given straight line. + Against would perhaps be the English word coming nearest to expressing all these meanings of πρός, but it would be intolerable as a translation.

+ + +

Todhunter points out that for the inference in this line Post. 4, that all right angles are equal, is necessary as well as the Common Notion that things which are equal to the same thing (or rather, here, to equal things) are equal. A similar remark applies to steps in the proofs of I. 15 and I. 28.

+ + +

The Greek expresses this by the future of the verb, δείξομεν, we shall prove, + which however would perhaps be misleading in English.

+ +
+
+ Proposition 15. +

If two straight lines cut one another, they make the vertical angles equal to one another.

+ +

For let the straight lines AB, CD cut one another at the point E;

+

I say that the angle AEC is equal to the angle DEB, and the angle CEB to the angle AED. +

+

For, since the straight line AE stands on the straight line CD, making the angles CEA, AED, the angles CEA, AED are equal to two right angles [I. 13] +

+

Again, since the straight line DE stands on the straight line AB, making the angles AED, DEB, the angles AED, DEB are equal to two right angles. [I. 13] +

+

But the angles CEA, AED were also proved equal to two right angles; therefore the angles CEA, AED are equal to the angles AED + DEB. [Post. 4 and C. N. 1] Let the angle AED be subtracted from each; therefore the remaining angle CEA is equal to the remaining angle BED. [C. N. 3]

+

Similarly it can be proved that the angles CEB, DEA are also equal.

+

Therefore etc. Q. E. D.

+ + +

[Porism. From this it is manifest that, if two straight lines cut one another, they will make the angles at the point of section equal to four right angles.

+ + +

The difference between adjacent angles (αἱ ἐφεξῆς γωνίαι) and vertical angles (αἱ κατὰ κορυφὴν γωνίαι) is thus explained by Proclus (p. 298, 14-24). The first term describes the angles made by two straight lines when one only is divided by the other, i.e. when one straight line meets another at a point which is not either of its extremities, but is not itself produced beyond the point of meeting. When the first straight line is produced, so that the lines cross at the point, they make two pairs of vertical angles (which are more clearly described as vertically opposite angles), and which are so called because their convergence is from opposite directions to one point (the intersection of the lines) as vertex (κορυφή).

+ + +

literally at the section, + πρὸς τῇ τομῆ.

+ +
+
+ Proposition 16. +

In any triangle, if one of the sides be produced, the exterior angle is greater than either of the interior and opposite angles.

+ +

Let ABC be a triangle, and let one side of it BC be produced to D;

+

I say that the exterior angle ACD is greater than either of the interior and opposite angles CBA, BAC.

+

Let AC be bisected at E [I. 10], and let BE be joined and produced in a straight line to F;

+

let EF be made equal to BE[I. 3], let FC be joined [Post. 1], and let AC be drawn through to G [Post. 2].

+

Then, since AE is equal to EC, and BE to EF, the two sides AE, EB are equal to the two sides CE, EF respectively; and the angle AEB is equal to the angle FEC, for they are vertical angles. [I. 15] Therefore the base AB is equal to the base FC, and the triangle ABE is equal to the triangle CFE, and the remaining angles are equal to the remaining angles respectively, namely those which the equal sides subtend; [I. 4] therefore the angle BAE is equal to the angle ECF. +

+

But the angle ECD is greater than the angle ECF; [C. N. 5] therefore the angle ACD is greater than the angle BAE.

+

Similarly also, if BC be bisected, the angle BCG, that is, the angle ACD [I. 15], can be proved greater than the angle ABC as well.

+

Therefore etc.

+ +

Q. E. D.

+ + +

literally the outside angle, + ἡ ἐκτὸς γωνία.

+ + +

τῶν ἐντὸς καὶ ἀπεναντίον γωνιῶν.

+ + +

The word is διήχθω, a variation on the more usual ἐκβεβλήσθω, let it be produced. +

+ + +

in the text FEC. +

+ +
+
+ Proposition 17. +

In any triangle two angles taken together in any manner are less than two right angles.

+ +

Let ABC be a triangle; I say that two angles of the triangle ABC taken together in any manner are less than two right angles.

+

For let BC be produced to D. [Post. 2]

+

Then, since the angle ACD is an exterior angle of the triangle ABC,

+

it is greater than the interior and opposite angle ABC. [I. 16] Let the angle ACB be added to each; therefore the angles ACD, ACB are greater than the angles ABC, BCA.

But the angles ACD, ACB are equal to two right angles. [I. 13]

+

Therefore the angles ABC, BCA are less than two right angles.

+

Similarly we can prove that the angles BAC, ACB are also less than two right angles, and so are the angles CAB, ABC as well.

+

Therefore etc.

+ +

Q. E. D.

+ + +

πάντη μεταλαμβανόμεναι, i.e. any pair added together.

+ +
+
+ + Proposition 18. +

In any triangle the greater side subtends the greater angle.

+ +

For let ABC be a triangle having the side AC greater than AB;

+

I say that the angle ABC is also greater than the angle BCA.

+

For, since AC is greater than AB, let AD be made equal to AB [I. 3], and let BD bejoined.

+

Then, since the angle ADB is an exterior angle of the triangle BCD,

+

it is greater than the interior and opposite angle DCB. [I. 16]

+

But the angle ADB is equal to the angle ABD, since the side AB is equal to AD; therefore the angle ABD is also greater than the angle ACB; therefore the angle ABC is much greater than the angle ACB.

+

Therefore etc.

+ +

Q. E. D.

+ + +

In the enunciation of this number we have ὑποτείνειν (subtend +) used with the simple accusative instead of the more usual ὑπό with accusative. The latter construction is used in the enunciation of I. 19, which otherwise only differs from that of I. 18 in the order of the words. The point to remember in order to distinguish the two is that the datum comes first and the quaesitum second, the datum being in this proposition the greater side and in the next the greater angle. Thus the enunciations are (I. 18) παντὸς τριγώνου ἡ μείζων πλευρὰ τὴν μείζονα γωνίαν ὑποτείνει and (I. 19) παντὸς τριγώνου ὑπὸ τὴν μείζονα γωνίαν ἡ μείζων πλευρὰ ὑποτείνει. In order to keep the proper order in English we must use the passive of the verb in I. 19. Aristotle quotes the result of I. 19, using the exact wording, ὑπὸ γὰρ τὴν μείζω γωνίαν ὑποτείνει (Meteorologica III. 5, 376 a 12).

+ +
+
+ Proposition 19. +

In any triangle the greater angle is subtended by the greater side.

+ +

Let ABC be a triangle having the angle ABC greater than the angle BCA;

+

I say that the side AC is also greater than the side AB.

+

For, if not, AC is either equal to AB or less.

+

Now AC is not equal to AB; for then the angle ABC would also have been equal to the angle ACB; [I. 5] but it is not;

+ therefore AC is not equal to AB.

+

Neither is AC less than AB, for then the angle ABC would also have been less than the angle ACB; [I. 18] but it is not; therefore AC is not less than AB.

+

And it was proved that it is not equal either. Therefore AC is greater than AB.

+

Therefore etc.

+ +

Q. E. D.

+ +
+
+ Proposition 20. +

In any triangle two sides taken together in any manner are greater than the remaining one.

+ +

For let ABC be a triangle; I say that in the triangle ABC two sides taken together in any manner are greater than the remaining one, namely BA, AC greater than BC, AB, BC greater than AC, BC, CA greater than AB.

+

For let BA be drawn through to the point D, let DA be made equal to CA, and let DC be joined.

+

Then, since DA is equal to AC, the angle ADC is also equal to the angle ACD; [I. 5]

+ therefore the angle BCD is greater than the angle ADC. [C.N. 5]

+

And, since DCB is a triangle having the angle BCD greater than the angle BDC, and the greater angle is subtended by the greater side, [I. 19] therefore DB is greater than BC. +

+

But DA is equal to AC; therefore BA, AC are greater than BC.

+

Similarly we can prove that AB, BC are also greater than CA, and BC, CA than AB.

+

Therefore etc.

+ +

Q. E. D.

+ + +

It was the habit of the Epicureans, says Proclus (p. 322), to ridicule this theorem as being evident even to an ass and requiring no proof, and their allegation that the theorem was known + (γνώριμον) even to an ass was based on the fact that, if fodder is placed at one angular point and the ass at another, he does not, in order to get to his food, traverse the two sides of the triangle but only the one side separating them (an argument which makes Savile exclaim that its authors were digni ipsi, qui cum Asino foenum essent, + p. 78). Proclus replies truly that a mere perception of the truth of the theorem is a different thing from a scientific proof of it and a knowledge of the reason why it is true. Moreover, as Simson says, the number of axioms should not be increased without necessity.

+ +
+
+ + Proposition 21. +

If on one of the sides of a triangle, from its extremities, there be constructed two straight lines meeting within the triangle, the straight lines so constructed will be less than the remaining two sides of the triangle, but will contain a greater angle.

+ +

On BC, one of the sides of the triangle ABC, from its extremities B, C, let the two straight lines BD, DC be constructed meeting within the triangle;

+

I say that BD, DC are less than the remaining two sides of the triangle BA, AC, but contain an angle BDC greater than the angle BAC.

+

For let BD be drawn through to E.

+

Then, since in any triangle two sides are greater than the remaining one, [I. 20] therefore, in the triangle ABE, the two sides AB, AE are greater than BE.

+

Let EC be added to each; therefore BA, AC are greater than BE, EC. +

+

Again, since, in the triangle CED, the two sides CE, ED are greater than CD, let DB be added to each; therefore CE, EB are greater than CD, DB.

+

But BA, AC were proved greater than BE, EC; therefore BA, AC are much greater than BD, DC.

+

Again, since in any triangle the exterior angle is greater than the interior and opposite angle, [I. 16] therefore, in the triangle CDE, the exterior angle BDC is greater than the angle CED. +

+

For the same reason, moreover, in the triangle ABE also, the exterior angle CEB is greater than the angle BAC. But the angle BDC was proved greater than the angle CEB; therefore the angle BDC is much greater than the angle BAC. +

+

Therefore etc.

+ +

Q. E. D.

+ + +

The word meeting + is not in the Greek, where the words are ἐντὸς συσταθῶσιν. συνίστασθαι is the word used of constructing two straight lines to a point (cf. I. 7) or so as to form a triangle; but it is necessary in English to indicate that they meet.

+ + +

Observe the elegant brevity of the Greek αἱ συσταθεῖσαι.

+ +
+
+ Proposition 22. +

Out of three straight lines, which are equal to three given straight lines, to construct a triangle: thus it is necessary that two of the straight lines taken together in any manner should be greater than the remaining one. [I. 20]

+ +

Let the three given straight lines be A, B, C, and of these let two taken together in any manner be greater than the remaining one, namely A, B greater than C, A, C greater than B, and B, C greater than A; thus it is required to construct a triangle out of straight lines equal to A, B, C.

+

Let there be set out a straight line DE, terminated at D but of infinite length in the direction of E, and let DF be made equal to A, FG equal to B, and GH equal to C. [I. 3]

+

With centre F and distance FD let the circle DKL be described; again, with centre G and distance GH let the circle KLH be described; and let KF, KG be joined;

+

I say that the triangle KFG has been constructed out of three straight lines equal to A, B, C.

+

For, since the point F is the centre of the circle DKL, FD is equal to FK.

+

But FD is equal to A; therefore KF is also equal to A.

+

Again, since the point G is the centre of the circle LKH, GH is equal to GK.

+

But GH is equal to C; therefore KG is also equal to C. And FG is also equal to B; therefore the three straight lines KF, FG, GK are equal to the three straight lines A, B, C.

+

Therefore out of the three straight lines KF, FG, GK, which are equal to the three given straight lines A, B, C, the triangle KFG has been constructed.

+ +

Q. E. F.

+ + +

This is the first case in the Elements of a διορισμός to a problem in the sense of a statement of the conditions or limits of the possibility of a solution. The criterion is of course supplied by the preceding proposition.

+ + +

This is usually translated (e.g. by Williamson and Simson) But it is necessary, + which is however inaccurate, since the Greek is not δεῖ δέ but δεῖ δή. The words are the same as those used to introduce the διορισμός in the other sense of the definition + or particular statement + of a construction to be effected. Hence, as in the latter case we say thus it is required + (e.g. to bisect the finite straight line AB, I. 10), we should here translate thus it is necessary. +

+ + +

To this enunciation all the MSS. and Boethius add, after the διορισμός, the words because in any triangle two sides taken together in any manner are greater than the remaining one. + But this explanation has the appearance of a gloss, and it is omitted by Proclus and Campanus. Moreover there is no corresponding addition to the διορισμός of VI. 28.

+ +
+
+ Proposition 23. +

On a given straight line and at a point on it to construct a rectilineal angle equal to a given rectilineal angle.

+ +

Let AB be the given straight line, A the point on it, and the angle DCE the given rectilineal angle;

+

thus it is required to construct on the given straight line AB, and at the point A on it, a rectilineal angle equal to the given rectilineal angle DCE.

+

On the straight lines CD, CE respectively let the points D, E be taken at random; let DE be joined, and out of three straight lines which are equal to the three straight lines CD, DE, CE let the triangle AFG be constructed in such a way that CD is equal to AF, CE to AG, and further DE to FG.

+

Then, since the two sides DC, CE are equal to the two sides FA, AG respectively, and the base DE is equal to the base FG, the angle DCE is equal to the angle FAG. [I. 8]

+

Therefore on the given straight line AB, and at the point A on it, the rectilineal angle FAG has been constructed equal to the given rectilineal angle DCE.

+ +

Q. E. F.

+ +
+
+ Proposition 24. +

If two triangles have the two sides equal to two sides respectively, but have the one of the angles contained by the equal straight lines greater than the other, they will also have the base greater than the base.

+ +

Let ABC, DEF be two triangles having the two sides AB, AC equal to the two sides DE, DF respectively, namely AB to DE, and AC to DF, and let the angle at A be greater than the angle at D;

+

I say that the base BC is also greater than the base EF.

+

For, since the angle BAC is greater than the angle EDF, let there be constructed, on the straight line DE, and at the point D on it, the angle EDG + equal to the angle BAC; [I. 23] let DG be made equal to either of the two straight lines AC, DF, and let EG, FG be joined.

+

+

Then, since AB is equal to DE, and AC to DG, the two sides BA, AC are equal to the two sides ED, DG, respectively; and the angle BAC is equal to the angle EDG; therefore the base BC is equal to the base EG. [I. 4]

+

Again, since DF is equal to DG, the angle DGF is also equal to the angle DFG; [I. 5] therefore the angle DFG is greater than the angle EGF.

+

Therefore the angle EFG is much greater than the angle EGF.

+

And, since EFG is a triangle having the angle EFG + greater than the angle EGF, and the greater angle is subtended by the greater side, [I. 19] the side EG is also greater than EF.

+

But EG is equal to BC. Therefore BC is also greater than EF. +

+

Therefore etc.

+ +

Q. E. D.

+ + +

I have naturally left out the well-known words added by Simson in order to avoid the necessity of considering three cases: Of the two sides DE, DF let DE be the side which is not greater than the other. + I doubt whether Euclid could have been induced to insert the words himself, even if it had been represented to him that their omission meant leaving two possible cases out of consideration. His habit and that of the great Greek geometers was, not to set out all possible cases, but to give as a rule one case, generally the most difficult, as here, and to leave the others to the reader to work out for himself. We have already seen one instance in I. 7.

+ +
+
+ Proposition 25. +

If two triangles have the two sides equal to two sides respectively, but have the base greater than the base, they will also have the one of the angles contained by the equal straight lines greater than the other.

+ +

Let ABC, DEF be two triangles having the two sides AB, AC equal to the two sides DE, DF respectively, namely AB to DE, and AC to DF; and let the base BC be greater than the base EF;

+

I say that the angle BAC is also greater than the angle EDF.

+

For, if not, it is either equal to it or less.

+

Now the angle BAC is not equal to the angle EDF; for then the base BC would also have been equal to the base EF, [I. 4] but it is not; therefore the angle BAC is not equal to the angle EDF.

+

Neither again is the angle BAC less than the angle EDF; for then the base BC would also have been less than the base EF, [I. 24] but it is not; therefore the angle BAC is not less than the angle EDF.

+

But it was proved that it is not equal either; therefore the angle BAC is greater than the angle EDF.

+

Therefore etc.

+ +

Q. E. D.

+ +
+
+ Proposition 26. +

If two triangles have the two angles equal to two angles respectively, and one side equal to one side, namely, either the side adjoining the equal angles, or that subtending one of the equal angles, they will also have the remaining sides equal to the remaining sides and the remaining angle to the remaining angle.

+ +

Let ABC, DEF be two triangles having the two angles ABC, BCA equal to the two angles DEF, EFD respectively, namely the angle ABC to the angle DEF, and the angle BCA to the angle EFD; and let them also have one side equal to one side, first that adjoining the equal angles, namely BC to EF;

+

I say that they will also have the remaining sides equal to the remaining sides respectively, namely AB to DE and AC to DF, and the remaining angle to the remaining angle, namely the angle BAC to the angle EDF.

+

For, if AB is unequal to DE, one of them is greater.

+

Let AB be greater, and let BG be made equal to DE; and let GC be joined.

+

Then, since BG is equal to DE, and BC to EF, the two sides GB, BC are equal to the two sides DE, EF respectively; and the angle GBC is equal to the angle DEF; therefore the base GC is equal to the base DF, and the triangle GBC is equal to the triangle DEF, and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend; [I. 4] therefore the angle GCB is equal to the angle DFE. But the angle DFE is by hypothesis equal to the angle BCA; therefore the angle BCG is equal to the angle BCA, the less to the greater: which is impossible. Therefore AB is not unequal to DE, and is therefore equal to it.

+

But BC is also equal to EF; therefore the two sides AB, BC are equal to the two sides DE, EF respectively, and the angle ABC is equal to the angle DEF; therefore the base AC is equal to the base DF, and the remaining angle BAC is equal to the remaining angle EDF. [I. 4] +

+

Again, let sides subtending equal angles be equal, as AB to DE;

+

I say again that the remaining sides will be equal to the remaining sides, namely AC to DF and BC to EF, and further the remaining angle BAC is equal to the remaining angle EDF.

+

For, if BC is unequal to EF, one of them is greater.

+

Let BC be greater, if possible, and let BH be made equal to EF; let AH be joined.

+

Then, since BH is equal to EF, and AB to DE, the two sides AB, BH are equal to the two sides DE, EF respectively, and they contain equal angles; therefore the base AH is equal to the base DF, and the triangle ABH is equal to the triangle DEF, and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend; [I. 4] therefore the angle BHA is equal to the angle EFD.

+

But the angle EFD is equal to the angle BCA; therefore, in the triangle AHC, the exterior angle BHA is equal to the interior and opposite angle BCA: which is impossible. [I. 16]

+

Therefore BC is not unequal to EF, and is therefore equal to it.

+

But AB is also equal to DE; therefore the two sides AB, BC are equal to the two sides DE, EF respectively, and they contain equal angles; therefore the base AC is equal to the base DF, the triangle ABC equal to the triangle DEF, and the remaining angle BAC equal to the remaining angle EDF. [I. 4]

+

Therefore etc.

+ +

Q. E. D.

+ + +

πλευρὰν τὴν πρὸς ταῖς ἴσαις γωνίαις.

+ + +

ὑπόκειται ἴση, according to the elegant Greek idiom. ὑπόκειμαι is used for the passive of ὑποτίθημι, as κεῖμαι is used for the passive of τίθημι, and so with the other compounds. Cf. προσκεῖσθαι, to be added. +

+ +
+
+ Proposition 27. +

If a straight line falling on two straight lines make the alternate angles equal to one another, the straight lines will be parallel to one another.

+ +

For let the straight line EF falling on the two straight lines AB, CD make the alternate angles AEF, EFD equal to one another;

+

I say that AB is parallel to CD.

+

For, if not, AB, CD when produced will meet either in the direction of B, D or towards A, C.

+

Let them be produced and meet, in the direction of B, D, at G.

+

Then, in the triangle GEF, the exterior angle AEF is equal to the interior and opposite angle EFG: which is impossible. [I. 16]

+

Therefore AB, CD when produced will not meet in the direction of B, D.

+

Similarly it can be proved that neither will they meet towards A, C.

+

But straight lines which do not meet in either direction are parallel; [Def. 23] therefore AB is parallel to CD.

+

Therefore etc.

+ +

Q. E. D.

+ + +

εὶς δύο εὐθείας ἐμπίπτουσα, the phrase being the same as that used in Post. 5, meaning a transversal.

+ + +

αἱ ἐναλλὰξ γωνίαι. Proclus (p. 357, 9) explains that Euclid uses the word alternate (or, more exactly, alternately, ἐναλλάξ) in two connexions, (1) of a certain transformation of a proportion, as in Book V. and the arithmetical Books, (2) as here, of certain of the angles formed by parallels with a straight line crossing them. Alternate angles are, according to Euclid as interpreted by Proclus, those which are not on the same side of the transversal, and are not adjacent, but are separated by the transversal, both being within the parallels but one above + and the other below. + The meaning is natural enough if we imagine the four internal angles to be taken in cyclic order and alternate angles to be any two of them not successive but separated by one angle of the four.

+ + +

literally towards the parts B, D or towards A, C, + ἐπὶ τὰ Β, Δ μέρη ἢ ἐπὶ τὰ Α Γ.

+ +
+
+ Proposition 28. +

If a straight line falling on two straight lines make the exterior angle equal to the interior and opposite angle on the same side, or the interior angles on the same side equal to two right angles, the straight lines will be parallel to one another.

+ +

For let the straight line EF falling on the two straight lines AB, CD make the exterior angle EGB equal to the interior and opposite angle GHD, or the interior angles on the same side, namely BGH, GHD, equal to two right angles;

+

I say that AB is parallel to CD.

+

For, since the angle EGB is equal to the angle GHD, while the angle EGB is equal to the angle AGH, [I. 15]

the angle AGH is also equal to the angle GHD; and they are alternate; therefore AB is parallel to CD. [I. 27]

+

Again, since the angles BGH, GHD are equal to two right angles, and the angles AGH, BGH are also equal to two right angles, [I. 13] the angles AGH, BGH are equal to the angles BGH, GHD.

+

Let the angle BGH be subtracted from each; therefore the remaining angle AGH is equal to the remaining angle GHD; and they are alternate; therefore AB is parallel to CD. [I. 27]

+

Therefore etc.

+ +

Q. E. D.

+ +
+
+ Proposition 29. +

A straight line falling on parallel straight lines makes the alternate angles equal to one another, the exterior angle equal to the interior and opposite angle, and the interior angles on the same side equal to two right angles.

+ +

For let the straight line EF fall on the parallel straight lines AB, CD;

+

I say that it makes the alternate angles AGH, GHD equal, the exterior angle EGB equal to the interior and opposite angle GHD, and the interior angles on the same side, namely BGH, GHD, equal to two right angles.

+

For, if the angle AGH is unequal to the angle GHD, one of them is greater.

+

Let the angle AGH be greater.

+

Let the angle BGH be added to each; therefore the angles AGH, BGH are greater than the angles BGH, GHD.

+

But the angles AGH, BGH are equal to two right angles; [I. 13] therefore the angles BGH, GHD are less than two right angles.

+

But straight lines produced indefinitely from angles less than two right angles meet; [Post. 5] therefore AB, CD, if produced indefinitely, will meet; but they do not meet, because they are by hypothesis parallel.

+

Therefore the angle AGH is not unequal to the angle GHD, and is therefore equal to it. +

+

Again, the angle AGH is equal to the angle EGB; [I. 15] therefore the angle EGB is also equal to the angle GHD. [C.N. 1]

+

Let the angle BGH be added to each; therefore the angles EGB, BGH are equal to the angles BGH, GHD. [C.N. 2]

+

But the angles EGB, BGH are equal to two right angles; [I. 13] therefore the angles BGH, GHD are also equal to two right angles.

+

Therefore etc.

+ +

Q. E. D.

+ + +

αἰ δὲ ἀπ· ὲλασσόνων ἢ δύο ὀρθῶν ἐκβαλλόμεναι εἰς ἄπειπον συμπίπτουσιν, a variation from the more explicit language of Postulate 5. A good deal is left to be understood, namely that the straight lines begin from points at which they meet a transversal, and make with it internal angles on the same side the sum of which is less than two right angles.

+ + +

literally because they are supposed parallel, + διὰ τὸ παραλλήλους αὐτὰς ὑποκεῖσθαι.

+ +
+
+ Proposition 30. +

Straight lines parallel to the same straight line are also parallel to one another.

+ +

Let each of the straight lines AB, CD be parallel to EF; I say that AB is also parallel to CD.

+

For let the straight line GK fall upon them;

+

Then, since the straight line GK has fallen on the parallel straight lines AB, EF, the angle AGK is equal to the angle GHF. [I. 29]

+

Again, since the straight line GK has fallen on the parallel straight lines EF, CD, the angle GHF is equal to the angle GKD. [I. 29] +

+

But the angle AGK was also proved equal to the angle GHF; therefore the angle AGK is also equal to the angle GKD; [C.N. 1] and they are alternate.

+

Therefore AB is parallel to CD.

+ +

Q. E. D.

+ + +

The usual conclusion in general terms (Therefore etc. +) repeating the enunciation is, curiously enough, wanting at the end of this proposition.

+ +
+
+ Proposition 31. +

Through a given point to draw a straight line parallel to a given straight line.

+ +

Let A be the given point, and BC the given straight line; thus it is required to draw through the point A a straight line parallel to the straight line BC.

+

Let a point D be taken at random on BC, and let AD be joined; on the straight line DA, and at the point A on it, let the angle DAE be constructed equal to the angle ADC [I. 23]; and let the straight line AF be produced in a straight line with EA.

+

Then, since the straight line AD falling on the two straight lines BC, EF has made the alternate angles EAD, ADC equal to one another, therefore EAF is parallel to BC. [I. 27]

+

Therefore through the given point A the straight line EAF has been drawn parallel to the given straight line BC.

+ +

Q. E. F.

+ +
+
+ Proposition 32. +

In any triangle, if one of the sides be produced, the exterior angle is equal to the two interior and opposite angles, and the three interior angles of the triangle are equal to two right angles.

+ +

Let ABC be a triangle, and let one side of it BC be produced to D;

+

I say that the exterior angle ACD is equal to the two interior and opposite angles CAB, ABC, and the three interior angles of the triangle ABC, BCA, CAB are equal to two right angles.

+

For let CE be drawn through the point C parallel to the straight line AB. [I. 31]

+

Then, since AB is parallel to CE, and AC has fallen upon them, the alternate angles BAC, ACE are equal to one another. [I. 29] +

+

Again, since AB is parallel to CE, and the straight line BD has fallen upon them, the exterior angle ECD is equal to the interior and opposite angle ABC. [I. 29]

+

But the angle ACE was also proved equal to the angle BAC; therefore the whole angle ACD is equal to the two interior and opposite angles BAC, ABC.

+

Let the angle ACB be added to each; therefore the angles ACD, ACB are equal to the three angles ABC, BCA, CAB.

+

But the angles ACD, ACB are equal to two right angles; [I. 13] therefore the angles ABC, BCA, CAB are also equal to two right angles.

+

Therefore etc.

+ +

Q. E. D.

+ +
+
+ Proposition 33. +

The straight lines joining equal and parallel straight lines (at the extremities which are) in the same directions (respectively) are themselves also equal and parallel.

+ +

Let AB, CD be equal and parallel, and let the straight lines AC, BD join them (at the extremities which are) in the same directions (respectively); I say that AC, BD are also equal and parallel.

+

Let BC be joined.

+

Then, since AB is parallel to CD, and BC has fallen upon them, the alternate angles ABC, BCD are equal to one another. [I. 29]

+

And, since AB is equal to CD, and BC is common, the two sides AB, BC are equal to the two sides DC, CB; and the angle ABC is equal to the angle BCD; therefore the base AC is equal to the base BD, and the griangle ABC is equal to the triangle DCB, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend; [I. 4] therefore the angle ACB is equal to the angle CBD.

+

And, since the straight line BC falling on the two straight lines AC, BD has made the alternate angles equal to one another, AC is parallel to BD. [I. 27]

+

And it was also proved equal to it.

+

Therefore etc.

+ +

Q. E. D.

+ + +

I have for clearness' sake inserted the words in brackets though they are not in the original Greek, which has joining...in the same directions + or on the same sides, + ἐπὶ τὰ αυτὰ μέρη ἐπιζευγνύουσαι. The expression tiwards the same parts, + though usage has sanctioned it, is perhaps not quite satisfactory.

+ + +

and 18. DCB. The Greek has BC, CD + and BCD + in these places respectively. Euclid is not always careful to write in corresponding order the letters denoting corresponding points in congruent figures. On the contrary, he evidently prefers the alphabetical order, and seems to disdain to alter it for the sake of beginners or others who might be confused by it. In the case of angles alteration is perhaps unnecessary; but in the case of triangles and pairs of corresponding sides I have ventured to alter the order to that which the mathematician of to-day expects.

+ +
+
+ Proposition 34. +

In parallelogrammic areas the opposite sides and angles are equal to one another, and the diameter bisects the areas.

+ +

Let ACDB be a parallelogrammic area, and BC its diameter; I say that the opposite sides and angles of the parallelogram ACDB are equal to one another, and the diameter BC bisects it.

+

For, since AB is parallel to CD, and the straight line BC has fallen upon them, the alternate angles ABC, BCD are equal to one another. [I. 29]

+

Again, since AC is parallel to BD, and BChas fallen upon them, the alternate angles ACB, CBD are equal to one another. [I. 29]

+

Therefore ABC, DCB are two triangles having the two angles ABC, BCA equal to the two angles DCB, CBD respectively, and one side equal to one side, namely that adjoining the equal angles and common to both of them, BC; therefore they will also have the remaining sides equal to the remaining sides respectively, and the remaining angle to the remaining angle; [I. 26] therefore the side AB is equal to CD, and AC to BD, and further the angle BAC is equal to the angle CDB.

+

And, since the angle ABC is equal to the angle BCD, and the angle CBD to the angle ACB, the whole angle ABD is equal to the whole angle ACD. [C.N. 2] + And the angle BAC was also proved equal to the angle CDB.

+

Therefore in parallelogrammic areas the opposite sides and angles are equal to one another.

+

I say, next, that the diameter also bisects the areas.

+

For, since AB is equal to CD, and BC is common, the two sides AB, BC are equal to the two sides DC, CB respectively; and the angle ABC is equal to the angle BCD; therefore the base AC is also equal to DB, and the triangle ABC is equal to the triangle DCB. [I. 4]

+

Therefore the diameter BC bisects the parallelogram ACDB.

+ +

Q. E. D.

+ + + +

It is to be observed that, when parallelograms have to be mentioned for the first time, Euclid calls them parallelogrammic areas + or, more exactly, parallelogram + areas (παραλληλόγραμμα χωρία). The meaning is simply areas bounded by parallel straight lines with the further limitation placed upon the term by Euclid that only four-sided figures are so called, although of course there are certain regular polygons which have opposite sides parallel, and which therefore might be said to be areas bounded by parallel straight lines. We gather from Proclus (p. 393) that the word parallelogram + was first introduced by Euclid, that its use was suggested by I. 33, and that the formation of the word παραλληλόγραμμος (parallel-lined) was analogous to that of εὐθύγραμμος (straight-lined or rectilineal).

+ + +

and 36. DC, CB. The Greek has in these places BCD + and CD, BC + respectively. Cf. note on I. 33, lines 15, 18.

+ +
+
+ Proposition 35. +

Parallelograms which are on the same base and in the same parallels are equal to one another.

+ +

Let ABCD, EBCF be parallelograms on the same base BC and in the same parallels AF, BC; I say that ABCD is equal to the parallelogram EBCF.

+

For, since ABCD is a parallelogram, AD is equal to BC. [I. 34] +

+

For the same reason also EF is equal to BC, so that AD is also equal to EF; [C.N. 1] and DE is common; therefore the whole AE is equal to the whole DF. [C.N. 2]

+

But AB is also equal to DC; [I. 34] therefore the two sides EA, AB are equal to the two sides FD, DC respectively,

+ and the angle FDC is equal to the angle EAB, the exterior to the interior; [I. 29] therefore the base EB is equal to the base FC, and the triangle EAB will be equal to the triangle FDC. [I. 4]

+

Let DGE be subtracted from each; therefore the trapezium ABGD which remains is equal to the trapezium EGCF which remains. [C.N. 3]

+

Let the triangle GBC be added to each; therefore the whole parallelogram ABCD is equal to the whole parallelogram EBCF. [C.N. 2]

+

Therefore etc.

+ +

Q. E. D.

+ + +

The text has DFC. +

+ + +

Euclid speaks of the triangle DGE without any explanation that, in the case which he takes (where AD, EF have no point in common), BE, CD must meet at a point G between the two parallels. He allows this to appear from the figure simply.

+ +
+
+ Proposition 36. +

Parallelograms which are on equal bases and in the same parallels are equal to one another.

+ +

Let ABCD, EFGH be parallelograms which are on equal bases BC, FG and in the same parallels AH, BG;

I say that the parallelogram ABCD is equal to EFGH.

+

For let BE, CH be joined.

+

Then, since BC is equal to FG while FG is equal to EH, BC is also equal to EH. [C.N. 1]

+

But they are also parallel.

+

And EB, HC join them; but straight lines joining equal and parallel straight lines (at the extremities which are) in the same directions (respectively) are equal and parallel. [I. 33]

+

Therefore EBCH is a parallelogram. [I. 34]

+

And it is equal to ABCD; for it has the same base BC with it, and is in the same parallels BC, AH with it. [I. 35]

+

For the same reason also EFGH is equal to the same EBCH; [I. 35] so that the parallelogram ABCD is also equal to EFGH. [C.N. 1]

+

Therefore etc. Q. E. D.

+ +
+
+ + Proposition 37. +

Triangles which are on the same base and in the same parallels are equal to one another.

+ +

Let ABC, DBC be triangles on the same base BC and in the same parallels AD, BC; I say that the triangle ABC is equal to the triangle DBC.

+

Let AD be produced in both directions to E, F; through B let BE be drawn parallel to CA, [I. 31] and through C let CF be drawn parallel to BD. [I. 31]

+

Then each of the figures EBCA, DBCF is a parallelogram; and they are equal,

+

for they are on the same base BC and in the same parallels BC, EF. [I. 35]

+

Moreover the triangle ABC is half of the parallelogram EBCA; for the diameter AB bisects it. [I. 34]

+

And the triangle DBC is half of the parallelogram DBCF; for the diameter DC bisects it. [I. 34]

+

[But the halves of equal things are equal to one another.]

+

Therefore the triangle ABC is equal to the triangle DBC.

+

Therefore etc.

+ +

Q. E. D.

+ + +

Here and in the next proposition Heiberg brackets the words But the halves of equal things are equal to one another + on the ground that, since the Common Notion which asserted this fact was interpolated at a very early date (before the time of Theon), it is probable that the words here were interpolated at the same time. Cf. note above (p. 224) on the interpolated Common Notion.

+ +
+
+ Proposition 38. +

Triangles which are on equal bases and in the same parallels are equal to one another.

+ +

Let ABC, DEF be triangles on equal bases BC, EF and in the same parallels BF, AD; I say that the triangle ABC is equal to the triangle DEF.

+

For let AD be produced in both directions to G, H; through B let BG be drawn parallel to CA, [I. 31] and through F let FH be drawn parallel to DE.

+

Then each of the figures GBCA, DEFH is a parallelogram; and GBCA is equal to DEFH;

+

for they are on equal bases BC, EF and in the same parallels BF, GH. [I. 36]

+

Moreover the triangle ABC is half of the parallelogram GBCA; for the diameter AB bisects it. [I. 34]

+

And the triangle FED is half of the parallelogram DEFH; for the diameter DF bisects it. [I. 34]

+

[But the halves of equal things are equal to one another.]

+

Therefore the triangle ABC is equal to the triangle DEF.

+

Therefore etc.

+ +

Q. E. D.

+ +
+
+ Proposition 39. +

Equal triangles which are on the same base and on the same side are also in the same parallels.

+ +

Let ABC, DBC be equal triangles which are on the same base BC and on the same side of it; [I say that they are also in the same parallels.]

+

And [For] let AD be joined; I say that AD is parallel to BC.

+

For, if not, let AE be drawn through the point A parallel to the straight line BC, [I. 31] and let EC be joined.

+

Therefore the triangle ABC is equal to the triangle EBC; for it is on the same base BC with it and in the same parallels. [I. 37]

+

But ABC is equal to DBC; therefore DBC is also equal to EBC, [C.N. 1] the greater to the less: which is impossible.

+

Therefore AE is not parallel to BC.

+

Similarly we can prove that neither is any other straight line except AD; therefore AD is parallel to BC.

+

Therefore etc.

+ +

Q. E. D.

+ + + +

[I say that they are also in the same parallels.] Heiberg has proved (Hermes, XXXVIII., 1903, p. 50) from a recently discovered papyrus-fragment (Fayūm towns and their papyri, p. 96, No. IX.) that these words are an interpolation by some one who did not observe that the words And let AD be joined + are part of the setting-out (ἔκθεσις), but took them as belonging to the construction (κατασκευή) and consequently thought that a διορισμός or definition + (of the thing to be proved) should precede. The interpolator then altered And + into For + in the next sentence.

+ +
+
+ [Proposition 40. +

Equal triangles which are on equal bases and on the same side are also in the same parallels.

+

Let ABC, CDE be equal triangles on equal bases BC, CE and on the same side.

+

I say that they are also in the same parallels.

+

For let AD be joined; I say that AD is parallel to BE.

+

For, if not, let AF be drawn through A parallel to BE [I. 31], and let FE be joined.

+

Therefore the triangle ABC is equal to the triangle FCE; for they are on equal bases BC, CE and in the same parallels BE, AF. [I. 38]

+

But the triangle ABC is equal to the triangle DCE; therefore the triangle DCE is also equal to the triangle FCE, [C.N. 1] the greater to the less: which is impossible. Therefore AF is not parallel to BE. +

+

Similarly we can prove that neither is any other straight line except AD; therefore AD is parallel to BE.

+

Therefore etc. Q. E. D.]

+
+
+ Proposition 41. +

If a parallelogram have the same base with a triangle and be in the same parallels, the parallelogram is double of the triangle.

+ +

For let the parallelogram ABCD have the same base BC with the triangle EBC, and let it be in the same parallels BC, AE;

+

I say that the parallelogram ABCD is double of the triangle BEC.

+

For let AC be joined.

+

Then the triangle ABC is equal to the triangle EBC; for it is on the same base BC with it and in the same parallels BC, AE. [I. 37]

+

But the parallelogram ABCD is double of the triangle ABC; for the diameter AC bisects it; [I. 34] so that the parallelogram ABCD is also double of the triangle EBC.

+

Therefore etc.

+ +

Q. E. D.

+ +
+
+ Proposition 42. +

To construct, in a given rectilineal angle, a parallelogram equal to a given triangle.

+ +

Let ABC be the given triangle, and D the given rectilineal angle; thus it is required to construct in the rectilineal angle D a parallelogram equal to the triangle ABC.

+

Let BC be bisected at E, and let AE be joined; on the straight line EC, and at the point E on it, let the angle CEF be constructed equal to the angle D; [I. 23] through A let AG be drawn parallel to EC, and [I. 31] through C let CG be drawn parallel to EF.

+

Then FECG is a parallelogram.

+

And, since BE is equal to EC, the triangle ABE is also equal to the triangle AEC, for they are on equal bases BE, EC and in the same parallels BC, AG; [I. 38] therefore the triangle ABC is double of the triangle AEC. +

+

But the parallelogram FECG is also double of the triangle AEC, for it has the same base with it and is in the same parallels with it; [I. 41] therefore the parallelogram FECG is equal to the triangle ABC.

+

And it has the angle CEF equal to the given angle D.

+

Therefore the parallelogram FECG has been constructed equal to the given triangle ABC, in the angle CEF which is equal to D. Q. E. F.

+ +
+
+ Proposition 43. +

In any parallelogram the complements of the parallelograms about the diameter are equal to one another.

+ +

Let ABCD be a parallelogram, and AC its diameter; and about AC let EH, FG be parallelograms, and BK, KD + the so-called complements;

+

I say that the complement BK is equal to the complement KD.

+

For, since ABCD is a parallelogram, and AC its diameter, the triangle ABC is equal to the triangle ACD. [I. 34]

+

Again, since EH is a parallelogram, and AK is its diameter, the triangle AEK is equal to the triangle AHK. +

+ For the same reason the triangle KFC is also equal to KGC.

+

Now, since the triangle AEK is equal to the triangle AHK, and KFC to KGC, + the triangle AEK together with KGC is equal to the triangle AHK together with KFC. [C.N. 2]

+

And the whole triangle ABC is also equal to the whole ADC; therefore the complement BK which remains is equal to the complement KD which remains. [C.N. 3]

+

Therefore etc.

+ +

Q. E. D.

+ + + +

complements, παραπληρώματα, the figures put in to fill up (interstices).

+ + +

Euclid's phraseology here and in the next proposition implies that the complements as well as the other parallelograms are about + the diagonal. The words are here περὶ δὲ τὴν ΑΓ παραλληλόγραμμα μὲν ἔστω τὰ ΕΘ, ΖΗ, τὰ δὲ λεγόμενα παραπληρώματα τὰ ΒΚ, ΚΔ. The expression the so-called complements + indicates that this technical use of παραπληρώματα was not new, though it might not be universally known.

+ +
+
+ Proposition 44. +

To a given straight line to apply, in a given rectilineal angle, a parallelogram equal to a given triangle.

+ +

Let AB be the given straight line, C the given triangle and D the given rectilineal angle; thus it is required to apply to the given straight line AB, in an angle equal to the angle D, a parallelogram equal to the given triangle C.

+

Let the parallelogram BEFG be constructed equal to the triangle C, in the angle EBG which is equal to D [I. 42]; let it be placed so that BE is in a straight line with AB; letFG be drawn through to H, and let AH be drawn through A parallel to either BG or EF. [I. 31]

+

Let HB be joined.

+

Then, since the straight line HF falls upon the parallels AH, EF, the angles AHF, HFE are equal to two right angles. [I. 29] Therefore the angles BHG, GFE are less than two right angles; and straight lines produced indefinitely from angles less than two right angles meet; [Post. 5] therefore HB, FE, when produced, will meet.

+

Let them be produced and meet at K; through the point K let KL be drawn parallel to either EA or FH, [I. 31] and let HA, GB be produced to the points L, M.

+

Then HLKF is a parallelogram, HK is its diameter, and AG, ME are parallelograms. and LB, BF the so-called complements, about HK; therefore LB is equal to BF. [I. 43]

+

But BF is equal to the triangle C; therefore LB is also equal to C. [C.N. 1]

+

And, since the angle GBE is equal to the angle ABM, [I. 15] while the angle GBE is equal to D, the angle ABM is also equal to the angle D.

+

Therefore the parallelogram LB equal to the given triangle C has been applied to the given straight line AB, in the angle ABM which is equal to D.

+ +

Q. E. F.

+ + +

The verb is in the aorist (ὲνέπεσεν) here and in similar expressions in the following propositions.

+ +
+
+ Proposition 45. +

To construct, in a given rectilineal angle, a parallelogram equal to a given rectilineal figure.

+ +

Let ABCD be the given rectilineal figure and E the given rectilineal angle; thus it is required to construct, in the given angle E, a parallelogram equal to the rectilineal figure ABCD.

+

Let DB be joined, and let the parallelogram FH be constructed equal to the triangle ABD, in the angle HKF which is equal to E; [I. 42] let the parallelogram GM equal to the triangle DBC be applied to the straight line GH, in the angle GHM which is equal to E. [I. 44]

+

Then, since the angle E is equal to each of the angles HKF, GHM, the angle HKF is also equal to the angle GHM. [C.N. 1] +

+

Let the angle KHG be added to each; therefore the angles FKH, KHG are equal to the angles KHG, GHM.

+

But the angles FKH, KHG are equal to two right angles; [I. 29] therefore the angles KHG, GHM are also equal to two right angles.

+

Thus, with a straight line GH, and at the point H on it, two straight lines KH, HM not lying on the same side make the adjacent angles equal to two right angles; therefore KH is in a straight line with HM. [I. 14]

+

And, since the straight line HG falls upon the parallels KM, FG, the alternate angles MHG, HGF are equal to one another. [I. 29]

+

Let the angle HGL be added to each; therefore the angles MHG, HGL are equal to the angles HGF, HGL. [C.N. 2]

+

But the angles MHG, HGL are equal to two right angles; [I. 29] therefore the angles HGF, HGL are also equal to two right angles. [C.N. 1] Therefore FG is in a straight line with GL. [I. 14]

+

And, since FK is equal and parallel to HG, [I. 34] and HG to ML also, + KF is also equal and parallel to ML; [C.N. 1; I. 30] and the straight lines KM, FL join them (at their extremities); therefore KM, FL are also equal and parallel. [I. 33] Therefore KFLM is a parallelogram.

+

And, since the triangle ABD is equal to the parallelogram FH, and DBC to GM, + the whole rectilineal figure ABCD is equal to the whole parallelogram KFLM.

+

Therefore the parallelogram KFLM has been constructed equal to the given rectilineal figure ABCD, in the angle FKM which is equal to the given angle E.

+ +

Q. E. F.

+ + +

rectilineal + simply, without figure, + εὐθύγραμμον being here used as a substantive, like the similarly formed παραλληλόγραμμον.

+ +
+
+ Proposition 46. +

On a given straight line to describe a square.

+ +

Let AB be the given straight line; thus it is required to describe a square on the straight line AB.

+

+

Let AC be drawn at right angles to the straight line AB from the point A on it [I. 11], and let AD be made equal to AB; through the point D let DE be drawn parallel to AB, and through the point B let BE be drawn parallel to AD. [I. 31]

+

Therefore ADEB is a parallelogram; therefore AB is equal to DE, and AD to BE. [I. 34]

+

But AB is equal to AD; therefore the four straight lines BA, AD, DE, EB are equal to one another; therefore the parallelogram ADEB is equilateral.

+

I say next that it is also right-angled.

+

For, since the straight line AD falls upon the parallels AB, DE, the angles BAD, ADE are equal to two right angles. [I. 29]

+

But the angle BAD is right; therefore the angle ADE is also right.

+

And in parallelogrammic areas the opposite sides and angles are equal to one another; [I. 34] therefore each of the opposite angles ABE, BED is also right. Therefore ADEB is right-angled.

+

And it was also proved equilateral.

+

Therefore it is a square; and it is described on the straight line AB.

+ +

Q. E. F.

+ + +

Proclus (p. 423, 18 sqq.) notes the difference between the word construct (συστἡσασθαι) applied by Euclid to the construction of a triangle (and, he might have added, of an angle) and the words describe on (ἀναγράφειν ἀπό) used of drawing a square on a given straight line as one side. The triangle (or angle) is, so to say, pieced together, while the describing of a square on a given straight line is the making of a figure from + one side, and corresponds to the multiplication of the number representing the side by itself.

+ +
+
+ Proposition 47. +

In right-angled triangles the square on the side subtending the right angle is equal to the squares on the sides containing the right angle.

+ +

Let ABC be a right-angled triangle having the angle BAC right;

+

I say that the square on BC is equal to the squares on BA, AC.

+

For let there be described on BC the square BDEC, and on BA, AC the squares GB, HC; [I. 46] through A let AL be drawn parallel to either BD or CE, and let AD, FC be joined.

+

+

Then, since each of the angles BAC, BAG is right, it follows that with a straight line BA, and at the point A on it, the two straight lines AC, AG not lying on the same side make the adjacent angles equal to two right angles; therefore CA is in a straight line with AG. [I. 14] +

+

For the same reason BA is also in a straight line with AH.

+

And, since the angle DBC is equal to the angle FBA: for each is right: let the angle ABC be added to each; therefore the whole angle DBA is equal to the whole angle FBC. [C.N. 2] +

+

And, since DB is equal to BC, and FB to BA, the two sides AB, BD are equal to the two sides FB, BC respectively, and the angle ABD is equal to the angle FBC; therefore the base AD is equal to the base FC, and the triangle ABD is equal to the triangle FBC. [I. 4]

+

Now the parallelogram BL is double of the triangle ABD, for they have the same base BD and are in the same parallels BD, AL. [I. 41]

+

And the square GB is double of the triangle FBC, for they again have the same base FB and are in the same parallels FB, GC. [I. 41]

+

[But the doubles of equals are equal to one another.] Therefore the parallelogram BL is also equal to the square GB.

+

Similarly, if AE, BK be joined, the parallelogram CL can also be proved equal to the square HC; therefore the whole square BDEC is equal to the two squares GB, HC. [C.N. 2]

+

And the square BDEC is described on BC, and the squares GB, HC on BA, AC.

+

Therefore the square on the side BC is equal to the squares on the sides BA, AC.

+

Therefore etc.

+ +

Q. E. D.

+ + +

, τὸ ἀπὸ...τετρἁγωνον, the word ἀναγραφέν or ἀναγεγραμμένον being understood.

+ + +

subtending the right angle. Here ὑποτεινούσης, subtending, + is used with the simple accusative (τὴν ὀρθὴν γωνίαν) instead of being followed by ὑπό and the accusative, which seems to be the original and more orthodox construction. Cf. I. 18, note.

+ + +

Euclid actually writes DB, BA, + and therefore the equal sides in the two triangles are not mentioned in corresponding order, though he adheres to the words ἑκατέρα ἑκατέρα respectively. + Here DB is equal to BC and BA to FB.

+ + +

[But the doubles of equals are equal to one another.] Heiberg brackets these words as an interpolation, since it quotes a Common Notion which is itself interpolated. Cf. notes on I. 37, p. 332, and on interpolated Common Notions, pp. 223-4.

+ +
+
+ Proposition 48. +

If in a triangle the square on one of the sides be equal to the squares on the remaining two sides of the triangle, the angle contained by the remaining two sides of the triangle is right.

+ +

For in the triangle ABC let the square on one side BC be equal to the squares on the sides BA, AC;

+

I say that the angle BAC is right.

+

For let AD be drawn from the point A at right angles to the straight line AC, let AD be made equal to BA, and let DC be joined.

+

Since DA is equal to AB, the square on DA is also equal to the square on AB.

+

Let the square on AC be added to each; therefore the squares on DA, AC are equal to the squares on BA, AC.

+

But the square on DC is equal to the squares on DA, AC, for the angle DAC is right; [I. 47] and the square on BC is equal to the squares on BA, AC, for this is the hypothesis; therefore the square on DC is equal to the square on BC, so that the side DC is also equal to BC.

+

And, since DA is equal to AB, and AC is common, the two sides DA, AC are equal to the two sides BA, AC; and the base DC is equal to the base BC; therefore the angle DAC is equal to the angle BAC. [I. 8] But the angle DAC is right; therefore the angle BAC is also right.

+

Therefore etc.

+ +

Q. E. D.

+ +
+
+
+
+ + BOOK II. +
+ DEFINITIONS. +
+ 1 +

Any rectangular parallelogram is said to be contained by the two straight lines containing the right angle.

+
+
+ 2 +

And in any parallelogrammic area let any one whatever of the parallelograms about its diameter with the two complements be called a gnomon.

+
+
+
+ + BOOK II. PROPOSITIONS. +
+ Proposition 1. +

If there be two straight lines, and one of them be cut into any number of segments whatever, the rectangle contained by the two straight lines is equal to the rectangles contained by the uncut straight line and each of the segments.

+

Let A, BC be two straight lines, and let BC be cut at random at the points D, E; I say that the rectangle contained by A, BC is equal to the rectangle contained by A, BD, that contained by A, DE and that contained by A, EC.

+

For let BF be drawn from B at right angles to BC; [I. 11] let BG be made equal to A, [I. 3] through G let GH be drawn parallel to BC, [I. 31] and through D, E, C let DK, EL, CH be drawn parallel to BG.

+

Then BH is equal to BK, DL, EH.

+

Now BH is the rectangle A, BC, for it is contained by GB, BC, and BG is equal to A;

+

BK is the rectangle A, BD, for it is contained by GB, BD, and BG is equal to A; and DL is the rectangle A, DE, for DK, that is BG [I. 34], is equal to A.

+

Similarly also EH is the rectangle A, EC.

+

Therefore the rectangle A, BC is equal to the rectangle A, BD, the rectangle A, DE and the rectangle A, EC.

+

Therefore etc. Q. E. D. + + +

the rectangle A, BC. From this point onward I shall translate thus in cases where Euclid leaves out the word contained (περιεχόμενον). Though the word rectangle + is also omitted in the Greek (the neuter article being sufficient to show that the rectangle is meant), it cannot be dispensed with in English. De Morgan advises the use of the expression the rectangle under two lines. + This does not seem to me a very good expression, and, if used in a translation from the Greek, it might suggest that ὑπό in τὸ ὑπό meant under, which it does not.

+

+
+
+ Proposition 2. +

If a straight line be cut at random, the rectangle contained by the whole and both of the segments is equal to the square on the whole.

+

For let the straight line AB be cut at random at the point C; I say that the rectangle contained by AB, BC together with the rectangle contained by BA, AC is equal to the square on AB.

+

For let the square ADEB be described on AB [I. 46], and let CF be drawn through C parallel to either AD or BE. [I. 31]

+

Then AE is equal to AF, CE.

+

Now AE is the square on AB;

+

AF is the rectangle contained by BA, AC, for it is contained by DA, AC, and AD is equal to AB; and CE is the rectangle AB, BC, for BE is equal to AB.

+

Therefore the rectangle BA, AC together with the rectangle AB, BC is equal to the square on AB.

+

Therefore etc. Q. E. D.

+
+
+ Proposition 3. +

If a straight line be cut at random, the rectangle contained by the whole and one of the segments is equal to the rectangle contained by the segments and the square on the aforesaid segment.

+

For let the straight line AB be cut at random at C; I say that the rectangle contained by AB, BC is equal to the rectangle contained by AC, CB together with the square on BC.

+

For let the square CDEB be described on CB; [I. 46] let ED be drawn through to F, and through A let AF be drawn parallel to either CD or BE. [I. 31]

+

Then AE is equal to AD, CE.

+

Now AE is the rectangle contained by AB, BC, for it is contained by AB, BE, and BE is equal to BC;

+

AD is the rectangle AC, CB, for DC is equal to CB; and DB is the square on CB. Therefore the rectangle contained by AB, BC is equal to the rectangle contained by AC, CB together with the square on BC.

+

Therefore etc. Q. E. D.

+
+
+ Proposition 4. +

If a straight line be cut at random, the square on the whole is equal to the squares on the segments and twice the rectangle contained by the segments.

+

For let the straight line AB be cut at random at C; I say that the square on AB is equal to the squares on AC, CB and twice the rectangle contained by AC, CB.

+

For let the square ADEB be described on AB, [I. 46] let BD be joined; through C let CF be drawn parallel to either AD or EB, and through G let HK be drawn parallel to either AB or DE. [I. 31]

+

+

Then, since CF is parallel to AD, and BD has fallen on them, the exterior angle CGB is equal to the interior and opposite angle ADB. [I. 29]

+

But the angle ADB is equal to the angle ABD, since the side BA is also equal to AD; [I. 5] therefore the angle CGB is also equal to the angle GBC, so that the side BC is also equal to the side CG. [I. 6]

+

But CB is equal to GK, and CG to KB; [I. 34] therefore GK is also equal to KB; therefore CGKB is equilateral.

+

I say next that it is also right-angled.

+

For, since CG is parallel to BK, the angles KBC, GCB are equal to two right angles. [I. 29]

+

But the angle KBC is right; therefore the angle BCG is also right, so that the opposite angles CGK, GKB are also right. [I. 34]

+

Therefore CGKB is right-angled; and it was also proved equilateral; therefore it is a square; and it is described on CB.

+

For the same reason HF is also a square; and it is described on HG, that is AC. [I. 34]

+

Therefore the squares HF, KC are the squares on AC, CB.

+

Now, since AG is equal to GE, and AG is the rectangle AC, CB, for GC is equal to CB, therefore GE is also equal to the rectangle AC, CB.

+

Therefore AG, GE are equal to twice the rectangle AC, CB.

+

But the squares HF, CK are also the squares on AC, CB; therefore the four areas HF, CK, AG, GE are equal to the squares on AC, CB and twice the rectangle contained by AC, CB.

+

But HF, CK, AG, GE are the whole ADEB, which is the square on AB.

+

Therefore the square on AB is equal to the squares on AC, CB and twice the rectangle contained by AC, CB.

+

Therefore etc. Q. E. D. + +

twice the rectangle contained by the segments. By a curious idiom this is in Greek the rectangle twice contained by the segments. + Similarly twice the rectangle contained by AC, CB + is expressed as the rectangle twice contained by AC, CB + (τὸ δὶς ὑπὸ τῶν ΑΓ, ΓΒ περιεχόμενον ὀρθογώνιον).

+ + +

described. 39, 45. the squares (before on +). These words are not in the Greek, which simply says that the squares are on + (εἰσἱν ἀπό) their respective sides.

+ + +

areas. It is necessary to supply some substantive (the Greek leaves it to be understood); and I prefer areas + to figures. +

+

+
+
+ Proposition 5. +

If a straight line be cut into equal and unequal segments, the rectangle contained by the unequal segments of the whole together with the square on the straight line between the points of section is equal to the square on the half.

+

For let a straight line AB be cut into equal segments at C and into unequal segments at D; I say that the rectangle contained by AD, DB together with the square on CD is equal to the square on CB.

+

For let the square CEFB be described on CB, [I. 46] and let BE be joined; through D let DG be drawn parallel to either CE or BF, through H again let KM be drawn parallel to either AB or EF, and again through A let AK be drawn parallel to either CL or BM. [I. 31]

+

Then, since the complement CH is equal to the complement HF, [I. 43] let DM be added to each; therefore the whole CM is equal to the whole DF.

+

But CM is equal to AL, since AC is also equal to CB; [I. 36] therefore AL is also equal to DF. Let CH be added to each; therefore the whole AH is equal to the gnomon NOP. +

+

But AH is the rectangle AD, DB, for DH is equal to DB, therefore the gnomon NOP is also equal to the rectangle AD, DB.

+

Let LG, which is equal to the square on CD, be added to each; therefore the gnomon NOP and LG are equal to the rectangle contained by AD, DB and the square on CD.

+

But the gnomon NOP and LG are the whole square CEFB, which is described on CB; therefore the rectangle contained by AD, DB together with the square on CD is equal to the square on CB.

+

Therefore etc. Q. E. D. + +

between the points of section, literally between the sections, + the word being the same (τομή) as that used of a conic section.

+

It will be observed that the gnomon is indicated in the figure by three separate letters and a dotted curve. This is no doubt a clearer way of showing what exactly the gnomon is than the method usual in our text-books. In this particular case the figure of the MSS. has two M's in it, the gnomon being MNΞ. I have corrected the lettering to avoid confusion.

+

+
+
+ Proposition 6. +

If a straight line be bisected and a straight line be added to it in a straight line, the rectangle contained by the whole with the added straight line and the added straight line together with the square on the half is equal to the square on the straight line made up of the half and the added straight line.

+

For let a straight line AB be bisected at the point C, and let a straight line BD be added to it in a straight line;

+

I say that the rectangle contained by AD, DB together with the square on CB is equal to the square on CD.

+

For let the square CEFD be described on CD, [I. 46] and let DE be joined; through the point B let BG be drawn parallel to either EC or DF, through the point H let KM be drawn parallel to either AB or EF, and further through A let AK be drawn parallel to either CL or DM. [I. 31]

+ Then, since AC is equal to CB, AL is also equal to CH. [I. 36] But CH is equal to HF. [I. 43] Therefore AL is also equal to HF.

+

Let CM be added to each; therefore the whole AM is equal to the gnomon NOP.

+

But AM is the rectangle AD, DB, for DM is equal to DB; therefore the gnomon NOP is also equal to the rectangle AD, DB.

+

Let LG, which is equal to the square on BC, be added to each; therefore the rectangle contained by AD, DB together with the square on CB is equal to the gnomon NOP and LG.

+

But the gnomon NOP and LG are the whole square CEFD, which is described on CD; therefore the rectangle contained by AD, DB together with the square on CB is equal to the square on CD.

+

Therefore etc. Q. E. D.

+
+
+ Proposition 7. +

If a straight line be cut at random, the square on the whole and that on one of the segments both together are equal to twice the rectangle contained by the whole and the said segment and the square on the remaining segment.

+

For let a straight line AB be cut at random at the point C;

+

I say that the squares on AB, BC are equal to twice the rectangle contained by AB, BC and the square on CA.

+

For let the square ADEB be described on AB, [I. 46] and let the figure be drawn.

+

Then, since AG is equal to GE, [I. 43] let CF be added to each; therefore the whole AF is equal to the whole CE.

+

Therefore AF, CE are double of AF.

+

But AF, CE are the gnomon KLM and the square CF; therefore the gnomon KLM and the square CF are double of AF.

+

But twice the rectangle AB, BC is also double of AF; for BF is equal to BC; therefore the gnomon KLM and the square CF are equal to twice the rectangle AB, BC.

+

Let DG, which is the square on AC, be added to each; therefore the gnomon KLM and the squares BG, GD are equal to twice the rectangle contained by AB, BC and the square on AC.

+

But the gnomon KLM and the squares BG, GD are the whole ADEB and CF, which are squares described on AB, BC; therefore the squares on AB, BC are equal to twice the rectangle contained by AB, BC together with the square on AC.

+

Therefore etc. Q. E. D.

+
+
+ Proposition 8. +

If a straight line be cut at random, four times the rectangle contained by the whole and one of the segments together with the square on the remaining segment is equal to the square described on the whole and the aforesaid segment as on one straight line.

+

For let a straight line AB be cut at random at the point C;

+

I say that four times the rectangle contained by AB, BC together with the square on AC is equal to the square described on AB, BC as on one straight line.

+

For let [the straight line] BD be produced in a straight line [with AB], and let BD be made equal to CB; let the square AEFD be described on AD, and let the figure be drawn double.

+

Then, since CB is equal to BD, while CB is equal to GK, and BD to KN, therefore GK is also equal to KN.

+

For the same reason QR is also equal to RP.

+

And, since BC is equal to BD, and GK to KN, therefore CK is also equal to KD, and GR to RN. [I. 36]

+

But CK is equal to RN, for they are complements of the parallelogram CP; [I. 43] therefore KD is also equal to GR; therefore the four areas DK, CK, GR, RN are equal to one another. Therefore the four are quadruple of CK.

+

Again, since CB is equal to BD, while BD is equal to BK, that is CG, and CB is equal to GK, that is GQ, therefore CG is also equal to GQ.

+

And, since CG is equal to GQ, and QR to RP, AG is also equal to MQ, and QL to RF. [I. 36]

+

But MQ is equal to QL, for they are complements of the parallelogram ML; [I. 43] therefore AG is also equal to RF; therefore the four areas AG, MQ, QL, RF are equal to one another. Therefore the four are quadruple of AG. But the four areas CK, KD, GR, RN were proved to be quadruple of CK; therefore the eight areas, which contain the gnomon STU, are quadruple of AK.

+

Now, since AK is the rectangle AB, BD, for BK is equal to BD, therefore four times the rectangle AB, BD is quadruple of AK.

+

But the gnomon STU was also proved to be quadruple of AK; therefore four times the rectangle AB, BD is equal to the gnomon STU.

+

Let OH, which is equal to the square on AC, be added to each; therefore four times the rectangle AB, BD together with the square on AC is equal to the gnomon STU and OH.

+

But the gnomon STU and OH are the whole square AEFD, which is described on AD. therefore four times the rectangle AB, BD together with the square on AC is equal to the square on AD

+

But BD is equal to BC; therefore four times the rectangle contained by AB, BC together with the square on AC is equal to the square on AD, that is to the square described on AB and BC as on one straight line.

+

Therefore etc. Q. E. D.

+
+
+ Proposition 9. +

If a straight line be cut into equal and unequal segments, the squares on the unequal segments of the whole are double of the square on the half and of the square on the straight line between the points of section.

+

For let a straight line AB be cut into equal segments at C, and into unequal segments at D;

+

I say that the squares on AD, DB are double of the squares on AC, CD.

+

For let CE be drawn from C at right angles to AB, and let it be made equal to either AC or CB; let EA, EB be joined, let DF be drawn through D parallel to EC, and FG through F parallel to AB, and let AF be joined.

+

Then, since AC is equal to CE, the angle EAC is also equal to the angle AEC.

+

And, since the angle at C is right, the remaining angles EAC, AEC are equal to one right angle. [I. 32]

+

And they are equal; therefore each of the angles CEA, CAE is half a right angle.

+

For the same reason each of the angles CEB, EBC is also half a right angle; therefore the whole angle AEB is right.

+

And, since the angle GEF is half a right angle, and the angle EGF is right, for it is equal to the interior and opposite angle ECB, [I. 29] the remaining angle EFG is half a right angle; [I. 32] therefore the angle GEF is equal to the angle EFG, so that the side EG is also equal to GF. [I. 6]

+

Again, since the angle at B is half a right angle, and the angle FDB is right, for it is again equal to the interior and opposite angle ECB, [I. 29] the remaining angle BFD is half a right angle; [I. 32] therefore the angle at B is equal to the angle DFB, so that the side FD is also equal to the side DB. [I. 6]

+

Now, since AC is equal to CE, the square on AC is also equal to the square on CE; therefore the squares on AC, CE are double of the square on AC.

+

But the square on EA is equal to the squares on AC, CE, for the angle ACE is right; [I. 47] therefore the square on EA is double of the square on AC.

+

Again, since EG is equal to GF, the square on EG is also equal to the square on GF; therefore the squares on EG, GF are double of the square on GF.

+

But the square on EF is equal to the squares on EG, GF; therefore the square on EF is double of the square on GF.

+

But GF is equal to CD; [I. 34] therefore the square on EF is double of the square on CD.

+

But the square on EA is also double of the square on AC; therefore the squares on AE, EF are double of the squares on AC, CD.

+

And the square on AF is equal to the squares on AE, EF, for the angle AEF is right; [I. 47] therefore the square on AF is double of the squares on AC, CD.

+

But the squares on AD, DF are equal to the square on AF, for the angle at D is right; [I. 47] therefore the squares on AD, DF are double of the squares on AC, CD.

+

And DF is equal to DB; therefore the squares on AD, DB are double of the squares on AC, CD.

+

Therefore etc. Q. E. D.

+
+
+ Proposition 10. +

If a straight line be bisected, and a straight line be added to it in a straight line, the square on the whole with the added straight line and the square on the added straight line both together are double of the square on the half and of the square described on the straight line made up of the half and the added straight line as on one straight line.

+

For let a straight line AB be bisected at C, and let a straight line BD be added to it in a straight line;

+

I say that the squares on AD, DB are double of the squares on AC, CD.

+

For let CE be drawn from the point C at right angles to AB [I. 11], and let it be made equal to either AC or CB [I. 3]; let EA, EB be joined; through E let EF be drawn parallel to AD, and through D let FD be drawn parallel to CE. [I. 31]

+

Then, since a straight line EF falls on the parallel straight lines EC, FD, the angles CEF, EFD are equal to two right angles; [I. 29] therefore the angles FEB, EFD are less than two right angles.

+

But straight lines produced from angles less than two right angles meet; [I. Post. 5] therefore EB, FD, if produced in the direction B, D, will meet.

+

Let them be produced and meet at G, and let AG be joined.

+

Then, since AC is equal to CE, the angle EAC is also equal to the angle AEC; [I. 5] and the angle at C is right; therefore each of the angles EAC, AEC is half a right angle. [I. 32]

+

For the same reason each of the angles CEB, EBC is also half a right angle; therefore the angle AEB is right.

+

And, since the angle EBC is half a right angle, the angle DBG is also half a right angle. [I. 15] But the angle BDG is also right, for it is equal to the angle DCE, they being alternate; [I. 29] therefore the remaining angle DGB is half a right angle; [I. 32] therefore the angle DGB is equal to the angle DBG, so that the side BD is also equal to the side GD. [I. 6]

+

Again, since the angle EGF is half a right angle, and the angle at F is right, for it is equal to the opposite angle, the angle at C, [I. 34] the remaining angle FEG is half a right angle; [I. 32] therefore the angle EGF is equal to the angle FEG, so that the side GF is also equal to the side EF. [I. 6]

+

Now, since the square on EC is equal to the square on CA, the squares on EC, CA are double of the square on CA.

+

But the square on EA is equal to the squares on EC, CA; [I. 47] therefore the square on EA is double of the square on AC. [C. N. 1]

+

Again, since FG is equal to EF, the square on FG is also equal to the square on FE; therefore the squares on GF, FE are double of the square on EF.

+

But the square on EG is equal to the squares on GF, FE; [I. 47] therefore the square on EG is double of the square on EF.

+

And EF is equal to CD; [I. 34] therefore the square on EG is double of the square on CD. But the square on EA was also proved double of the square on AC; therefore the squares on AE, EG are double of the squares on AC, CD.

+

And the square on AG is equal to the squares on AE, EG; [I. 47] therefore the square on AG is double of the squares on AC, CD. But the squares on AD, DG are equal to the square on AG; [I. 47] therefore the squares on AD, DG are double of the squares on AC, CD.

+

And DG is equal to DB; therefore the squares on AD, DB are double of the squares on AC, CD.

+

Therefore etc. Q. E. D.

+
+
+ Proposition 11. +

To cut a given straight line so that the rectangle contained by the whole and one of the segments is equal to the square on the remaining segment.

+

Let AB be the given straight line; thus it is required to cut AB so that the rectangle contained by the whole and one of the segments is equal to the square on the remaining segment.

+

For let the square ABDC be described on AB; [I. 46] let AC be bisected at the point E, and let BE be joined; let CA be drawn through to F, and let EF be made equal to BE; let the square FH be described on AF, and let GH be drawn through to K.

+

I say that AB has been cut at H so as to make the rectangle contained by AB, BH equal to the square on AH.

+

For, since the straight line AC has been bisected at E, and FA is added to it, the rectangle contained by CF, FA together with the square on AE is equal to the square on EF. [II. 6]

+

But EF is equal to EB; therefore the rectangle CF, FA together with the square on AE is equal to the square on EB.

+

But the squares on BA, AE are equal to the square on EB, for the angle at A is right; [I. 47] therefore the rectangle CF, FA together with the square on AE is equal to the squares on BA, AE.

+

Let the square on AE be subtracted from each; therefore the rectangle CF, FA which remains is equal to the square on AB. +

+

Now the rectangle CF, FA is FK, for AF is equal to FG; and the square on AB is AD; therefore FK is equal to AD.

+

Let AK be subtracted from each; therefore FH which remains is equal to HD.

+

And HD is the rectangle AB, BH, for AB is equal to BD; and FH is the square on AH; therefore the rectangle contained by AB, BH is equal to the square on HA. therefore the given straight line AB has been cut at H so as to make the rectangle contained by AB, BH equal to the square on HA. Q. E. F.

+
+
+ Proposition 12. +

In obtuse-angled triangles the square on the side subtending the obtuse angle is greater than the squares on the sides containing the obtuse angle by twice the rectangle contained by one of the sides about the obtuse angle, namely that on which the + perpendicular falls, and the straight line cut off outside by the perpendicular towards the obtuse angle.

+

Let ABC be an obtuse-angled triangle having the angle BAC obtuse, and let BD be drawn from the point B perpendicular to CA produced;

+

I say that the square on BC is greater than the squares on BA, AC by twice the rectangle contained by CA, AD.

+

For, since the straight line CD has been cut at random at the point A, the square on DC is equal to the squares on CA, AD and twice the rectangle contained by CA, AD. [II. 4]

+

Let the square on DB be added to each; therefore the squares on CD, DB are equal to the squares on CA, AD, DB and twice the rectangle CA, AD.

+

But the square on CB is equal to the squares on CD, DB, for the angle at D is right; [I. 47] and the square on AB is equal to the squares on AD, DB; [I. 47] therefore the square on CB is equal to the squares on CA, AB and twice the rectangle contained by CA, AD; so that the square on CB is greater than the squares on CA, AB by twice the rectangle contained by CA, AD.

+

Therefore etc. Q. E. D.

+
+
+ Proposition 13. +

In acute-angled triangles the square on the side subtending the acute angle is less than the squares on the sides containing the acute angle by twice the rectangle contained by one of the sides about the acute angle, namely that on which the perpendicular falls, and the straight line cut off within by the perpendicular towards the acutc angle.

+

Let ABC be an acute-angled triangle having the angle at B acute, and let AD be drawn from the point A perpendicular to BC;

+

I say that the square on AC is less than the squares on CB, BA by twice the rectangle contained by CB, BD.

+

For, since the straight line CB has been cut at random at D, the squares on CB, BD are equal to twice the rectangle contained by CB, BD and the square on DC. [II. 7] +

+

Let the square on DA be added to each; therefore the squares on CB, BD, DA are equal to twice the rectangle contained by CB, BD and the squares on AD, DC.

+

But the square on AB is equal to the squares on BD, DA, for the angle at D is right; [I. 47] and the square on AC is equal to the squares on AD, DC; therefore the squares on CB, BA are equal to the square on AC and twice the rectangle CB, BD,

+

so that the square on AC alone is less than the squares on CB, BA by twice the rectangle contained by CB, BD.

+

Therefore etc. Q. E. D.

+
+
+ Proposition 14. +

To construct a square equal to a given rectilineal figure.

+

Let A be the given rectilineal figure; thus it is required to construct a square equal to the rectilineal figure A.

+

+

For let there be constructed the rectangular parallelogram BD equal to the rectilineal figure A. [I. 45]

+

Then, if BE is equal to ED, that which was enjoined will have been done; for a square BD has been constructed equal to the rectilineal figure A.

+

But, if not, one of the straight lines BE, ED is greater.

+

Let BE be greater, and let it be produced to F; let EF be made equal to ED, and let BF be bisected at G.

+

With centre G and distance one of the straight lines GB, GF let the semicircle BHF be described; let DE be produced to H, and let GH be joined.

+

Then, since the straight line BF has been cut into equal segments at G, and into unequal segments at E, the rectangle contained by BE, EF together with the square on EG is equal to the square on GF. [II. 5] +

+

But GF is equal to GH; therefore the rectangle BE, EF together with the square on GE is equal to the square on GH.

+

But the squares on HE, EG are equal to the square on GH; [I. 47] therefore the rectangle BE, EF together with the square on GE is equal to the squares on HE, EG.

+

Let the square on GE be subtracted from each; therefore the rectangle contained by BE, EF which remains is equal to the square on EH. +

+

But the rectangle BE, EF is BD, for EF is equal to ED; therefore the parallelogram BD is equal to the square on HE.

+

And BD is equal to the rectilineal figure A.

+

Therefore the rectilineal figure A is also equal to the square which can be described on EH.

+

Therefore a square, namely that which can be described on EH, has been constructed equal to the given rectilineal figure A. Q. E. F.

+ +

that which was enjoined will have been done, literally would have been done, + γεγονὸς ἂν εἴη τὸ ἐπιταχθέν.

+ + +

which can be described, expressed by the future passive participle, ἀναγραφησομἑνῳ, ἀναγραφησόμενον.

+ +
+
+
+ +
+ + BOOK III. +
+ DEFINITIONS. +
+ 1 +

Equal circles are those the diameters of which are equal, or the radii of which are equal.

+
+
+ 2 +

A straight line is said to touch a circle which, meeting the circle and being produced, does not cut the circle.

+
+
+ 3 +

Circles are said to touch one another which, meeting one another, do not cut one another.

+
+
+ 4 +

In a circle straight lines are said to be equally distant from the centre when the perpendiculars drawn to them from the centre are equal.

+
+
+5 +

And that straight line is said to be at a greater distance on which the greater perpendicular falls.

+
+
+6 +

A segment of a circle is the figure contained by a straight line and a circumference of a circle.

+
+
+7 +

An angle of a segment is that contained by a straight line and a circumference of a circle.

+
+
+8 +

An angle in a segment is the angle which, when a point is taken on the circumference of the segment and straight lines are joined from it to the extremities of the straight line which is the base of the segment, is contained by the straight lines so joined.

+
+
+9 +

And, when the straight lines containing the angle cut off a circumference, the angle is said to stand upon that circumference.

+
+
+10 + +

A sector of a circle is the figure which, when an angle is constructed at the centre of the circle, is contained by the straight lines containing the angle and the circumference cut off by them.

+
+
+11 +

Similar segments of circles are those which admit equal angles, or in which the angles are equal to one another.

+
+
+
+ + BOOK III. PROPOSITIONS. +
+ PROPOSITION 1. +

To find the centre of a given circle.

+

Let ABC be the given circle; thus it is required to find the centre of the circle ABC.

+

Let a straight line AB be drawn through it at random, and let it be bisected at the point D; from D let DC be drawn at right angles to AB and let it be drawn through to E; let CE be bisected at F; I say that F is the centre of the circle ABC.

+

+

For suppose it is not, but, if possible, let G be the centre, and let GA, GD, GB be joined.

+

Then, since AD is equal to DB, and DG is common, the two sides AD, DG are equal to the two sides BD, DG respectively; and the base GA is equal to the base GB, for they are radii; therefore the angle ADG is equal to the angle GDB. [I. 8] +

+

But, when a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right; [I. Def. 10] therefore the angle GDB is right. +

+

But the angle FDB is also right; therefore the angle FDB is equal to the angle GDB, the greater to the less: which is impossible. Therefore G is not the centre of the circle ABC. +

+

Similarly we can prove that neither is any other point except F. Therefore the point F is the centre of the circle ABC. +

+ +

Porism. From this it is manifest that, if in a circle a straight line cut a straight line into two equal parts and at right angles, the centre of the circle is on the cutting straight line. Q. E. F.

+ + +

12. For suppose it is not. This is expressed in the Greek by the two words Μὴ γάρ, but such an elliptical phrase is impossible in English.

+ + +

the two sides AD, DG are equal to the two sides BD, DG respectively. As before observed, Euclid is not always careful to put the equals in corresponding order. The text here has GD, DB. +

+ +
+
+ PROPOSITION 2. +

If on the circumference of a circle two points be taken at random, the straight line joining the points will fall within the circle.

+

Let ABC be a circle, and let two points A, B be taken at random on its circumference; I say that the straight line joined from A to B will fall within the circle.

+

+

For suppose it does not, but, if possible, let it fall outside, as AEB; let the centre of the circle ABC be taken [III. 1], and let it be D; let DA, DB be joined, and let DFE be drawn through.

+

Then, since DA is equal to DB, the angle DAE is also equal to the angle DBE. [I. 5] And, since one side AEB of the triangle DAE is produced, the angle DEB is greater than the angle DAE. [I. 16] +

+

But the angle DAE is equal to the angle DBE; therefore the angle DEB is greater than the angle DBE. And the greater angle is subtended by the greater side; [I. 19] therefore DB is greater than DE. But DB is equal to DF; + therefore DF is greater than DE, + the less than the greater : which is impossible. +

+

Therefore the straight line joined from A to B will not fall outside the circle.

+

Similarly we can prove that neither will it fall on the circumference itself; therefore it will fall within. +

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 3. +

If in a circle a straight line through the centre bisect a straight line not through the centre, it also cuts it at right angles; and if it cut it at right angles, it also bisects it.

+

Let ABC be a circle, and in it let a straight line CD + through the centre bisect a straight line AB not through the centre at the point F;

I say that it also cuts it at right angles.

+

For let the centre of the circle ABC + be taken, and let it be E; let EA, EB be joined.

+

Then, since AF is equal to FB, and FE is common, two sides are equal to two sides; + and the base EA is equal to the base EB; therefore the angle AFE is equal to the angle BFE. [I. 8] +

+

But, when a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right; [I. Def. 10] therefore each of the angles AFE, BFE is right. +

+

Therefore CD, which is through the centre, and bisects AB which is not through the centre, also cuts it at right angles.

+

Again, let CD cut AB at right angles; I say that it also bisects it. that is, that AF is equal to FB.

+

For, with the same construction, since EA is equal to EB, the angle EAF is also equal to the angle EBF. [I. 5]

+

But the right angle AFE is equal to the right angle BFE, therefore EAF, EBF are two triangles having two angles equal to two angles and one side equal to one side, namely EF, which is common to them, and subtends one of the equal angles; therefore they will also have the remaining sides equal to the remaining sides; [I. 26] therefore AF is equal to FB. +

+

Therefore etc. Q. E. D. + +

with the same construction, τῶν αὐτῶν κατασκευασθέντων.

+

+
+
+ PROPOSITION 4. +

If in a circle two straight lines cut one another which are not through the centre, they do not bisect one another.

+

Let ABCD be a circle, and in it let the two straight lines AC, BD, which are not through the centre, cut one another at E;

I say that they do not bisect one another.

+

For, if possible, let them bisect one another, so that AE is equal to EC, and BE to ED; let the centre of the circle ABCD be taken [III. 1], and let it be F; let FE be joined.

+

Then, since a straight line FE through the centre bisects a straight line AC not through the centre, it also cuts it at right angles; [III. 3] therefore the angle FEA is right. +

+

Again, since a straight line FE bisects a straight line BD, it also cuts it at right angles; [III. 3] therefore the angle FEB is right. +

+

But the angle FEA was also proved right; therefore the angle FEA is equal to the angle FEB, the less to the greater: which is impossible. +

+

Therefore AC, BD do not bisect one another.

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 5. +

If two circles cut one another, they will not have the same centre.

+

For let the circles ABC, CDG cut one another at the points B, C; I say that they will not have the same centre.

+

+

For, if possible, let it be E; let EC be joined, and let EFG be drawn through at random.

+

Then, since the point E is the centre of the circle ABC, EC is equal to EF. [I. Def. 15] +

+

Again, since the point E is the centre of the circle CDG, EC is equal to EG. +

+

But EC was proved equal to EF also; therefore EF is also equal to EG, the less to the greater : which is impossible. +

+

Therefore the point E is not the centre of the circles ABC, CDG.

+

Therefore etc. Q. E. D.

+
+
+ + PROPOSITION 6. +

If two circles touch one another, they will not have the same centre.

+

For let the two circles ABC, CDE touch one another at the point C; I say that they will not have the same centre.

+

+

For, if possible, let it be F; let FC be joined, and let FEB be drawn through at random.

+

Then, since the point F is the centre of the circle ABC, FC is equal to FB. +

+

Again, since the point F is the centre of the circle CDE, FC is equal to FE. +

+

But FC was proved equal to FB; therefore FE is also equal to FB, the less to the greater: which is impossible. +

+

Therefore F is not the centre of the circles ABC, CDE.

+

Therefore etc. Q. E. D.

+
+
+ + PROPOSITION 7. +

If on the diameter of a circle a point be taken which is not the centre of the circle, and from the point straight lines fall upon the circle, that will be greatest on which the centre is, the remainder of the same diameter will be least, and of the rest + the nearer to the straight line through the centre is always greater than the more remote, and only two equal straight lines will fall from the point on the circle, one on each side of the least straight line.

+

Let ABCD be a circle, and let AD be a diameter of it; on AD let a point F be taken which is not the centre of the circle, let E be the centre of the circle, and from F let straight lines FB, FC, FG fall upon the circle ABCD; I say that FA is greatest, FD is least, and of the rest FB is greater than FC, and FC than FG.

+

For let BE, CE, GE be joined.

+

+

Then, since in any triangle two sides are greater than the remaining one, [I. 20] EB, EF are greater than BF. +

+

But AE is equal to BE; therefore AF is greater than BF. +

+

Again, since BE is equal to CE, and FE is common, the two sides BE, EF are equal to the two sides CE, EF.

+

But the angle BEF is also greater than the angle CEF; therefore the base BF is greater than the base CF. [I. 24]

+

For the same reason CF is also greater than FG. +

+

Again, since GF, FE are greater than EG, and EG is equal to ED, GF, FE are greater than ED. +

+

Let EF be subtracted from each; therefore the remainder GF is greater than the remainder FD. +

+

Therefore FA is greatest, FD is least, and FB is greater than FC, and FC than FG.

+

I say also that from the point F only two equal straight lines will fall on the circle ABCD, one on each side of the least FD.

+

For on the straight line EF, and at the point E on it, let the angle FEH be constructed equal to the angle GEF [I. 23], and let FH be joined.

+

Then, since GE is equal to EH, and EF is common, the two sides GE, EF are equal to the two sides HE, EF; and the angle GEF is equal to the angle HEF; + therefore the base FG is equal to the base FH. [I. 4] +

+

I say again that another straight line equal to FG will not fall on the circle from the point F.

+

For, if possible, let FK so fall.

+

Then, since FK is equal to FG, and FH to FG, FK is also equal to FH, the nearer to the straight line through the centre being thus equal to the more remote: which is impossible. +

+

Therefore another straight line equal to GF will not fall from the point F upon the circle; therefore only one straight line will so fall. +

+

Therefore etc. Q. E. D. + +

of the same diameter. I have inserted these words for clearness' sake. The text has simply ἐλαχίστη δὲ ἡ λοιπή, and the remaining (straight line) least. +

+ + +

one on each side. The word one + is not in the Greek, but is necessary to give the force of ἐφ̓ ἑκάτερα τῆς ἐλαχίστης, literally on both sides, + or on each of the two sides, of the least. +

+

+
+
+ PROPOSITION 8. +

If a point be taken outside a circle and from the point straight lines be drawn through to the circle, one of which is through the centre and the others are drawn at random, then, of the straight lines which fall on the concave circumference, that through the centre is greatest, while of the rest + the nearer to that through the centre is always greater than the more remote, but, of the straight lines falling on the convex circumference, that between the point and the diameter is least, while of the rest the nearer to the least is always less than the more remote, and only two equal straight lines will fall on the circle from the point, one on each side of the least.

+

Let ABC be a circle, and let a point D be taken outside ABC; let there be drawn through from it straight lines DA, DE, DF, DC, and let DA be through the centre; I say that, of the straight lines falling on the concave circumference AEFC, the straight line DA through the centre is greatest,

while DE is greater than DF and DF than DC.; but, of the straight lines falling on the convex circumference HLKG, the straight line DG between the point and the diameter AG is least; and the nearer to the least DG is always less than the more remote, namely DK than DL, and DL than DH.

+

For let the centre of the circle ABC be taken [III. 1], and let it be M; let ME, MF, MC, MK, ML, MH be joined.

+

Then, since AM is equal to EM, let MD be added to each; therefore AD is equal to EM, MD. +

+

But EM, MD are greater than ED; [I. 20] therefore AD is also greater than ED. +

+

Again, since ME is equal to MF, and MD is common, therefore EM, MD are equal to FM, MD; and the angle EMD is greater than the angle FMD; + therefore the base ED is greater than the base FD. [I. 24] +

+

Similarly we can prove that FD is greater than CD; therefore DA is greatest, while DE is greater than DF, and DF than DC.

+

Next, since MK, KD are greater than MD, [I. 20] and MG is equal to MK, therefore the remainder KD is greater than the remainder GD, so that GD is less than KD. +

+

And, since on MD, one of the sides of the triangle MLD, two straight lines MK, KD were constructed meeting within the triangle, therefore MK, KD are less than ML, LD; [I. 21] and MK is equal to ML; therefore the remainder DK is less than the remainder DL. +

+

Similarly we can prove that DL is also less than DH; therefore DG is least, while DK is less than DL, and DL than DH. +

+

I say also that only two equal straight lines will fall from the point D on the circle, one on each side of the least DG.

+

On the straight line MD, and at the point M on it, let the angle DMB be constructed equal to the angle KMD, and let DB be joined.

+

Then, since MK is equal to MB, and MD is common, the two sides KM, MD are equal to the two sides BM, MD respectively; and the angle KMD is equal to the angle BMD; therefore the base DK is equal to the base DB. [I. 4] +

+

I say that no other straight line equal to the straight line DK will fall on the circle from the point D.

+

For, if possible, let a straight line so fall, and let it be DN. Then, since DK is equal to DN, +

+

while DK is equal to DB, DB is also equal to DN, that is, the nearer to the least DG equal to the more remote: which was proved impossible.

+

Therefore no more than two equal straight lines will fall on the circle ABC from the point D, one on each side of DG the least.

+

Therefore etc.

+
+
+ PROPOSITION 9. +

If a point be taken within a circle, and more than two equal straight lines fall from the point on the circle, the point taken is the centre of the circle.

+

Let ABC be a circle and D a point within it, and from D let more than two equal straight lines, namely DA, DB, DC, fall on the circle ABC; I say that the point D is the centre of the circle ABC.

+

+

For let AB, BC be joined and bisected at the points E, F, and let ED, FD be joined and drawn through to the points G, K, H, L.

+

Then, since AE is equal to EB, and ED is common, the two sides AE, ED are equal to the two sides BE, ED; and the base DA is equal to the base DB; therefore the angle AED is equal to the angle BED. [I. 8] +

+

Therefore each of the angles AED, BED is right; [I. Def. 10] therefore GK cuts AB into two equal parts and at right angles.

+

And since, if in a circle a straight line cut a straight line into two equal parts and at right angles, the centre of the circle is on the cutting straight line, [III. 1, Por.] the centre of the circle is on GK. +

+

For the same reason the centre of the circle ABC is also on HL. +

+

And the straight lines GK, HL have no other point common but the point D; therefore the point D is the centre of the circle ABC. +

+

Therefore etc. Q. E. D.

+
+
+ + PROPOSITION 10. +

A circle does not cut a circle at more points than two.

+

For, if possible, let the circle ABC cut the circle DEF at more points than two, namely B, C, F, H;

+

let BH, BG be joined and bisected at the points K, L, and from K, L let KC, LM be drawn at right angles to BH, BG and carried through to the points A, E.

+

+

Then, since in the circle ABC a straight line AC cuts a straight line BH into two equal parts and at right angles, the centre of the circle ABC is on AC. [III. 1, Por.] +

+

Again, since in the same circle ABC a straight line NO cuts a straight line BG into two equal parts and at right angles, the centre of the circle ABC is on NO. +

+

But it was also proved to be on AC, and the straight lines AC, NO meet at no point except at P; therefore the point P is the centre of the circle ABC. +

+

Similarly we can prove that P is also the centre of the circle DEF; therefore the two circles ABC, DEF which cut one another have the same centre P: which is impossible. [III. 5] +

+

Therefore etc. Q. E. D. + +

The word circle (κύκλος) is here employed in the unusual sense of the circumference (περιφέρεια) of a circle. Cf. note on I. Def. 15.

+

+
+
+ PROPOSITION 11. +

If two circles touch one another internally, and their centres be taken, the straight line joining their centres, if it be also produced, will fall on the point of contact of the circles.

+

For let the two circles ABC, ADE touch one another internally at the point A, and let the centre F of the circle ABC, and the centre G of ADE, be taken; I say that the straight line joined from G to F and produced will fall on A.

+

+

For suppose it does not, but, if possible, let it fall as FGH, and let AF, AG be joined.

+

Then, since AG, GF are greater than FA, that is, than FH,

+

let FG be subtracted from each; therefore the remainder AG is greater than the remainder GH.

+

But AG is equal to GD; therefore GD is also greater than GH, the less than the greater: which is impossible. +

+

Therefore the straight line joined from F to G will not fall outside; therefore it will fall at A on the point of contact. +

+

Therefore etc. Q. E. D. + +

the straight line joining their centres, literally the straight line joined to their centres + (ἡ ἐπὶ τὰ κέντρα αὐτῶν ἐπιζευγνυμένη εὐθεῖα).

+ + +

point of contact is here συναφἡ, and in the enunciation of the next proposition ἐπαφἡ.

+

+
+
+ PROPOSITION 12. +

If two circles touch one another externally, the straight line joining their centres will pass through the point of contact.

+

For let the two circles ABC, ADE touch one another externally at the point A, and let the centre F of ABC, and the centre G of ADE, be taken; I say that the straight line joined from F to G will pass through the point of contact at A.

+

For suppose it does not, but, if possible, let it pass as FCDG, and let AF, AG be joined.

+

+

Then, since the point F is the centre of the circle ABC, FA is equal to FC. +

+

Again, since the point G is the centre of the circle ADE, GA is equal to GD. +

+

But FA was also proved equal to FC; therefore FA, AG are equal to FC, GD, so that the whole FG is greater than FA, AG; but it is also less [I. 20]: which is impossible.

+

Therefore the straight line joined from F to G will not fail to pass through the point of contact at A; therefore it will pass through it. +

+

Therefore etc. Q. E. D.] + +

will not fail to pass. The Greek has the double negative, οὐκ ἄρα ἡ...εὐθεῖα... οὐκ ἐλεύσεται, literally the straight line...will not not-pass.... +

+

+
+
+ PROPOSITION 13. +

A circle does not touch a circle at more points than one, whether it touch it internally or externally.

+

For, if possible, let the circle ABDC touch the circle EBFD, first internally, at more points than one, namely D, B.

+

Let the centre G of the circle ABDC, and the centre H of EBFD, be taken.

+

Therefore the straight line

+ joined from G to H will fall on B, D. [III. 11]

+

Let it so fall, as BGHD.

+

Then, since the point G is the centre of the circle ABCD, BG is equal to GD; + therefore BG is greater than HD; therefore BH is much greater than HD. +

+

Again, since the point H is the centre of the circle EBFD, BH is equal to HD; + but it was also proved much greater than it: which is impossible. +

+

Therefore a circle does not touch a circle internally at more points than one.

+

I say further that neither does it so touch it externally.

+

For, if possible, let the circle ACK touch the circle ABDC at more points than one, namely A, C, and let AC be joined.

+

Then, since on the circumference of each of the circles ABDC, ACK two points A, C have been taken at random, the straight line joining the points will fall within each circle; [III. 2] but it fell within the circle ABCD and outside ACK [III. Def. 3]: which is absurd. +

+

Therefore a circle does not touch a circle externally at more points than one.

+

And it was proved that neither does it so touch it internally.

+

Therefore etc. Q. E. D. + +

ABDC. Euclid writes ABCD (here and in the next proposition), notwithstanding the order in which the points are placed in the figure.

+ + +

does it so touch it. It is necessary to supply these words which the Greek (ὅτι οὐδὲ ἐκτός and ὅτι οὐδὲ ἐντός) leaves to be understood.

+

+
+
+ PROPOSITION 14. +

In a circle equal straight lines are equally distant from the centre, and those which are equally distant from the centre are equal to one another.

+

Let ABDC be a circle, and let AB, CD be equal straight lines in it; I say that AB, CD are equally distant from the centre.

+

For let the centre of the circle ABDC be taken [III. 1], and let it be E; from E let EF, EG be drawn perpendicular to AB, CD, and let AE, EC be joined.

+

Then, since a straight line EF through the centre cuts a straight line AB not through the centre at right angles, it also bisects it. [III. 3]

+ Therefore AF is equal to FB; therefore AB is double of AF. +

+

For the same reason CD is also double of CG; and AB is equal to CD; therefore AF is also equal to CG. +

+

And, since AE is equal to EC, the square on AE is also equal to the square on EC. But the squares on AF, EF are equal to the square on AE, for the angle at F is right; and the squares on EG, GC are equal to the square on EC, for the angle at G is right; [I. 47] therefore the squares on AF, FE are equal to the squares on CG, GE, of which the square on AF is equal to the square on CG, for AF is equal to CG; therefore the square on FE which remains is equal to the square on EG, therefore EF is equal to EG. +

+

But in a circle straight lines are said to be equally distant from the centre when the perpendiculars drawn to them from the centre are equal; [III. Def. 4] therefore AB, CD are equally distant from the centre. +

+

Next, let the straight lines AB, CD be equally distant from the centre; that is, let EF be equal to EG.

+

I say that AB is also equal to CD.

+

For, with the same construction, we can prove, similarly, that AB is double of AF, and CD of CG.

+

And, since AE is equal to CE, the square on AE is equal to the square on CE. But the squares on EF, FA are equal to the square on AE, and the squares on EG, GC equal to the square on CE. [I. 47]

+

Therefore the squares on EF, FA are equal to the squares on EG, GC, of which the square on EF is equal to the square on EG, for EF is equal to EG; therefore the square on AF which remains is equal to the square on CG; therefore AF is equal to CG. And AB is double of AF, and CD double of CG; therefore AB is equal to CD. +

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 15. +

Of straight lines in a circle the diameter is greatest, and of the rest the nearer to the centre is always greater than the more remote.

+

Let ABCD be a circle, let AD be its diameter and E the centre; and let BC be nearer to the diameter AD, and FG more remote; I say that AD is greatest and BC greater than FG.

+

For from the centre E let EH, EK +

be drawn perpendicular to BC, FG.

+

Then, since BC is nearer to the centre and FG more remote, EK is greater than EH. [III. Def. 5]

+

Let EL be made equal to EH, through L let LM be drawn at right angles to EK and carried through to N, and let ME, EN, FE, EG be joined.

+

Then, since EH is equal to EL, BC is also equal to MN. [III. 14] +

+

Again, since AE is equal to EM, and ED to EN, AD is equal to ME, EN. +

+

But ME, EN are greater than MN, [I. 20] and MN is equal to BC; therefore AD is greater than BC. +

+

And, since the two sides ME, EN are equal to the two sides FE, EG, and the angle MEN greater than the angle FEG, therefore the base MN is greater than the base FG. [I. 24]

+

But MN was proved equal to BC.

+

Therefore the diameter AD is greatest and BC greater than FG.

+

Therefore etc. Q. E. D. + +

Of straight lines. The Greek leaves these words to be understood.

+ + +

Nearer to the diameter AD. As BC, FG are not in general parallel to AD, Euclid should have said nearer to the centre. +

+

+
+
+ PROPOSITION 16. +

The straight line drawn at right angles to the diameter of a circle from its extremity will fall outside the circle, and into the space between the straight line and the circumference another straight line cannot be interposed; further the angle of the semicircle is greater, and the remaining angle less, than any acute rectilineal angle.

+

Let ABC be a circle about D as centre and AB as diameter; I say that the straight line drawn from A at right angles to AB from its extremity will fall outside the circle.

+

For suppose it does not, but, if possible, let it fall within as CA, and let DC be joined.

+

Since DA is equal to DC,

+ the angle DAC is also equal to the angle ACD. [I. 5] +

+

But the angle DAC is right; therefore the angle ACD is also right: thus, in the triangle ACD, the two angles DAC, ACD are equal to two right angles: which is impossible. [I. 17]

+

Therefore the straight line drawn from the point A at right angles to BA will not fall within the circle.

+

Similarly we can prove that neither will it fall on the circumference; therefore it will fall outside. +

+

Let it fall as AE; I say next that into the space between the straight line AE and the circumference CHA another straight line cannot be interposed.

+

For, if possible, let another straight line be so interposed, as FA, and let DG be drawn from the point D perpendicular to FA.

+

Then, since the angle AGD is right, and the angle DAG is less than a right angle, AD is greater than DG. [I. 19] +

+

But DA is equal to DH; therefore DH is greater than DG, the less than the greater: which is impossible. +

+

Therefore another straight line cannot be interposed into the space between the straight line and the circumference.

+

I say further that the angle of the semicircle contained by the straight line BA and the circumference CHA is greater than any acute rectilineal angle, and the remaining angle contained by the circumference CHA and the straight line AE is less than any acute rectilineal angle.

+

For, if there is any rectilineal angle greater than the angle contained by the straight line BA and the circumference CHA, and any rectilineal angle less than the angle contained by the circumference CHA and the straight line AE, then into the space between the circumference and the straight line AE a straight line will be interposed such as will make an angle contained by straight lines which is greater than the angle contained by the straight line BA and the circumference CHA, and another angle contained by straight lines which is less than the angle contained by the circumference CHA and the straight line AE.

+

But such a straight line cannot be interposed;

+

therefore there will not be any acute angle contained by straight lines which is greater than the angle contained by the straight line BA and the circumference CHA, nor yet any acute angle contained by straight lines which is less than the angle contained by the circumference CHA and the straight line AE.—

+ +

Porism. From this it is manifest that the straight line drawn at right angles to the diameter of a circle from its extremity touches the circle. Q. E. D.

+ + +

cannot be interposed, literally will not fall in between + (οὐ παρεμπεσεῖται).

+ +
+
+ PROPOSITION 17. +

From a given point to draw a straight line touching a given circle.

+

Let A be the given point, and BCD the given circle; thus it is required to draw from the point A a straight line touching the circle BCD.

+

For let the centre E of the circle be taken; [III. 1] let AE be joined, and with centre E and distance EA let the circle AFG be described;

from D let DF be drawn at right angles to EA, and let EF, AB be joined; I say that AB has been drawn from the point A touching the circle BCD.

+

For, since E is the centre of the circles BCD, AFG, EA is equal to EF, and ED to EB; therefore the two sides AE, EB are equal to the two sides FE, ED: and they contain a common angle, the angle at E; therefore the base DF is equal to the base AB, and the triangle DEF is equal to the triangle BEA, and the remaining angles to the remaining angles; [I. 4] therefore the angle EDF is equal to the angle EBA. +

+

But the angle EDF is right; therefore the angle EBA is also right. +

+

Now EB is a radius; and the straight line drawn at right angles to the diameter of a circle, from its extremity, touches the circle; [III. 16, Por.] therefore AB touches the circle BCD. +

+

Therefore from the given point A the straight line AB has been drawn touching the circle BCD.

+
+
+ PROPOSITION 18. +

If a straight line touch a circle, and a straight line be joined from the centre to the point of contact, the straight line so joined will be perpendicular to the tangent.

+

For let a straight line DE touch the circle ABC at the point C, let the centre F of the circle ABC be taken, and let FC be joined from F to C; I say that FC is perpendicular to DE.

+

+

For, if not, let FG be drawn from F perpendicular to DE.

+

Then, since the angle FGC is right, the angle FCG is acute; [I. 17] and the greater angle is subtended by the greater side; [I. 19] therefore FC is greater than FG. +

+

But FC is equal to FB; therefore FB is also greater than FG, the less than the greater: which is impossible. +

+

Therefore FG is not perpendicular to DE.

+

Similarly we can prove that neither is any other straight line except FC; therefore FC is perpendicular to DE. +

+

Therefore etc. Q. E. D. + + +

the tangent, ἡ ἐφαπτομένη.

+ +

+
+
+ PROPOSITION 19. +

If a straight line touch a circle, and from the point of contact a straight line be drawn at right angles to the tangent, the centre of the circle will be on the straight line so drawn.

+

For let a straight line DE touch the circle ABC at the point C, and from C let CA be drawn at right angles to DE; I say that the centre of the circle is on AC.

+

For suppose it is not, but, if

possible, let F be the centre, and let CF be joined.

+

Since a straight line DE touches the circle ABC, and FC has been joined from the centre to the point of contact, FC is perpendicular to DE; [III. 18] therefore the angle FCE is right. +

+

But the angle ACE is also right; therefore the angle FCE is equal to the angle ACE, the less to the greater: which is impossible. +

+

Therefore F is not the centre of the circle ABC.

+

Similarly we can prove that neither is any other point except a point on AC.

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 20. +

In a circle the angle at the centre is double of the angle at the circumference, when the angles have the same circumference as base.

+

Let ABC be a circle, let the angle BEC be an angle at its centre, and the angle BAC an angle at the circumference, and let them have the same circumference BC as base; I say that the angle BEC is double of the angle BAC.

+

+

For let AE be joined and drawn through to F.

+

Then, since EA is equal to EB, the angle EAB is also equal to the angle EBA; [I. 5] therefore the angles EAB, EBA are double of the angle EAB.

+

But the angle BEF is equal to the angles EAB, EBA; [I. 32] therefore the angle BEF is also double of the angle EAB.

+

For the same reason the angle FEC is also double of the angle EAC. +

+

Therefore the whole angle BEC is double of the whole angle BAC.

+

Again let another straight line be inflected, and let there be another angle BDC; let DE be joined and produced to G.

+

Similarly then we can prove that the angle GEC is double of the angle EDC, of which the angle GEB is double of the angle EDB; therefore the angle BEC which remains is double of the angle BDC. +

+

Therefore etc. Q. E. D. + +

let another straight line be inflected, κεκλάσθω δὴ πάλιν (without εὐθεῖα). The verb κλάω (to break off) was the regular technical term for drawing from a point a (broken) straight line which first meets another straight line or curve and is then bent back from it to another point, or (in other words) for drawing straight lines from two points meeting at a point on a curve or another straight line. κεκλάσθαι is one of the geometrical terms the definition of which must according to Aristotle be assumed (Anal. Post. I. 10, 76 b 9).

+

+
+
+ PROPOSITION 21. +

In a circle the angles in the same segment are equal to one another.

+

Let ABCD be a circle, and let the angles BAD, BED be angles in the same segment BAED; I say that the angles BAD, BED are equal to one another.

+

For let the centre of the circle ABCD be taken, and let it be F; let

+ BF, FD be joined.

+

Now, since the angle BFD is at the centre, and the angle BAD at the circumference, and they have the same circumference BCD as base, therefore the angle BFD is double of the angle BAD. [III. 20] +

+

For the same reason the angle BFD is also double of the angle BED; therefore the angle BAD is equal to the angle BED. +

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 22. +

The opposite angles of quadrilaterals in circles are equal to two right angles.

+

Let ABCD be a circle, and let ABCD be a quadrilateral in it; I say that the opposite angles are equal to two right angles.

+

Let AC, BD be joined.

+

Then, since in any triangle the three angles are equal to two right angles, [I. 32] the three angles CAB, ABC, BCA of the triangle ABC are equal to two right angles. +

+

But the angle CAB is equal to the angle BDC, for they are in the same segment BADC; [III. 21] and the angle ACB is equal to the angle

+ ADB, for they are in the same segment ADCB; therefore the whole angle ADC is equal to the angles BAC, ACB.

+

Let the angle ABC be added to each; therefore the angles ABC, BAC, ACB are equal to the angles ABC, ADC. But the angles ABC, BAC, ACB are equal to two right angles; therefore the angles ABC, ADC are also equal to two right angles.

+

Similarly we can prove that the angles BAD, DCB are also equal to two right angles.

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 23. +

On the same straight line there cannot be constructed two similar and unequal segments of circles on the same side.

+

For, if possible, on the same straight line AB let two similar and unequal segments of circles ACB, ADB be constructed on the same side;

let ACD be drawn through, and let CB, DB be joined.

+

Then, since the segment ACB is similar to the segment ADB, and similar segments of circles are those which admit equal angles, [III. Def. 11] the angle ACB is equal to the angle ADB, the exterior to the interior: which is impossible. [I. 16] +

+

Therefore etc. Q. E. D. + +

cannot be constructed, οὐ συσταθήσεται, the same phrase as in 1. 7.

+

+
+
+ PROPOSITION 24. +

Similar segments of circles on equal straight lines are equal to one another.

+

For let AEB, CFD be similar segments of circles on equal straight lines AB, CD; I say that the segment AEB is equal to the segment CFD.

+

For, if the segment AEB be applied to CFD, and if the point A be placed on C and the straight line AB on CD, the point B will also coincide with the point D, because AB is equal to CD; + and, AB coinciding with CD, the segment AEB will also coincide with CFD. +

+

+

For, if the straight line AB coincide with CD but the segment AEB do not coincide with CFD, it will either fall with it, or outside it; or it will fall awry, as CGD, and a circle cuts a circle at more points than two : which is impossible. [III. 10]

+

Therefore, if the straight line AB be applied to CD, the segment AEB will not fail to coincide with CFD also; therefore it will coincide with it and will be equal to it. +

+

Therefore etc. Q. E. D. + +

fall awry, παραλλάξει, the same word as used in the like case in I. 8. The word implies that the applied figure will partly fall short of, and partly overlap, the figure to which it is applied.

+

+
+
+ PROPOSITION 25. +

Given a segment of a circle, to describe the complete circle of which it is a segment.

+

Let ABC be the given segment of a circle; thus it is required to describe the complete circle belonging to the segment ABC, that is, of which it is a segment.

+

For let AC be bisected at D, let DB be drawn from the point D at right angles to AC, and let AB. be joined; the angle ABD is then greater than, equal to, or less than the angle BAD. +

+

First let it be greater; and on the straight line BA, and at the point A on it, let the angle BAE be constructed equal to the angle ABD; let DB be drawn through to E, and let EC be joined.

+

Then, since the angle ABE is equal to the angle BAE,

+ the straight line EB is also equal to EA. [I. 6] +

+

And, since AD is equal to DC, and DE is common, the two sides AD, DE are equal to the two sides CD, DE respectively; and the angle ADE is equal to the angle CDE, for each is right; therefore the base AE is equal to the base CE. +

+

But AE was proved equal to BE; therefore BE is also equal to CE; therefore the three straight lines AE, EB, EC are equal to one another.

+

Therefore the circle drawn with centre E and distance one of the straight lines AE, EB, EC will also pass through the remaining points and will have been completed. [III. 9]

+

Therefore, given a segment of a circle, the complete circle has been described.

+

And it is manifest that the segment ABC is less than a semicircle, because the centre E happens to be outside it.

+

Similarly, even if the angle ABD be equal to the angle BAD, AD being equal to each of the two BD, DC, the three straight lines DA, DB, DC will be equal to one another,

+ D will be the centre of the completed circle, and ABC will clearly be a semicircle. +

+

But, if the angle ABD be less than the angle BAD, and if we construct, on the straight line BA and at the point A on it, an angle equal to the angle ABD, the centre will fall on DB within the segment ABC, and the segment

+ ABC will clearly be greater than a semicircle.

+

Therefore, given a segment of a circle, the complete circle has been described. Q. E. F. + +

to describe the complete circle, προσαναγράψαι τὸν κύκλον, literally “to describe the circle on to it.’

+

+
+
+ PROPOSITION 26. +

In equal circles equal angles stand on equal circumferences, whether they stand at the centres or at the circumferences.

+

Let ABC, DEF be equal circles, and in them let there be equal angles, namely at the centres the angles BGC, EHF, and at the circumferences the angles BAC, EDF; I say that the circumference BKC is equal to the circumference ELF.

+

+

For let BC, EF be joined.

+

Now, since the circles ABC, DEF are equal, the radii are equal. +

+

Thus the two straight lines BG, GC are equal to the two straight lines EH, HF; and the angle at G is equal to the angle at H; therefore the base BC is equal to the base EF. [I. 4] +

+

And, since the angle at A is equal to the angle at D, the segment BAC is similar to the segment EDF; [III. Def. 11] and they are upon equal straight lines.

+

But similar segments of circles on equal straight lines are equal to one another; [III. 24] therefore the segment BAC is equal to EDF. But the whole circle ABC is also equal to the whole circle DEF; therefore the circumference BKC which remains is equal to the circumference ELF.

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 27. +

In equal circles angles standing on equal circumferences are equal to one another, whether they stand at the centres or at the circumferences.

+

For in equal circles ABC, DEF, on equal circumferences BC, EF, let the angles BGC, EHF stand at the centres G, H, and the angles BAC, EDF at the circumferences; I say that the angle BGC is equal to the angle EHF, and the angle BAC is equal to the angle EDF.

+

+

For, if the angle BGC is unequal to the angle EHF, one of them is greater. Let the angle BGC be greater : and on the straight line BG, and at the point G on it, let the angle BGK be constructed equal to the angle EHF. [I. 23]

+

Now equal angles stand on equal circumferences, when they are at the centres; [III. 26] therefore the circumference BK is equal to the circumference EF. +

+

But EF is equal to BC; therefore BK is also equal to BC, the less to the greater : which is impossible. +

+

Therefore the angle BGC is not unequal to the angle EHF; therefore it is equal to it. +

+

And the angle at A is half of the angle BGC, and the angle at D half of the angle EHF; [III. 20] therefore the angle at A is also equal to the angle at D.

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 28. +

In equal circles equal straight lines cut off equal circumferences, the greater equal to the greater and the less to the less.

+

Let ABC, DEF be equal circles, and in the circles let AB, DE be equal straight lines cutting off ACB, DFE as greater circumferences and AGB, DHE as lesser; I say that the greater circumference ACB is equal to the greater circumference DFE, and the less circumference AGB to DHE.

+

+

For let the centres K, L of the circles be taken, and let AK, KB, DL, LE be joined.

+

Now, since the circles are equal, the radii are also equal; therefore the two sides AK, KB are equal to the two sides DL, LE; and the base AB is equal to the base DE; therefore the angle AKB is equal to the angle DLE. [I. 8] +

+

But equal angles stand on equal circumferences, when they are at the centres; [III. 26] therefore the circumference AGB is equal to DHE. +

+

And the whole circle ABC is also equal to the whole circle DEF; therefore the circumference ACB which remains is also equal to the circumference DFE which remains.

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 29. +

In equal circles equal circumferences are subtended by equal straight lines.

+

Let ABC, DEF be equal circles, and in them let equal circumferences BGC, EHF be cut off; and let the straight lines BC, EF be joined; I say that BC is equal to EF.

+

+

For let the centres of the circles be taken, and let them be K, L; let BK, KC, EL, LF be joined.

+

Now, since the circumference BGC is equal to the circumference EHF, the angle BKC is also equal to the angle ELF. [III. 27] +

+

And, since the circles ABC, DEF are equal, the radii are also equal; therefore the two sides BK, KC are equal to the two sides EL, LF; and they contain equal angles; therefore the base BC is equal to the base EF. [I. 4] +

+

Therefore etc.

+
+
+ PROPOSITION 30. +

To bisect a given circumference.

+

Let ADB be the given circumference; thus it is required to bisect the circumference ADB.

+

Let AB be joined and bisected at C; from the point C let CD be drawn at right angles to the straight line AB,

and let AD, DB be joined.

+

Then, since AC is equal to CB, and CD is common, the two sides AC, CD are equal to the two sides BC, CD; and the angle ACD is equal to the angle BCD, for each is right; therefore the base AD is equal to the base DB. [I. 4] +

+

But equal straight lines cut off equal circumferences, the greater equal to the greater, and the less to the less; [III. 28] and each of the circumferences AD, DB is less than a semicircle; therefore the circumference AD is equal to the circumference DB. +

+

Therefore the given circumference has been bisected at the point D. Q. E. F.

+
+
+ PROPOSITION 31. +

In a circle the angle in the semicircle is right, that in a greater segment less than a right angle, and that in a less segment greater than a right angle; and further the angle of the greater segment is greater than a right angle, and the angle of the less segment less than a right angle.

+

Let ABCD be a circle, let BC be its diameter, and E its centre, and let BA, AC, AD, DC be joined; I say that the angle BAC in the semicircle BAC is right, the angle ABC in the segment ABC greater than the semicircle is less

than a right angle, and the angle ADC in the segment ADC less than the semicircle is greater than a right angle.

+

Let AE be joined, and let BA be carried through to F.

+

Then, since BE is equal to EA, the angle ABE is also equal to the angle BAE. [I. 5] +

+

Again, since CE is equal to EA, the angle ACE is also equal to the angle CAE. [I. 5] +

+

Therefore the whole angle BAC is equal to the two angles ABC, ACB.

+

But the angle FAC exterior to the triangle ABC is also equal to the two angles ABC, ACB; [I. 32] therefore the angle BAC is also equal to the angle FAC; therefore each is right; [I. Def. 10] therefore the angle BAC in the semicircle BAC is right. +

+

Next, since in the triangle ABC the two angles ABC, BAC are less than two right angles, [I. 17] and the angle BAC is a right angle, the angle ABC is less than a right angle; and it is the angle in the segment ABC greater than the semicircle.

+

Next, since ABCD is a quadrilateral in a circle, and the opposite angles of quadrilaterals in circles are equal to two right angles, [III. 22] while the angle ABC is less than a right angle, therefore the angle ADC which remains is greater than a right angle; and it is the angle in the segment ADC less than the semicircle.

+

I say further that the angle of the greater segment, namely that contained by the circumference ABC and the straight line AC, is greater than a right angle; and the angle of the less segment, namely that contained by the circumference ADC and the straight line AC, is less than a right angle.

+

This is at once manifest. For, since the angle contained by the straight lines BA, AC is right, the angle contained by the circumference ABC and the straight line AC is greater than a right angle. +

+

Again, since the angle contained by the straight lines AC, AF is right, the angle contained by the straight line CA and the circumference ADC is less than a right angle. +

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 32. +

If a straight line touch a circle, and from the point of contact there be drawn across, in the circle, a straight line cutting the circle, the angles which it makes with the tangent will be equal to the angles in the alternate segments of the circle.

+

For let a straight line EF touch the circle ABCD at the point B, and from the point B let there be drawn across, in the circle ABCD, a straight line BD cutting it; I say that the angles which BD makes with the tangent EF will be equal to the angles in the alternate segments of the circle, that is, that the angle FBD is equal to the angle constructed in the segment BAD, and the angle EBD is equal to the angle constructed in the segment DCB.

+

For let BA be drawn from B at right angles to EF, let a point C be taken at random on the circumference BD,

and let AD, DC, CB be joined.

+

Then, since a straight line EF touches the circle ABCD at B, and BA has been drawn from the point of contact at right angles to the tangent, the centre of the circle ABCD is on BA. [III. 19]

+

Therefore BA is a diameter of the circle ABCD; therefore the angle ADB, being an angle in a semicircle, is right. [III. 31] +

+

Therefore the remaining angles BAD, ABD are equal to one right angle. [I. 32]

+

But the angle ABF is also right; therefore the angle ABF is equal to the angles BAD, ABD.

+

Let the angle ABD be subtracted from each; therefore the angle DBF which remains is equal to the angle BAD in the alternate segment of the circle.

+

Next, since ABCD is a quadrilateral in a circle, its opposite angles are equal to two right angles. [III. 22]

+

But the angles DBF, DBE are also equal to two right angles; therefore the angles DBF, DBE are equal to the angles BAD, BCD, of which the angle BAD was proved equal to the angle DBF; therefore the angle DBE which remains is equal to the angle DCB in the alternate segment DCB of the circle. +

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 33. +

On a given straight line to describe a segment of a circle admitting an angle equal to a given rectilineal angle.

+

Let AB be the given straight line, and the angle at C the given rectilineal angle; thus it is required to describe on the given straight line AB a segment of a circle admitting an angle equal to the angle at C.

+

+

The angle at C is then acute, or right, or obtuse.

+

First let it be acute, and, as in the first figure, on the straight line AB, and at the point A, let the angle BAD be constructed equal to the angle at C; therefore the angle BAD is also acute. +

+

Let AE be drawn at right angles to DA, let AB be bisected at F, let FG be drawn from the point F at right angles to AB, and let GB be joined.

+

Then, since AF is equal to FB, and FG is common, the two sides AF, FG are equal to the two sides BF, FG; and the angle AFG is equal to the angle BFG; therefore the base AG is equal to the base BG. [I. 4] +

+

Therefore the circle described with centre G and distance GA will pass through B also.

+

Let it be drawn, and let it be ABE; let EB be joined.

+

Now, since AD is drawn from A, the extremity of the diameter AE, at right angles to AE, therefore AD touches the circle ABE. [III. 16, Por.] +

+

Since then a straight line AD touches the circle ABE, and from the point of contact at A a straight line AB is drawn across in the circle ABE, the angle DAB is equal to the angle AEB in the alternate segment of the circle. [III. 32] +

+

But the angle DAB is equal to the angle at C; therefore the angle at C is also equal to the angle AEB.

+

Therefore on the given straight line AB the segment AEB of a circle has been described admitting the angle AEB equal to the given angle, the angle at C.

+

Next let the angle at C be right;

and let it be again required to describe on AB a segment of a circle admitting an angle equal to the right angle at C.

+

Let the angle BAD be constructed equal to the right angle at C, as is the case in the second figure; let AB be bisected at F, and with centre F and distance either FA or FB let the circle AEB be described.

+

Therefore the straight line AD touches the circle ABE, because the angle at A is right. [III. 16, Por.]

+

And the angle BAD is equal to the angle in the segment AEB, for the latter too is itself a right angle, being an angle in a semicircle. [III. 31]

+

But the angle BAD is also equal to the angle at C.

+

Therefore the angle AEB is also equal to the angle at C.

+

Therefore again the segment AEB of a circle has been described on AB admitting an angle equal to the angle at C.

+

Next, let the angle at C be obtuse;

and on the straight line AB, and at the point A, let the angle BAD be constructed equal to it, as is the case in the third figure; let AE be drawn at right angles to AD, let AB be again bisected at F, let FG be drawn at right angles to AB, and let GB be joined.

+

Then, since AF is again equal to FB, and FG is common, the two sides AF, FG are equal to the two sides BF, FG; and the angle AFG is equal to the angle BFG; therefore the base AG is equal to the base BG. [I. 4] +

+

Therefore the circle described with centre G and distance GA will pass through B also; let it so pass, as AEB.

+

Now, since AD is drawn at right angles to the diameter AE from its extremity, AD touches the circle AEB. [III. 16, Por.] +

+

And AB has been drawn across from the point of contact at A; therefore the angle BAD is equal to the angle constructed in the alternate segment AHB of the circle. [III. 32] +

+

But the angle BAD is equal to the angle at C.

+

Therefore the angle in the segment AHB is also equal to the angle at C.

+

Therefore on the given straight line AB the segment AHB of a circle has been described admitting an angle equal to the angle at C. Q. E. F.

+
+
+ PROPOSITION 34. +

From a given circle to cut off a segment admitting an angle equal to a given rectilineal angle.

+

Let ABC be the given circle, and the angle at D the given rectilineal angle; thus it is required to cut off from the circle ABC a segment admitting an angle equal to the given rectilineal angle, the angle at D.

+

Let EF be drawn touching ABC at the point B, and on the straight line FB, and at the point B on it, let the angle FBC be constructed equal to the angle at D. [I. 23]

+

Then, since a straight line EF touches the circle ABC, and BC has been drawn across from the point of contact at B, the angle FBC is equal to the angle constructed in the alternate segment BAC. [III. 32]

+

+

But the angle FBC is equal to the angle at D; therefore the angle in the segment BAC is equal to the angle at D. +

+

Therefore from the given circle ABC the segment BAC. has been cut off admitting an angle equal to the given rectilineal angle, the angle at D. Q. E. F.

+
+
+ PROPOSITION 35. +

If in a circle two straight lines cut one another, the rectangle contained by the segments of the one is equal to the rectangle contained by the segments of the other.

+

For in the circle ABCD let the two straight lines AC, BD cut one another at the point E; I say that the rectangle contained by AE, EC is equal to the rectangle contained by DE, EB.

+

+

If now AC, BD are through the centre, so that E is the centre of the circle ABCD, it is manifest that, AE, EC, DE, EB being equal, the rectangle contained by AE, EC is also equal to the rectangle contained by DE, EB.

+

Next let AC, DB not be through the centre; let the centre of ABCD be taken, and let it be F; from F let FG, FH be drawn perpendicular to the straight lines AC, DB, and let FB, FC, FE be joined.

+

+

Then, since a straight line GF through the centre cuts a straight line AC not through the centre at right angles, it also bisects it; [III. 3] therefore AG is equal to GC. +

+

Since, then, the straight line AC has been cut into equal parts at G and into unequal parts at E, the rectangle contained by AE, EC together with the square on EG is equal to the square on GC; [II. 5]

+

Let the square on GF be added; therefore the rectangle AE, EC together with the squares on GE, GF is equal to the squares on CG, GF.

+

But the square on FE is equal to the squares on EG, GF, and the square on FC is equal to the squares on CG, GF; [I. 47] therefore the rectangle AE, EC together with the square on FE is equal to the square on FC. +

+

And FC is equal to FB; therefore the rectangle AE, EC together with the square on EF is equal to the square on FB.

+

For the same reason, also, the rectangle DE, EB together with the square on FE is equal to the square on FB.

+

But the rectangle AE, EC together with the square on FE was also proved equal to the square on FB; therefore the rectangle AE, EC together with the square on FE is equal to the rectangle DE, EB together with the square on FE.

+

Let the square on FE be subtracted from each; therefore the rectangle contained by AE, EC which remains is equal to the rectangle contained by DE, EB.

+

Therefore etc.

+
+
+ PROPOSITION 36. +

If a point be taken outside a circle and from it there fall on the circle two straight lines, and if one of them cut the circle and the other touch it, the rectangle contained by the whole of the straight line which cuts the circle and the straight line intercepted on it outside between the point and the convex circumference will be equal to the square on the tangent.

+

For let a point D be taken outside the circle ABC, and from D let the two straight lines DCA, DB fall on the circle ABC; let DCA cut the circle ABC and let BD touch it; I say that the rectangle contained by AD, DC is equal to the square on DB.

+

+

Then DCA is either through the centre or not through the centre.

+

First let it be through the centre, and let F be the centre of the circle ABC; let FB be joined; therefore the angle FBD is right. [III. 18] +

+

And, since AC has been bisected at F, and CD is added to it, the rectangle AD, DC together with the square on FC is equal to the square on FD. [II. 6]

+

But FC is equal to FB; therefore the rectangle AD, DC together with the square on FB is equal to the square on FD.

+

And the squares on FB, BD are equal to the square on FD; [I. 47] therefore the rectangle AD, DC together with the square on FB is equal to the squares on FB, BD.

+

Let the square on FB be subtracted from each; therefore the rectangle AD, DC which remains is equal to the square on the tangent DB.

+

Again, let DCA not be through the centre of the circle ABC; let the centre E be taken, and from E let EF be drawn perpendicular to AC; let EB, EC, ED be joined.

+

Then the angle EBD is right. [III. 18]

+

+

And, since a straight line EF through the centre cuts a straight line AC not through the centre at right angles, it also bisects it; [III. 3] therefore AF is equal to FC. +

+

Now, since the straight line AC has been bisected at the point F, and CD is added to it, the rectangle contained by AD, DC together with the square on FC is equal to the square on FD. [II. 6]

+

Let the square on FE be added to each; therefore the rectangle AD, DC together with the squares on CF, FE is equal to the squares on FD, FE.

+

But the square on EC is equal to the squares on CF, FE, for the angle EFC is right; [I. 47] and the square on ED is equal to the squares on DF, FE; therefore the rectangle AD, DC together with the square on EC is equal to the square on ED.

+

And EC is equal to EB; therefore the rectangle AD, DC together with the square on EB is equal to the square on ED.

+

But the squares on EB, BD are equal to the square on ED, for the angle EBD is right; [I. 47] therefore the rectangle AD, DC together with the square on EB is equal to the squares on EB, BD.

+

Let the square on EB be subtracted from each; therefore the rectangle AD, DC which remains is equal to the square on DB.

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 37. +

If a point be taken outside a circle and from the point there fall on the circle two straight lines, if one of them cut the circle, and the other fall on it, and if further the rectangle contained by the whole of the straight line which cuts the circle and the straight line intercepted on it outside between the point and the convex circumference be equal to the square on the straight line which falls on the circle, the straight line which falls on it will touch the circle.

+

For let a point D be taken outside the circle ABC; from D let the two straight lines DCA, DB fall on the circle ACB; let DCA cut the circle and DB fall on it; and let the rectangle AD, DC be equal to the square on DB.

+

+

I say that DB touches the circle ABC.

+

For let DE be drawn touching ABC; let the centre of the circle ABC be taken, and let it be F; let FE, FB, FD be joined.

+

Thus the angle FED is right. [III. 18]

+

Now, since DE touches the circle ABC, and DCA cuts it, the rectangle AD, DC is equal to the square on DE. [III. 36]

+

But the rectangle AD, DC was also equal to the square on DB; therefore the square on DE is equal to the square on DB; therefore DE is equal to DB. +

+

And FE is equal to FB; therefore the two sides DE, EF are equal to the two sides DB, BF; and FD is the common base of the triangles; therefore the angle DEF is equal to the angle DBF. [I. 8] +

+

But the angle DEF is right; therefore the angle DBF is also right. +

+

And FB produced is a diameter; and the straight line drawn at right angles to the diameter of a circle, from its extremity, touches the circle; [III. 16, Por.] therefore DB touches the circle. +

+

Similarly this can be proved to be the case even if the centre be on AC.

+

Therefore etc. Q. E. D.

+
+
+
+
+ + BOOK IV. +
+ DEFINITIONS. +
+1 +

A rectilineal figure is said to be inscribed in a rectilineal figure when the respective angles of the inscribed figure lie on the respective sides of that in which it is inscribed.

+
+
+2 +

Similarly a figure is said to be circumscribed about a figure when the respective sides of the circumscribed figure pass through the respective angles of that about which it is circumscribed.

+
+
+3 +

A rectilineal figure is said to be inscribed in a circle when each angle of the inscribed figure lies on the circumference of the circle.

+
+
+4 +

A rectilineal figure is said to be circumscribed about a circle, when each side of the circumscribed figure touches the circumference of the circle.

+
+
+5 +

Similarly a circle is said to be inscribed in a figure when the circumference of the circle touches each side of the figure in which it is inscribed.

+
+
+6 +

A circle is said to be circumscribed about a figure when the circumference of the circle passes through each angle of the figure about which it is circumscribed.

+
+
+7 +

A straight line is said to be fitted into a circle when its extremities are on the circumference of the circle.

+
+
+
+ + BOOK IV. PROPOSITIONS +
+ PROPOSITION 1. +

Into a given circle to fit a straight line equal to a given straight line which is not greater than the diameter of the circle.

+

Let ABC be the given circle, and D the given straight line not greater than the diameter of the circle; thus it is required to fit into the circle ABC a straight line equal to the straight line D.

+

+

Let a diameter BC of the circle ABC be drawn.

+

Then, if BC is equal to D, that which was enjoined will have been done; for BC has been fitted into the circle ABC equal to the straight line D.

+

But, if BC is greater than D, let CE be made equal to D, and with centre C and distance CE let the circle EAF be described; let CA be joined.

+

Then, since the point C is the centre of the circle EAF, CA is equal to CE. +

+

But CE is equal to D; therefore D is also equal to CA. +

+

Therefore into the given circle ABC there has been fitted CA equal to the given straight line D.

+
+
+ PROPOSITION 2. +

In a given circle to inscribe a triangle equiangular with a given triangle.

+

Let ABC be the given circle, and DEF the given triangle; thus it is required to inscribe in the circle ABC a triangle equiangular with the triangle DEF.

+

Let GH be drawn touching the circle ABC at A [III. 16, Por.]; on the straight line AH, and at the point A on it, let the angle HAC be constructed equal to the angle DEF, and on the straight line AG, and at the point A on it, let the angle GAB be constructed equal to the angle DFE; [I. 23] let BC be joined.

+

+

Then, since a straight line AH touches the circle ABC, and from the point of contact at A the straight line AC is drawn across in the circle, therefore the angle HAC is equal to the angle ABC in the alternate segment of the circle. [III. 32]

+

But the angle HAC is equal to the angle DEF; therefore the angle ABC is also equal to the angle DEF.

+

For the same reason the angle ACB is also equal to the angle DFE; therefore the remaining angle BAC is also equal to the remaining angle EDF. [I. 32]

+

Therefore in the given circle there has been inscribed a triangle equiangular with the given triangle. Q. E. F.

+
+
+ PROPOSITION 3. +

About a given circle to circumscribe a triangle equiangular with a given triangle.

+

Let ABC be the given circle, and DEF the given triangle; thus it is required to circumscribe about the circle ABC a triangle equiangular with the triangle DEF.

+

+

Let EF be produced in both directions to the points G, H, let the centre K of the circle ABC be taken [III. 1], and let the straight line KB be drawn across at random; on the straight line KB, and at the point K on it, let the angle BKA be constructed equal to the angle DEG, and the angle BKC equal to the angle DFH; [I. 23] and through the points A, B, C let LAM, MBN, NCL be drawn touching the circle ABC. [III. 16, Por.]

+

Now, since LM, MN, NL touch the circle ABC at the points A, B, C, and KA, KB, KC have been joined from the centre K to the points A, B, C, therefore the angles at the points A, B, C are right. [III. 18]

+

And, since the four angles of the quadrilateral AMBK are equal to four right angles, inasmuch as AMBK is in fact divisible into two triangles, and the angles KAM, KBM are right, + therefore the remaining angles AKB, AMB are equal to two right angles.

+

But the angles DEG, DEF are also equal to two right angles; [I. 13] therefore the angles AKB, AMB are equal to the angles DEG, DEF, of which the angle AKB is equal to the angle DEG; therefore the angle AMB which remains is equal to the angle DEF which remains. +

+

Similarly it can be proved that the angle LNB is also equal to the angle DFE; therefore the remaining angle MLN is equal to the angle EDF. [I. 32] +

+

Therefore the triangle LMN is equiangular with the triangle DEF; and it has been circumscribed about the circle ABC.

+

Therefore about a given circle there has been circumscribed a triangle equiangular with the given triangle. Q. E. F. + +

at random, literally as it may chance, + ὡς ἕτυχεν. The same expression is used in III. 1 and commonly.

+ + +

is in fact divisible, καὶ διαιρεῖται, literally is actually divided. +

+

+
+
+ PROPOSITION 4. +

In a given triangle to inscribe a circle.

+

Let ABC be the given triangle; thus it is required to inscribe a circle in the triangle ABC.

+

Let the angles ABC, ACB + be bisected by the straight lines BD, CD [I. 9], and let these meet one another at the point D; from D let DE, DF, DG be drawn perpendicular to the straight lines AB, BC, CA.

+

+

Now, since the angle ABD is equal to the angle CBD, and the right angle BED is also equal to the right angle BFD, EBD, FBD are two triangles having two angles equal to two angles and one side equal to one side, namely that subtending one of the equal angles, which is BD common to the triangles; therefore they will also have the remaining sides equal to the remaining sides; [I. 26] + therefore DE is equal to DF. +

+

For the same reason DG is also equal to DF. +

+

Therefore the three straight lines DE, DF, DG are equal to one another; therefore the circle described with centre D and distance one of the straight lines DE, DF, DG will pass also through the remaining points, and will touch the straight lines AB, BC, CA, because the angles at the points E, F, G + are right. +

+

For, if it cuts them, the straight line drawn at right angles to the diameter of the circle from its extremity will be found to fall within the circle : which was proved absurd; [III. 16] therefore the circle described with centre D and distance one of the straight lines DE, DF, DG will not cut the straight lines AB, BC, CA; + therefore it will touch them, and will be the circle inscribed in the triangle ABC. [IV. Def. 5] +

+

Let it be inscribed, as FGE.

+

Therefore in the given triangle ABC the circle EFG has been inscribed. Q. E. F. + +

and distance one of the (straight lines D)E, (D)F, (D)G. The words and letters here shown in brackets are put in to fill out the rather careless language of the Greek. Here and in several other places in Book IV. Euclid says literally and with distance one of the (points) E, F, G + (καὶ διαστήματι ὲνὶ τῶν E, Z, H) and the like. In one case (IV. 13) he actually has with distance one of the points G, H, K, L, M + (διαστήματι ὲνὶ τῶν Η, Θ, Κ, Λ, Μ σημείων). Heiberg notes Graecam locutionem satis miram et negligentem, + but, in view of its frequent occurrence in good MSS., does not venture to correct it.

+

+
+
+ PROPOSITION 5. +

About a given triangle to circumscribe a circle.

+

Let ABC be the given triangle; thus it is required to circumscribe a circle about the given triangle ABC.

+

+

Let the straight lines AB, AC be bisected at the points D, E [I. 10], and from the points D, E let DF, EF be drawn at right angles to AB, AC; they will then meet within the triangle ABC, or on the straight line BC, or outside BC.

+

First let them meet within at F, and let FB, FC, FA be joined.

+

Then, since AD is equal to DB, and DF is common and at right angles, therefore the base AF is equal to the base FB. [I. 4]

+

Similarly we can prove that CF is also equal to AF; so that FB is also equal to FC; therefore the three straight lines FA, FB, FC are equal to one another. +

+

Therefore the circle described with centre F and distance one of the straight lines FA, FB, FC will pass also through the remaining points, and the circle will have been circumscribed about the triangle ABC.

+

Let it be circumscribed, as ABC.

+

Next, let DF, EF meet on the straight line BC at F, as is the case in the second figure; and let AF be joined.

+

Then, similarly, we shall prove that the point F is the centre of the circle circumscribed about the triangle ABC.

+

Again, let DF, EF meet outside the triangle ABC at F, as is the case in the third figure, and let AF, BF, CF be joined.

+

Then again, since AD is equal to DB, and DF is common and at right angles, therefore the base AF is equal to the base BF. [I. 4]

+

Similarly we can prove that CF is also equal to AF; so that BF is also equal to FC; therefore the circle described with centre F and distance one of the straight lines FA, FB, FC will pass also through the remaining points, and will have been circumscribed about the triangle ABC.

+

Therefore about the given triangle a circle has been circumscribed. Q. E. F.

+

And it is manifest that, when the centre of the circle falls within the triangle, the angle BAC, being in a segment greater than the semicircle, is less than a right angle; when the centre falls on the straight line BC, the angle BAC, being in a semicircle, is right; and when the centre of the circle falls outside the triangle, the angle BAC, being in a segment less than the semicircle, is greater than a right angle. [III. 31]

+
+
+ + PROPOSITION 6. +

In a given circle to inscribe a square.

+

Let ABCD be the given circle; thus it is required to inscribe a square in the circle ABCD.

+

Let two diameters AC, BD of the circle ABCD be drawn at right angles to one another, and let AB, BC, CD, DA be joined.

+

+

Then, since BE is equal to ED, for E is the centre, and EA is common and at right angles, therefore the base AB is equal to the base AD. [I. 4]

+

For the same reason each of the straight lines BC, CD is also equal to each of the straight lines AB, AD; therefore the quadrilateral ABCD is equilateral. +

+

I say next that it is also right-angled.

+

For, since the straight line BD is a diameter of the circle ABCD, therefore BAD is a semicircle; therefore the angle BAD is right. [III. 31] +

+

For the same reason each of the angles ABC, BCD, CDA is also right; therefore the quadrilateral ABCD is right-angled. +

+

But it was also proved equilateral; therefore it is a square; [I. Def. 22] and it has been inscribed in the circle ABCD.

+

Therefore in the given circle the square ABCD has been inscribed. Q. E. F.

+
+
+ PROPOSITION 7. +

About a given circle to circumscribe a square.

+

Let ABCD be the given circle; thus it is required to circumscribe a square about the circle ABCD.

+

Let two diameters AC, BD of the circle ABCD be drawn at right angles to one another, and through the points A, B, C, D let FG, GH, HK, KF be drawn touching the circle ABCD. [III. 16, Por.]

+

+

Then, since FG touches the circle ABCD, and EA has been joined from the centre E to the point of contact at A, therefore the angles at A are right. [III. 18] +

+

For the same reason the angles at the points B, C, D are also right. +

+

Now, since the angle AEB is right, and the angle EBG is also right, therefore GH is parailel to AC. [I. 28] +

+

For the same reason AC is also parallel to FK, so that GH is also parallel to FK. [I. 30] +

+

Similarly we can prove that each of the straight lines GF, HK is parallel to BED. +

+

Therefore GK, GC, AK, FB, BK are parallelograms; therefore GF is equal to HK, and GH to FK. [I. 34]

+

And, since AC is equal to BD, and AC is also equal to each of the straight lines GH, FK, while BD is equal to each of the straight lines GF, HK, [I. 34] therefore the quadrilateral FGHK is equilateral. +

+

I say next that it is also right-angled.

+

For, since GBEA is a parallelogram, and the angle AEB is right, therefore the angle AGB is also right. [I. 34]

+

Similarly we can prove that the angles at H, K, F are also right. +

+

Therefore FGHK is right-angled.

+

But it was also proved equilateral; therefore it is a square; and it has been circumscribed about the circle ABCD.

+

Therefore about the given circle a square has been circumscribed. Q. E. F.

+
+
+ + PROPOSITION 8. +

In a given square to inscribe a circle.

+

Let ABCD be the given square; thus it is required to inscribe a circle in the given square ABCD.

+

+

Let the straight lines AD, AB be bisected at the points E, F respectively [I. 10], through E let EH be drawn parallel to either AB or CD, and through F let FK be drawn parallel to either AD or BC; [I. 31] therefore each of the figures AK, KB, AH, HD, AG, GC, BG, GD is a parallelogram, and their opposite sides are evidently equal. [I. 34]

+

Now, since AD is equal to AB, and AE is half of AD, and AF half of AB, therefore AE is equal to AF, so that the opposite sides are also equal; + therefore FG is equal to GE. +

+

Similarly we can prove that each of the straight lines GH, GK is equal to each of the straight lines FG, GE; therefore the four straight lines GE, GF, GH, GK are equal to one another. +

+

Therefore the circle described with centre G and distance one of the straight lines GE, GF, GH, GK will pass also through the remaining points.

+

And it will touch the straight lines AB, BC, CD, DA, because the angles at E, F, H, K are right.

+

For, if the circle cuts AB, BC, CD, DA, the straight line drawn at right angles to the diameter of the circle from its extremity will fall within the circle : which was proved absurd; [III. 16] therefore the circle described with centre G and distance one of the straight lines GE, GF, GH, GK will not cut the straight lines AB, BC, CD, DA.

+

Therefore it will touch them, and will have been inscribed in the square ABCD.

+

Therefore in the given square a circle has been inscribed. Q. E. F.

+
+
+ PROPOSITION 9. +

About a given square to circumscribe a circle.

+

Let ABCD be the given square; thus it is required to circumscribe a circle about the square ABCD.

+

+

For let AC, BD be joined, and let them cut one another at E.

+

Then, since DA is equal to AB, and AC is common, therefore the two sides DA, AC are equal to the two sides BA, AC; and the base DC is equal to the base BC; therefore the angle DAC is equal to the angle BAC. [I. 8] +

+

Therefore the angle DAB is bisected by AC.

+

Similarly we can prove that each of the angles ABC, BCD, CDA is bisected by the straight lines AC, DB.

+

Now, since the angle DAB is equal to the angle ABC, and the angle EAB is half the angle DAB, and the angle EBA half the angle ABC, therefore the angle EAB is also equal to the angle EBA; so that the side EA is also equal to EB. [I. 6]

+

Similarly we can prove that each of the straight lines EA, EB is equal to each of the straight lines EC, ED.

+

Therefore the four straight lines EA, EB, EC, ED are equal to one another.

+

Therefore the circle described with centre E and distance one of the straight lines EA, EB, EC, ED will pass also through the remaining points; and it will have been circumscribed about the square ABCD.

+

Let it be circumscribed, as ABCD.

+

Therefore about the given square a circle has been circumscribed. Q. E. F.

+
+
+ + PROPOSITION 10. +

To construct an isosceles triangle having each of the angles at the base double of the remaining one.

+

Let any straight line AB be set out, and let it be cut at the point C so that the rectangle contained by AB, BC is equal to the square on CA; [II. 11] with centre A and distance AB let the circle BDE be described, and let there be fitted in the circle BDE the straight line BD equal to the straight line AC which is not greater than the diameter of the circle BDE. [IV. 1]

+

+

Let AD, DC be joined, and let the circle ACD be circumscribed about the triangle ACD. [IV. 5]

+

Then, since the rectangle AB, BC is equal to the square on AC, and AC is equal to BD, therefore the rectangle AB, BC is equal to the square on BD.

+

And, since a point B has been taken outside the circle ACD, and from B the two straight lines BA, BD have fallen on the circle ACD, and one of them cuts it, while the other falls on it, and the rectangle AB, BC is equal to the square on BD, therefore BD touches the circle ACD. [III. 37] +

+

Since, then, BD touches it, and DC is drawn across from the point of contact at D, therefore the angle BDC is equal to the angle DAC in the alternate segment of the circle. [III. 32]

+

Since, then, the angle BDC is equal to the angle DAC, let the angle CDA be added to each; therefore the whole angle BDA is equal to the two angles CDA, DAC.

+

But the exterior angle BCD is equal to the angles CDA, DAC; [I. 32] therefore the angle BDA is also equal to the angle BCD.

+

But the angle BDA is equal to the angle CBD, since the side AD is also equal to AB; [I. 5] so that the angle DBA is also equal to the angle BCD.

+

Therefore the three angles BDA, DBA, BCD are equal to one another.

+

And, since the angle DBC is equal to the angle BCD, the side BD is also equal to the side DC. [I. 6] +

+

But BD is by hypothesis equal to CA; therefore CA is also equal to CD, so that the angle CDA is also equal to the angle DAC; [I. 5] therefore the angles CDA, DAC are double of the angle DAC.

+

But the angle BCD is equal to the angles CDA, DAC; therefore the angle BCD is also double of the angle CAD.

+

But the angle BCD is equal to each of the angles BDA, DBA; therefore each of the angles BDA, DBA is also double of the angle DAB.

+

Therefore the isosceles triangle ABD has been constructed having each of the angles at the base DB double of the remaining one. Q. E. F.

+
+
+ PROPOSITION 11. +

In a given circle to inscribe an equilateral and equiangular pentagon.

+

Let ABCDE be the given circle; thus it is required to inscribe in the circle ABCDE an equilateral and equiangular pentagon.

+

Let the isosceles triangle FGH be set out having each of the angles at G, H double of the angle at F; [IV. 10]

let there be inscribed in the circle ABCDE the triangle ACD equiangular with the triangle FGH, so that the angle CAD is equal to the angle at F and the angles at G, H respectively equal to the angles ACD, CDA; [IV. 2] therefore each of the angles ACD, CDA is also double of the angle CAD.

+

Now let the angles ACD, CDA be bisected respectively by the straight lines CE, DB [I. 9], and let AB, BC, DE, EA be joined.

+

Then, since each of the angles ACD, CDA is double of the angle CAD, and they have been bisected by the straight lines CE, DB, therefore the five angles DAC, ACE, ECD, CDB, BDA are equal to one another.

+

But equal angles stand on equal circumferences; [III. 26] therefore the five circumferences AB, BC, CD, DE, EA are equal to one another.

+

But equal circumferences are subtended by equal straight lines; [III. 29] therefore the five straight lines AB, BC, CD, DE, EA are equal to one another; therefore the pentagon ABCDE is equilateral. +

+

I say next that it is also equiangular.

+

For, since the circumference AB is equal to the circumference DE, let BCD be added to each; therefore the whole circumference ABCD is equal to the whole circumference EDCB. +

+

And the angle AED stands on the circumference ABCD, and the angle BAE on the circumference EDCB; therefore the angle BAE is also equal to the angle AED. [III. 27] +

+

For the same reason each of the angles ABC, BCD, CDE is also equal to each of the angles BAE, AED; therefore the pentagon ABCDE is equiangular. +

+

But it was also proved equilateral; therefore in the given circle an equilateral and equiangular pentagon has been inscribed. Q. E. F.

+
+
+ PROPOSITION 12. +

About a given circle to circumscribe an equilateral and equiangular pentagon.

+

Let ABCDE be the given circle; thus it is required to circumscribe an equilateral and equiangular pentagon about the circle ABCDE.

+

Let A, B, C, D, E be conceived to be the angular points of the inscribed pentagon, so that the circumferences AB, BC, CD, DE, EA are equal; [IV. 11]

through A, B, C, D, E let GH, HK, KL, LM, MG be drawn touching the circle; [III. 16, Por.] let the centre F of the circle ABCDE be taken [III. 1], and let FB, FK, FC, FL, FD be joined.

+

Then, since the straight line KL touches the circle ABCDE at C, and FC has been joined from the centre F to the point of contact at C, therefore FC is perpendicular to KL; [III. 18] therefore each of the angles at C is right. +

+

For the same reason the angles at the points B, D are also right. +

+

And, since the angle FCK is right, therefore the square on FK is equal to the squares on FC, CK.

+

For the same reason [I. 47] the square on FK is also equal to the squares on FB, BK; so that the squares on FC, CK are equal to the squares on FB, BK, of which the square on FC is equal to the square on FB; therefore the square on CK which remains is equal to the square on BK.

+

Therefore BK is equal to CK.

+

And, since FB is equal to FC, and FK common, the two sides BF, FK are equal to the two sides CF, FK; and the base BK equal to the base CK; therefore the angle BFK is equal to the angle KFC, [I. 8] and the angle BKF to the angle FKC. Therefore the angle BFC is double of the angle KFC, and the angle BKC of the angle FKC. +

+

For the same reason the angle CFD is also double of the angle CFL, and the angle DLC of the angle FLC. +

+

Now, since the circumference BC is equal to CD, the angle BFC is also equal to the angle CFD. [III. 27]

+

And the angle BFC is double of the angle KFC, and the angle DFC of the angle LFC; therefore the angle KFC is also equal to the angle LFC. +

+

But the angle FCK is also equal to the angle FCL; therefore FKC, FLC are two triangles having two angles equal to two angles and one side equal to one side, namely FC which is common to them; therefore they will also have the remaining sides equal to the remaining sides, and the remaining angle to the remaining angle; [I. 26] therefore the straight line KC is equal to CL, and the angle FKC to the angle FLC. +

+

And, since KC is equal to CL, therefore KL is double of KC.

+

For the same reason it can be proved that HK is also double of BK. +

+

And BK is equal to KC; therefore HK is also equal to KL. +

+

Similarly each of the straight lines HG, GM, ML can also be proved equal to each of the straight lines HK, KL; therefore the pentagon GHKLM is equilateral. +

+

I say next that it is also equiangular.

+

For, since the angle FKC is equal to the angle FLC, and the angle HKL was proved double of the angle FKC, and the angle KLM double of the angle FLC, therefore the angle HKL is also equal to the angle KLM. +

+

Similarly each of the angles KHG, HGM, GML can also be proved equal to each of the angles HKL, KLM; therefore the five angles GHK, HKL, KLM, LMG, MGH are equal to one another.

+

Therefore the pentagon GHKLM is equiangular.

+

And it was also proved equilateral; and it has been circumscribed about the circle ABCDE. Q. E. F.

+
+
+ PROPOSITION 13. +

In a given pentagon, which is equilateral and equiangular, to inscribe a circle.

+

Let ABCDE be the given equilateral and equiangular pentagon; thus it is required to inscribe a circle in the pentagon ABCDE.

+

For let the angles BCD, CDE be bisected by the straight lines CF, DF respectively; and from the point F, at which the straight lines CF, DF meet one another, let the straight lines FB, FA, FE be joined.

+

Then, since BC is equal to CD, and CF common, the two sides BC, CF are equal to the two sides DC, CF;

and the angle BCF is equal to the angle DCF; therefore the base BF is equal to the base DF, and the triangle BCF is equal to the triangle DCF, and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend. [I. 4]

+

Therefore the angle CBF is equal to the angle CDF.

+

And, since the angle CDE is double of the angle CDF, and the angle CDE is equal to the angle ABC, while the angle CDF is equal to the angle CBF; therefore the angle CBA is also double of the angle CBF; therefore the angle ABF is equal to the angle FBC; therefore the angle ABC has been bisected by the straight line BF.

+

Similarly it can be proved that the angles BAE, AED have also been bisected by the straight lines FA, FE respectively.

+

Now let FG, FH, FK, FL, FM be drawn from the point F perpendicular to the straight lines AB, BC, CD, DE, EA.

+

Then, since the angle HCF is equal to the angle KCF, and the right angle FHC is also equal to the angle FKC, FHC, FKC are two triangles having two angles equal to two angles and one side equal to one side, namely FC which is common to them and subtends one of the equal angles; therefore they will also have the remaining sides equal to the remaining sides; [I. 26] therefore the perpendicular FH is equal to the perpendicular FK.

+

Similarly it can be proved that each of the straight lines FL, FM, FG is also equal to each of the straight lines FH, FK; therefore the five straight lines FG, FH, FK, FL, FM are equal to one another.

+

Therefore the circle described with centre F and distance one of the straight lines FG, FH, FK, FL, FM will pass also through the remaining points; and it will touch the straight lines AB, BC, CD, DE, EA, because the angles at the points G, H, K, L, M are right.

+

For, if it does not touch them. but cuts them, it will result that the straight line drawn at right angles to the diameter of the circle from its extremity falls within the circle: which was proved absurd. [III. 16]

+

Therefore the circle described with centre F and distance one of the straight lines FG, FH, FK, FL, FM will not cut the straight lines AB, BC, CD, DE, EA; therefore it will touch them. +

+

Let it be described, as GHKLM.

+

Therefore in the given pentagon, which is equilateral and equiangular, a circle has been inscribed. Q. E. F.

+
+
+ PROPOSITION 14. +

About a given pentagon, which is equilateral and equiangular, to circumscribe a circle.

+

Let ABCDE be the given pentagon, which is equilateral and equiangular; thus it is required to circumscribe a circle about the pentagon ABCDE.

+

Let the angles BCD, CDE be bisected by the straight lines CF, DF respectively, and from the point F, at which the straight lines meet, let the straight lines FB, FA, FE be joined to the points B, A, E.

+

+

Then in manner similar to the preceding it can be proved that the angles CBA, BAE, AED have also been bisected by the straight lines FB, FA, FE respectively.

+

Now, since the angle BCD is equal to the angle CDE, and the angle FCD is half of the angle BCD, and the angle CDF half of the angle CDE, therefore the angle FCD is also equal to the angle CDF, so that the side FC is also equal to the side FD. [I. 6] +

+

Similarly it can be proved that each of the straight lines FB, FA, FE is also equal to each of the straight lines FC, FD; therefore the five straight lines FA, FB, FC, FD, FE are equal to one another.

+

Therefore the circle described with centre F and distance one of the straight lines FA, FB, FC, FD, FE will pass also through the remaining points, and will have been circumscribed.

+

Let it be circumscribed, and let it be ABCDE.

+

Therefore about the given pentagon, which is equilateral and equiangular, a circle has been circumscribed. Q. E. F.

+
+
+ PROPOSITION 15. +

In a given circle to inscribe an equilateral and equiangular hexagon.

+

Let ABCDEF be the given circle; thus it is required to inscribe an equilateral and equiangular hexagon in the circle ABCDEF.

+

Let the diameter AD of the circle ABCDEF be drawn; let the centre G of the circle be taken, and with centre D and distance DG let the circle EGCH be described;

let EG, CG be joined and carried through to the points B, F, and let AB, BC, CD, DE, EF, FA be joined.

+

I say that the hexagon ABCDEF is equilateral and equiangular.

+

For, since the point G is the centre of the circle ABCDEF, GE is equal to GD. +

+

Again, since the point D is the centre of the circle GCH, DE is equal to DG. +

+

But GE was proved equal to GD; therefore GE is also equal to ED; therefore the triangle EGD is equilateral; and therefore its three angles EGD, GDE, DEG are equal to one another, inasmuch as, in isosceles triangles, the angles at the base are equal to one another. [I. 5]

+

And the three angles of the triangle are equal to two right angles; [I. 32] therefore the angle EGD is one-third of two right angles. +

+

Similarly, the angle DGC can also be proved to be onethird of two right angles.

+

And, since the straight line CG standing on EB makes the adjacent angles EGC, CGB equal to two right angles, therefore the remaining angle CGB is also one-third of two right angles.

+

Therefore the angles EGD, DGC, CGB are equal to one another; so that the angles vertical to them, the angles BGA, AGF, FGE are equal. [I. 15]

+

Therefore the six angles EGD, DGC, CGB, BGA, AGF, FGE are equal to one another.

+

But equal angles stand on equal circumferences; [III. 26] therefore the six circumferences AB, BC, CD, DE, EF, FA are equal to one another.

+

And equal circumferences are subtended by equal straight lines; [III. 29] therefore the six straight lines are equal to one another; therefore the hexagon ABCDEF is equilateral. +

+

I say next that it is also equiangular.

+

For, since the circumference FA is equal to the circumference ED, let the circumference ABCD be added to each; therefore the whole FABCD is equal to the whole EDCBA; + and the angle FED stands on the circumference FABCD, and the angle AFE on the circumference EDCBA; therefore the angle AFE is equal to the angle DEF. [III. 27] +

+

Similarly it can be proved that the remaining angles of the hexagon ABCDEF are also severally equal to each of the angles AFE, FED; therefore the hexagon ABCDEF is equiangular. +

+

But it was also proved equilateral; and it has been inscribed in the circle ABCDEF.

+

Therefore in the given circle an equilateral and equiangular hexagon has been inscribed. Q. E. F.

+ +

Porism. From this it is manifest that the side of the hexagon is equal to the radius of the circle.

+

And, in like manner as in the case of the pentagon, if through the points of division on the circle we draw tangents to the circle, there will be circumscribed about the circle an equilateral and equiangular hexagon in conformity with what was explained in the case of the pentagon.

+

And further by means similar to those explained in the case of the pentagon we can both inscribe a circle in a given hexagon and circumscribe one about it. Q. E. F.

+ +
+
+ + PROPOSITION 16. +

In a given circle to inscribe a fifteen-angled figure which shall be both equilateral and equiangular.

+

Let ABCD be the given circle; thus it is required to inscribe in the circle ABCD a fifteenangled figure which shall be both equilateral and equiangular.

+

+

In the circle ABCD let there be inscribed a side AC of the equilateral triangle inscribed in it, and a side AB of an equilateral pentagon; therefore, of the equal segments of which there are fifteen in the circle ABCD, there will be five in the circumference ABC which is one-third of the circle, and there will be three in the circumference AB which is one-fifth of the circle; therefore in the remainder BC there will be two of the equal segments. +

+

Let BC be bisected at E; [III. 30] therefore each of the circumferences BE, EC is a fifteenth of the circle ABCD.

+

If therefore we join BE, EC and fit into the circle ABCD straight lines equal to them and in contiguity, a fifteen-angled figure which is both equilateral and equiangular will have been inscribed in it. Q. E. F.

+

And, in like manner as in the case of the pentagon, if through the points of division on the circle we draw tangents to the circle, there will be circumscribed about the circle a fifteen-angled figure which is equilateral and equiangular.

+

And further, by proofs similar to those in the case of the pentagon, we can both inscribe a circle in the given fifteenangled figure and circumscribe one about it. Q. E. F.

+
+
+
+
+ + BOOK V. +
+ DEFINITIONS. +
+1 +

A magnitude is a part of a magnitude, the less of the greater, when it measures the greater.

+
+
+2 +

The greater is a multiple of the less when it is measured by the less.

+
+
+3 + +

A ratio is a sort of relation in respect of size between two magnitudes of the same kind.

+
+
+4 +

Magnitudes are said to have a ratio to one another which are capable, when multiplied, of exceeding one another.

+
+
+5 +

Magnitudes are said to be in the same ratio, the first to the second and the third to the fourth, when, if any equimultiples whatever be taken of the first and third, and any equimultiples whatever of the second and fourth, the former equimultiples alike exceed, are alike equal to, or alike fall short of, the latter equimultiples respectively taken in corresponding order.

+
+
+6 +

Let magnitudes which have the same ratio be called proportional.

+
+
+7 +

When, of the equimultiples, the multiple of the first magnitude exceeds the multiple of the second, but the multiple of the third does not exceed the multiple of the fourth, then the first is said to have a greater ratio to the second than the third has to the fourth.

+
+
+8 +

A proportion in three terms is the least possible.

+
+
+9 +

When three magnitudes are proportional, the first is said to have to the third the duplicate ratio of that which it has to the second.

+
+
+10 +

When four magnitudes are ltcontinuouslygt proportional, the first is said to have to the fourth the triplicate ratio of that which it has to the second, and so on continually, whatever be the proportion.

+
+
+11 +

The term corresponding magnitudes is used of antecedents in relation to antecedents, and of consequents in relation to consequents.

+
+
+12 +

Alternate ratio means taking the antecedent in relation to the antecedent and the consequent in relation to the consequent.

+
+
+13 +

Inverse ratio means taking the consequent as antecedent in relation to the antecedent as consequent.

+
+
+14 + +

Composition of a ratio means taking the antecedent together with the consequent as one in relation to the consequent by itself.

+
+
+15 +

Separation of a ratio means taking the excess by which the antecedent exceeds the consequent in relation to the consequent by itself.

+
+
+16 +

Conversion of a ratio means taking the antecedent in relation to the excess by which the antecedent exceeds the consequent.

+
+
+17 +

A ratio ex aequali arises when, there being several magnitudes and another set equal to them in multitude which taken two and two are in the same proportion, as the first is to the last among the first magnitudes, so is the first to the last among the second magnitudes;

+

Or, in other words, it means taking the extreme terms by virtue of the removal of the intermediate terms.

+
+
+18 +

A perturbed proportion arises when, there being three magnitudes and another set equal to them in multitude, as antecedent is to consequent among the first magnitudes, so is antecedent to consequent among the second magnitudes, while, as the consequent is to a third among the first magnitudes, so is a third to the antecedent among the second magnitudes.

+
+
+
+ + BOOK V. PROPOSITIONS. +
+ PROPOSITION I. +

If there be any number of magnitudes whatever which are, respectively, equimultiples of any magnitudes equal in multitude, then, whatever multiple one of the magnitudes is of one, that multiple also will all be of all.

+

Let any number of magnitudes whatever AB, CD be respectively equimultiples of any magnitudes E, F equal in multitude; I say that, whatever multiple AB is of E, that multiple will AB, CD also be of E, F.

+

+

For, since AB is the same multiple of E that CD is of F, as many magnitudes as there are in AB equal to E, so many also are there in CD equal to F.

+

Let AB be divided into the magnitudes AG, GB equal to E, and CD into CH, HD equal to F; then the multitude of the magnitudes AG, GB will be equal to the multitude of the magnitudes CH, HD.

+

Now, since AG is equal to E, and CH to F, therefore AG is equal to E, and AG, CH to E, F.

+

For the same reason

+

GB is equal to E, and GB, HD to E, F; therefore, as many magnitudes as there are in AB equal to E, so many also are there in AB, CD equal to E, F; therefore, whatever multiple AB is of E, that multiple will AB, CD also be of E, F.

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 2. +

If a first magnitude be the same multiple of a second that a third is of a fourth, and a fifth also be the same multiple of the second that a sixth is of the fourth, the sum of the first and fifth will also be the same multiple of the second that the sum of the third and sixth is of the fourth.

+

Let a first magnitude, AB, be the same multiple of a second, C, that a third, DE, is of a fourth, F, and let a fifth, BG, also be the same multiple of the second, C, that a sixth, EH, is of the fourth F;

I say that the sum of the first and fifth, AG, will be the same multiple of the second, C, that the sum of the third and sixth, DH, is of the fourth, F.

+

For, since AB is the same multiple of C that DE is of F, therefore, as many magnitudes as there are in AB equal to C, so many also are there in DE equal to F.

+

For the same reason also, as many as there are in BG equal to C, so many are there also in EH equal to F; therefore, as many as there are in the whole AG equal to C, so many also are there in the whole DH equal to F.

+

Therefore, whatever multiple AG is of C, that multiple also is DH of F.

+

Therefore the sum of the first and fifth, AG, is the same multiple of the second, C, that the sum of the third and sixth, DH, is of the fourth, F.

+

Therefore etc. Q.E.D.

+
+
+ PROPOSITION 3. +

If a first magnitude be the same multiple of a second that a third is of a fourth, and if equimultiples be taken of the first and third, then also ex aequali the magnitudes taken will be equimultiples respectively, the one of the second and the other of the fourth.

+

Let a first magnitude A be the same multiple of a second B that a third C is of a fourth D, and let equimultiples EF, GH be taken of A, C; I say that EF is the same multiple of B that GH is of D.

+

For, since EF is the same multiple of A that GH is of C, therefore, as many magnitudes as there are in EF equal to A, so many also are there in GH equal to C.

+

Let EF be divided into the magnitudes EK, KF equal to A, and GH into the magnitudes GL, LH equal to C; then the multitude of the magnitudes EK, KF will be equal to the multitude of the magnitudes GL, LH.

+

+

And, since A is the same multiple of B that C is of D, while EK is equal to A, and GL to C, therefore EK is the same multiple of B that GL is of D.

+

For the same reason

+

KF is the same multiple of B that LH is of D.

+

Since, then, a first magnitude EK is the same multiple of a second B that a third GL is of a fourth D, and a fifth KF is also the same multiple of the second B that a sixth LH is of the fourth D, therefore the sum of the first and fifth, EF, is also the same multiple of the second B that the sum of the third and sixth, GH, is of the fourth D. [V. 2]

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 4. +

If a first magnitude have to a second the same ratio as a third to a fourth, any equimultiples whatever of the first and third will also have the same ratio to any equimultiples whatever of the second and fourth respectively, taken in corresponding order.

+

For let a first magnitude A have to a second B the same ratio as a third C to a fourth D; and let equimultiples E, F be taken of A, C, and G, H other, chance, equimultiples of B, D; I say that, as E is to G, so is F to H.

+

+

For let equimultiples K, L be taken of E, F, and other, chance, equimultiples M, N of G, H.

+

Since E is the same multiple of A that F is of C, and equimultiples K, L of E, F have been taken, therefore K is the same multiple of A that L is of C. [V. 3]

+

For the same reason M is the same multiple of B that N is of D. +

+

And, since, as A is to B, so is C to D, and of A, C equimultiples K, L have been taken, and of B, D other, chance, equimultiples M, N, therefore, if K is in excess of M, L also is in excess of N, if it is equal, equal, and if less, less. [V. Def. 5]

+

And K, L are equimultiples of E, F, and M, N other, chance, equimultiples of G, H; therefore, as E is to G, so is F to H. [V. Def. 5]

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 5. +

If a magnitude be the same multiple of a magnitude that a part subtracted is of a part subtracted, the remainder will also be the same multiple of the remainder that the whole is of the whole.

+

For let the magnitude AB be the same multiple of the magnitude CD that the part AE subtracted is of the part CF subtracted; I say that the remainder EB is also the same multiple of the remainder FD that the whole AB is of the whole CD.

+

+

For, whatever multiple AE is of CF, let EB be made that multiple of CG.

+

Then, since AE is the same multiple of CF that EB is of GC, therefore AE is the same multiple of CF that AB is of GF. [V. 1]

+

But, by the assumption, AE is the same multiple of CF that AB is of CD.

+

Therefore AB is the same multiple of each of the magnitudes GF, CD; therefore GF is equal to CD. +

+

Let CF be subtracted from each; therefore the remainder GC is equal to the remainder FD.

+

And, since AE is the same multiple of CF that EB is of GC, and GC is equal to DF, therefore AE is the same multiple of CF that EB is of FD.

+

But, by hypothesis,

+

AE is the same multiple of CF that AB is of CD; therefore EB is the same multiple of FD that AB is of CD.

+

That is, the remainder EB will be the same multiple of the remainder FD that the whole AB is of the whole CD.

+

Therefore etc. Q. E. D. + +

let EB be made that multiple of CG, τοσαυταπλάσιον γεγονέτω καὶ τὸ ΕΒ τοῦ ΓΗ. From this way of stating the construction one might suppose that CG was given and EB had to be found equal to a certain multiple of it. But in fact EB is what is given and CG has to be found, i.e. CG has to be constructed as a certain submultiple of EB.

+

+
+
+ PROPOSITION 6. +

If two magnitudes be equimultiples of two magnitudes, and any magnitudes subtracted from them be equimultiples of the same, the remainders also are either equal to the same or equimultiples of them.

+

For let two magnitudes AB, CD be equimultiples of two magnitudes E, F, and let AG, CH subtracted from them be equimultiples of the same two E, F;

I say that the remainders also, GB, HD, are either equal to E, F or equimultiples of them.

+

For, first, let GB be equal to E; I say that HD is also equal to F.

+

For let CK be made equal to F.

+

Since AG is the same multiple of E that CH is of F, while GB is equal to E and KC to F, therefore AB is the same multiple of E that KH is of F. [V. 2]

+

But, by hypothesis, AB is the same multiple of E that CD is of F; therefore KH is the same multiple of F that CD is of F.

+

Since then each of the magnitudes KH, CD is the same multiple of F, therefore KH is equal to CD. +

+

Let CH be subtracted from each; therefore the remainder KC is equal to the remainder HD.

+

But F is equal to KC; therefore HD is also equal to F.

+

Hence, if GB is equal to E, HD is also equal to F.

+

Similarly we can prove that, even if GB be a multiple of E, HD is also the same multiple of F.

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 7. +

Equal magnitudes have to the same the same ratio, as also has the same to equal magnitudes.

+

Let A, B be equal magnitudes and C any other, chance, magnitude; I say that each of the magnitudes A, B has the same ratio to C, and C has the same ratio to each of the magnitudes A, B.

+

+

For let equimultiples D, E of A, B be taken, and of C another, chance, multiple F.

+

Then, since D is the same multiple of A that E is of B, while A is equal to B, therefore D is equal to E. +

+

But F is another, chance, magnitude.

+

If therefore D is in excess of F, E is also in excess of F, if equal to it, equal; and, if less, less.

+

And D, E are equimultiples of A, B, while F is another, chance, multiple of C; therefore, as A is to C, so is B to C. [V. Def. 5] +

+

I say next that C also has the same ratio to each of the magnitudes A, B.

+

For, with the same construction, we can prove similarly that D is equal to E; and F is some other magnitude.

+

If therefore F is in excess of D, it is also in excess of E, if equal, equal; and, if less, less.

+

And F is a multiple of C, while D, E are other, chance, equimultiples of A, B; therefore, as C is to A, so is C to B. [V. Def. 5] +

+

Therefore etc.

+ +

Porism. From this it is manifest that, if any magnitudes are proportional, they will also be proportional inversely. Q. E. D.

+ +
+
+ PROPOSITION 8. +

Of unequal magnitudes, the greater has to the same a greater ratio than the less has; and the same has to the less a greater ratio than it has to the greater.

+

Let AB, C be unequal magnitudes, and let AB be greater; let D be another, chance, magnitude; I say that AB has to D a greater ratio than C has to D, and D has to C a greater ratio than it has to AB.

+

+

For, since AB is greater than C, let BE be made equal to C; then the less of the magnitudes AE, EB, if multiplied, will sometime be greater than D. [V. Def. 4]

+

[Case I.]

+

First, let AE be less than EB; let AE be multiplied, and let FG be a multiple of it which is greater than D; then, whatever multiple FG is of AE, let GH be made the same multiple of EB and K of C; and let L be taken double of D, M triple of it, and successive multiples increasing by one, until what is taken is a multiple of D and the first that is greater than K. Let it be taken, and let it be N which is quadruple of D and the first multiple of it that is greather than K.

+

Then, since K is less than N first, therefore K is not less than M.

+

And, since FG is the same multiple of AE that GH is of EB, therefore FG is the same multiple of AE that FH is of AB. [V. 1]

+

But FG is the same multiple of AE that K is of C; therefore FH is the same multiple of AB that K is of C; therefore FH, K are equimultiples of AB, C. +

+

Again, since GH is the same multiple of EB that K is of C, and EB is equal to C, therefore GH is equal to K. +

+

But K is not less than M; therefore neither is GH less than M. +

+

And FG is greater than D; therefore the whole FH is greater than D, M together.

+

But D, M together are equal to N, inasmuch as M is triple of D, and M, D together are quadruple of D, while N is also quadruple of D; whence M, D together are equal to N.

+

But FH is greater than M, D; therefore FH is in excess of N, while K is not in excess of N.

+

And FH, K are equimultiples of AB, C, while N is another, chance, multiple of D; therefore AB has to D a greater ratio than C has to D. [V. Def. 7] +

+

I say next, that D also has to C a greater ratio than D has to AB.

+

For, with the same construction, we can prove similarly that N is in excess of K, while N is not in excess of FH.

+

And N is a multiple of D, while FH, K are other, chance, equimultiples of AB, C; therefore D has to C a greater ratio than D has to AB. [V. Def. 7] +

+

[Case 2.]

+

Again, let AE be greater than EB.

+

Then the less, EB, if multiplied, will sometime be greater than D. [V. Def. 4]

+

Let it be multiplied, and let GH be a multiple of EB and greater than D; and, whatever multiple GH is of EB, let FG be made the same multiple of AE, and K +

of C.

+

Then we can prove similarly that FH, K are equimultiples of AB, C; and, similarly, let N be taken a multiple of D but the first that is greater than FG, so that FG is again not less than M.

+

But GH is greater than D; therefore the whole FH is in excess of D, M, that is, of N.

+

Now K is not in excess of N, inasmuch as FG also, which is greater than GH, that is, than K, is not in excess of N.

+

And in the same manner, by following the above argument, we complete the demonstration.

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 9. +

Magnitudes which have the same ratio to the same are equal to one another; and magnitudes to which the same has the same ratio are equal.

+

For let each of the magnitudes A, B have the same ratio to C; I say that A is equal to B.

+

+

For, otherwise, each of the magnitudes A, B would not have had the same ratio to C; [V. 8] but it has; therefore A is equal to B. +

+

Again, let C have the same ratio to each of the magnitudes A, B; I say that A is equal to B.

+

For, otherwise, C would not have had the same ratio to each of the magnitudes A, B; [V. 8] but it has; therefore A is equal to B. +

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 10. +

Of magnitudes which have a ratio to the same, that which has a greater ratio is greater; and that to which the same has a greater ratio is less.

+

For let A have to C a greater ratio than B has to C; I say that A is greater than B.

+

+

For, if not, A is either equal to B or less.

+

Now A is not equal to B; for in that case each of the magnitudes A, B would have had the same ratio to C; [V. 7] but they have not; therefore A is not equal to B. +

+

Nor again is A less than B; for in that case A would have had to C a less ratio than B has to C; [V. 8] but it has not; therefore A is not less than B. +

+

But it was proved not to be equal either; therefore A is greater than B. +

+

Again, let C have to B a greater ratio than C has to A; I say that B is less than A.

+

For, if not, it is either equal or greater.

+

Now B is not equal to A; for in that case C would have had the same ratio to each of the magnitudes A, B; [V. 7] but it has not; therefore A is not equal to B. +

+

Nor again is B greater than A; for in that case C would have had to B a less ratio than it has to A; [V. 8] but it has not; therefore B is not greater than A. +

+

But it was proved that it is not equal either; therefore B is less than A. +

+

Therefore etc. Q. E. D.

+
+
+ + PROPOSITION 11. +

Ratios which are the same with the same ratio are also the same with one another.

+

For, as A is to B, so let C be to D, and, as C is to D, so let E be to F; I say that, as A is to B, so is E to F.

+

+

For of A, C, E let equimultiples G, H, K be taken, and of B, D, F other, chance, equimultiples L, M, N.

+

Then since, as A is to B, so is C to D, and of A, C equimultiples G, H have been taken, and of B, D other, chance, equimultiples L, M, therefore, if G is in excess of L, H is also in excess of M, if equal, equal, and if less, less.

+

Again, since, as C is to D, so is E to F, and of C, E equimultiples H, K have been taken, and of D, F other, chance, equimultiples M, N, therefore, if H is in excess of M, K is also in excess of N, if equal, equal, and if less, less.

+

But we saw that, if H was in excess of M, G was also in excess of L; if equal, equal; and if less, less; so that, in addition, if G is in excess of L, K is also in excess of N, if equal, equal, and if less, less.

+

And G, K are equimultiples of A, E, while L, N are other, chance, equimultiples of B, F; therefore, as A is to B, so is E to F. +

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 12. +

If any number of magnitudes be proportional, as one of the antecedents is to one of the consequents, so will all the antecedents be to all the consequents.

+

Let any number of magnitudes A, B, C, D, E, F be proportional, so that, as A is to B, so is C to D and E to F; I say that, as A is to B, so are A, C, E to B, D, F.

+

+

For of A, C, E let equimultiples G, H, K be taken, and of B, D, F other, chance, equimultiples L, M, N.

+

Then since, as A is to B, so is C to D, and E to F, and of A, C, E equimultiples G, H, K have been taken, and of B, D, F other, chance, equimultiples L, M, N, therefore, if G is in excess of L, H is also in excess of M, and K of N, if equal, equal, and if less, less; so that, in addition, if G is in excess of L, then G, H, K are in excess of L, M, N, if equal, equal, and if less, less.

+

Now G and G, H, K are equimultiples of A and A, C, E, since, if any number of magnitudes whatever are respectively equimultiples of any magnitudes equal in multitude, whatever multiple one of the magnitudes is of one, that multiple also will all be of all. [V. 1]

+

For the same reason L and L, M, N are also equimultiples of B and B, D, F; therefore, as A is to B, so are A, C, E to B, D, F. [V. Def. 5] +

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 13. +

If a first magnitude have to a second the same ratio as a third to a fourth, and the third have to the fourth a greater ratio than a fifth has to a sixth, the first will also have to the second a greater ratio than the fifth to the sixth.

+

For let a first magnitude A have to a second B the same ratio as a third C has to a fourth D, and let the third C have to the fourth D a greater ratio than a fifth E has to a sixth F; I say that the first A will also have to the second B a greater ratio than the fifth E to the sixth F.

+

+

For, since there are some equimultiples of C, E, and of D, F other, chance, equimultiples, such that the multiple of C is in excess of the multiple of D, while the multiple of E is not in excess of the multiple of F, [V. Def. 7] let them be taken, and let G, H be equimultiples of C, E, and K, L other, chance, equimultiples of D, F, so that G is in excess of K, but H is not in excess of L; and, whatever multiple G is of C, let M be also that multiple of A, and, whatever multiple K is of D, let N be also that multiple of B.

+

Now, since, as A is to B, so is C to D, and of A, C equimultiples M, G have been taken, and of B, D other, chance, equimultiples N, K, therefore, if M is in excess of N, G is also in excess of K, if equal, equal, and if less, less. [V. Def. 5]

+

But G is in excess of K; therefore M is also in excess of N.

+

But H is not in excess of L; and M, H are equimultiples of A, E, and N, L other, chance, equimultiples of B, F; therefore A has to B a greater ratio than E has to F. [V. Def. 7] +

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 14. +

If a first magnitude have to a second the same ratio as a third has to a fourth, and the first be greater than the third, the second will also be greater than the fourth; if equal, equal; and if less, less.

+

For let a first magnitude A have the same ratio to a second B as a third C has to a fourth D; and let A be greater than C; I say that B is also greater than D.

+

+

For, since A is greater than C, and B is another, chance, magnitude, therefore A has to B a greater ratio than C has to B. [V. 8]

+

But, as A is to B, so is C to D; therefore C has also to D a greater ratio than C has to B. [V. 13] +

+

But that to which the same has a greater ratio is less; [V. 10] therefore D is less than B; so that B is greater than D. +

+

Similarly we can prove that, if A be equal to C, B will also be equal to D; and, if A be less than C, B will also be less than D.

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 15. +

Parts have the same ratio as the same multiples of them taken in corresponding order.

+

For let AB be the same multiple of C that DE is of F; I say that, as C is to F, so is AB to DE.

+

+

For, since AB is the same multiple of C that DE is of F, as many magnitudes as there are in AB equal to C, so many are there also in DE equal to F.

+

Let AB be divided into the magnitudes AG, GH, HB equal to C, and DE into the magnitudes DK, KL, LE equal to F; then the multitude of the magnitudes AG, GH, HB will be equal to the multitude of the magnitudes DK, KL, LE.

+

And, since AG. GH, HB are equal to one another, and DK, KL, LE are also equal to one another, therefore, as AG is to DK, so is GH to KL, and HB to LE. [V. 7]

+

Therefore, as one of the antecedents is to one of the consequents, so will all the antecedents be to all the consequents; [V. 12] therefore, as AG is to DK, so is AB to DE. +

+

But AG is equal to C and DK to F; therefore, as C is to F, so is AB to DE. +

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 16. +

If four magnitudes be proportional, they will also be proportional alternately.

+

Let A, B, C, D be four proportional magnitudes, so that, as A is to B, so is C to D; I say that they will also be so alternately, that is, as A is to C, so is B to D.

+

+

For of A, B let equimultiples E, F be taken, and of C, D other, chance, equimultiples G, H.

+

Then, since E is the same multiple of A that F is of B, and parts have the same ratio as the same multiples of them, [V. 15] therefore, as A is to B, so is E to F.

+

But as A is to B, so is C to D; therefore also, as C is to D, so is E to F. [V. 11]

+

Again, since G, H are equimultiples of C, D, therefore, as C is to D, so is G to H. [V. 15]

+

But, as C is to D, so is E to F; therefore also, as E is to F, so is G to H. [V. 11]

+

But, if four magnitudes be proportional, and the first be greater than the third, the second will also be greater than the fourth; if equal, equal; and if less, less. [V. 14]

+

Therefore, if E is in excess of G, F is also in excess of H, if equal, equal, and if less, less.

+

Now E, F are equimultiples of A, B, and G, H other, chance, equimultiples of C, D; therefore, as A is to C, so is B to D. [V. Def. 5] +

+

Therefore etc. Q. E. D. + +

Let A, B, C, D be four proportional magnitudes, so that, as A is to B, so is C to D. + In a number of expressions like this it is absolutely necessary, when translating into English, to interpolate words which are not in the Greek. Thus the Greek here is: Ἕστω τέσσαρα μεγέθη ἀνάλογον τὰ Α, Β, Γ, Δ, ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς τὸ Δ, literally Let A, B, C, D be four proportional magnitudes, as A to B, so C to D. + The same remark applies to the corresponding expressions in the next propositions, V. 17, 18, and to other forms of expression in V. 20-23 and later propositions: e.g. in V. 20 we have a phrase meaning literally Let there be magnitudes...which taken two and two are in the same ratio, as A to B, so D to E, + etc.: in V. 21 (magnitudes)...which taken two and two are in the same ratio, and let the proportion of them be perturbed, as A to B, so E to F, + etc. In all such cases (where the Greek is so terse as to be almost ungrammatical) I shall insert the words necessary in English, without further remark.

+

+
+
+ PROPOSITION 17. +

If magnitudes be proportional + componendo, they will also be proportional + separando.

+

Let AB, BE, CD, DF be magnitudes proportional componendo, so that, as AB is to BE, so is CD to DF; I say that they will also be proportional separando, that is, as AE is to EB, so is CF to DF.

+

For of AE, EB, CF, FD let equimultiples GH, HK, LM, MN be taken, and of EB, FD other, chance, equimultiples, KO, NP.

+

Then, since GH is the same multiple of AE that HK is of EB, therefore GH is the same multiple of AE that GK is of AB. [V. 1]

+

But GH is the same multiple of AE that LM is of CF; therefore GK is the same multiple of AB that LM is of CF.

+

+

Again, since LM is the same multiple of CF that MN is of FD, therefore LM is the same multiple of CF that LN is of CD. [V. 1]

+

But LM was the same multiple of CF that GK is of AB; therefore GK is the same multiple of AB that LN is of CD.

+

Therefore GK, LN are equimultiples of AB, CD.

+

Again, since HK is the same multiple of EB that MN is of FD, and KO is also the same multiple of EB that NP is of FD, therefore the sum HO is also the same multiple of EB that MP is of FD. [V. 2] +

+

And, since, as AB is to BE, so is CD to DF, and of AB, CD equimultiples GK, LN have been taken, and of EB, FD equimultiples HO, MP, therefore, if GK is in excess of HO, LN is also in excess of MP, if equal, equal, and if less, less.

+

Let GK be in excess of HO; then, if HK be subtracted from each, GH is also in excess of KO. +

+

But we saw that, if GK was in excess of HO, LN was also in excess of MP; therefore LN is also in excess of MP, + and, if MN be subtracted from each, LM is also in excess of NP; so that, if GH is in excess of KO, LM is also in excess of NP.

+

Similarly we can prove that, if GH be equal to KO, LM will also be equal to NP, and if less, less.

+

And GH, LM are equimultiples of AE, CF, while KO, NP are other, chance, equimultiples of EB, FD; therefore, as AE is to EB, so is CF to FD. +

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 18. +

If magnitudes be proportional + separando, they will also be proportional + componendo.

+

Let AE, EB, CF, FD be magnitudes proportional separando, so that, as AE is to EB, so is CF to FD;

I say that they will also be proportional componendo, that is, as AB is to BE, so is CD to FD.

+

For, if CD be not to DF as AB to BE, then, as AB is to BE, so will CD be either to some magnitude less than DF or to a greater.

+

First, let it be in that ratio to a less magnitude DG.

+

Then, since, as AB is to BE, so is CD to DG, they are magnitudes proportional componendo; so that they will also be proportional separando. [V. 17] +

+

Therefore, as AE is to EB, so is CG to GD.

+

But also, by hypothesis, as AE is to EB, so is CF to FD. +

+

Therefore also, as CG is to GD, so is CF to FD. [V. 11]

+

But the first CG is greater than the third CF; therefore the second GD is also greater than the fourth FD. [V. 14] +

+

But it is also less: which is impossible.

+

Therefore, as AB is to BE, so is not CD to a less magnitude than FD.

+

Similarly we can prove that neither is it in that ratio to a greater; it is therefore in that ratio to FD itself. +

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 19 +

If, as a whole is to a whole, so is a part subtracted to a part subtracted, the remainder will also be to the remainder as whole to whole.

+

For, as the whole AB is to the whole CD, so let the part AE subtracted be to the part CF subtracted; I say that the remainder EB will also be to the remainder FD as the whole AB to the whole CD.

+

+

For since, as AB is to CD, so is AE to CF, alternately also, as BA is to AE, so is DC to CF. [V. 16]

+

And, since the magnitudes are proportional componendo, they will also be proportional separando, [V. 17] that is, as BE is to EA, so is DF to CF, and, alternately, as BE is to DF, so is EA to FC. [V. 16] +

+

But, as AE is to CF, so by hypothesis is the whole AB to the whole CD.

+

Therefore also the remainder EB will be to the remainder FD as the whole AB is to the whole CD. [V. 11]

+

Therefore etc. +[

+ +

Porism. From this it is manifest that, if magnitudes be proportional componendo, they will also be proportional convertendo.

+ +

] Q. E. D.

+
+
+ PROPOSITION 20. +

If there be three magnitudes, and others equal to them in multitude, which taken two and two are in the same ratio, and if + ex aequali + the first be greater than the third, the fourth will also be greater than the sixth; if equal, equal; and, if less, less.

+

Let there be three magnitudes A, B, C, and others D, E, F equal to them in multitude, which taken two and two are in the same ratio, so that, as A is to B, so is D to E, and as B is to C, so is E to F; and let A be greater than C ex aequali; I say that D will also be greater than F; if A is equal to C, equal; and, if less, less.

+

+

For, since A is greater than C, and B is some other magnitude, and the greater has to the same a greater ratio than the less has, [V. 8] therefore A has to B a greater ratio than C has to B.

+

But, as A is to B, so is D to E, and, as C is to B, inversely, so is F to E; therefore D has also to E a greater ratio than F has to E. [V. 13]

+

But, of magnitudes which have a ratio to the same, that which has a greater ratio is greater; [V. 10] therefore D is greater than F. +

+

Similarly we can prove that, if A be equal to C, D will also be equal to F; and if less, less.

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 21. +

If there be three magnitudes, and others equal to them in multitude, which taken two and two together are in the same ratio, and the proportion of them be perturbed, then, if ex aequali + the first magnitude is greater than the third, the fourth will also be greater than the sixth; if equal, equal; and if less, less.

+

Let there be three magnitudes A, B, C, and others D, E, F equal to them in multitude, which taken two and two are in the same ratio, and let the proportion of them be perturbed, so that, as A is to B, so is E to F, and, as B is to C, so is D to E, and let A be greater than C + ex aequali; I say that D will also be greater than F; if A is equal to C, equal; and if less, less.

+

+

For, since A is greater than C, and B is some other magnitude, therefore A has to B a greater ratio than C has to B. [V. 8]

+

But, as A is to B, so is E to F, and, as C is to B, inversely, so is E to D. Therefore also E has to F a greater ratio than E has to D. [V. 13]

+

But that to which the same has a greater ratio is less; [V. 10] therefore F is less than D; therefore D is greater than F. +

+

Similarly we can prove that, if A be equal to C, D will also be equal to F; and if less, less.

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 22. +

If there be any number of magnitudes whatever, and others equal to them in multitude, which taken two and two together are in the same ratio, they will also be in the same ratio ex aequali.

+

Let there be any number of magnitudes A, B, C, and others D, E, F equal to them in multitude, which taken two and two together are in the same ratio, so that, as A is to B, so is D to E, and, as B is to C, so is E to F; I say that they will also be in the same ratio ex aequali, ltthat is, as A is to C, so is D to Fgt. +

+

For of A, D let equimultiples G, H be taken, and of B, E other, chance, equimultiples K, L; and, further, of C, F other, chance, equimultiples M, N.

+

+

Then, since, as A is to B, so is D to E, and of A, D equimultiples G, H have been taken, and of B, E other, chance, equimultiples K, L, therefore, as G is to K, so is H to L. [V. 4] +

+

For the same reason also, as K is to M, so is L to N. +

+

Since, then, there are three magnitudes G, K, M, and others H, L, N equal to them in multitude, which taken two and two together are in the same ratio, therefore, ex aequali, if G is in excess of M, H is also in excess of N; if equal, equal; and if less, less. [V. 20]

+

And G, H are equimultiples of A, D, and M, N other, chance, equimultiples of C, F. +

+

Therefore, as A is to C, so is D to F. [V. Def. 5]

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 23. +

If there be three magnitudes, and others equal to them in multitude, which taken two and two together are in the same ratio, and the proportion of them be perturbed, they will also be in the same ratio + ex aequali.

+

Let there be three magnitudes A, B, C, and others equal to them in multitude, which, taken two and two together, are in the same proportion, namely D, E, F; and let the proportion of them be perturbed, so that, as A is to B, so is E to F, and, as B is to C, so is D to E; I say that, as A is to C, so is D to F.

+

+

Of A, B, D let equimultiples G, H, K be taken, and of C, E, F other, chance, equimultiples L, M, N.

+

Then, since G, H are equimultiples of A, B, and parts have the same ratio as the same multiples of them, [V. 15] therefore, as A is to B, so is G to H. +

+

For the same reason also, as E is to F, so is M to N. And, as A is to B, so is E to F; therefore also, as G is to H, so is M to N. [V. 11] +

+

Next, since, as B is to C, so is D to E, alternately, also, as B is to D, so is C to E. [V. 16]

+

And, since H, K are equimultiples of B, D, and parts have the same ratio as their equimultiples, therefore, as B is to D, so is H to K. [V. 15] +

+

But, as B is to D, so is C to E; therefore also, as H is to K, so is C to E. [V. 11] +

+

Again, since L, M are equimultiples of C, E, therefore, as C is to E, so is L to M. [V. 15] +

+

But, as C is to E, so is H to K; therefore also, as H is to K, so is L to M, [V. 11] and, alternately, as H is to L, so is K to M. [V. 16]

+

But it was also proved that, as G is to H, so is M to N. +

+

Since, then, there are three magnitudes G, H, L, and others equal to them in multitude K, M, N, which taken two and two together are in the same ratio, and the proportion of them is perturbed, therefore, ex aequali, if G is in excess of L, K is also in excess of N; if equal, equal; and if less, less. [V. 21]

+

And G, K are equimultiples of A, D, and L, N of C, F.

+

Therefore, as A is to C, so is D to F.

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 24. +

If a first magnitude have to a second the same ratio as a third has to a fourth, and also a fifth have to the second the same ratio as a sixth to the fourth, the first and fifth added together will have to the second the same ratio as the third and sixth have to the fourth.

+

Let a first magnitude AB have to a second C the same ratio as a third DE has to a fourth F; and let also a fifth BG have to

the second C the same ratio as a sixth EH has to the fourth F; I say that the first and fifth added together, AG, will have to the second C the same ratio as the third and sixth, DH, has to the fourth F.

+

For since, as BG is to C, so is EH to F, inversely, as C is to BG, so is F to EH.

+

Since, then, as AB is to C, so is DE to F, and, as C is to BG, so is F to EH, therefore, ex aequali, as AB is to BG, so is DE to EH. [V. 22]

+

And, since the magnitudes are proportional separando, they will also be proportional componendo; [V. 18] therefore, as AG is to GB, so is DH to HE. +

+

But also, as BG is to C, so is EH to F; therefore, ex aequali, as AG is to C, so is DH to F. [V. 22]

+

Therefore etc. Q. E. D.

+
+
+ + PROPOSITION 25. +

If four magnitudes be proportional, the greatest and the least are greater than the remaining two.

+

Let the four magnitudes AB, CD, E, F be proportional so that, as AB is to CD, so is E to F, and let AB be the greatest of them and F the least; I say that AB, F are greater than CD, E.

+

+

For let AG be made equal to E, and CH equal to F.

+

Since, as AB is to CD, so is E to F, and E is equal to AG, and F to CH, therefore, as AB is to CD, so is AG to CH. +

+

And since, as the whole AB is to the whole CD, so is the part AG subtracted to the part CH subtracted, the remainder GB will also be to the remainder HD as the whole AB is to the whole CD. [V. 19] +

+

But AB is greater than CD; therefore GB is also greater than HD. +

+

And, since AG is equal to E, and CH to F, therefore AG, F are equal to CH, E.

+

And if, GB, HD being unequal, and GB greater, AG, F be added to GB and CH, E be added to HD, it follows that AB, F are greater than CD, E. +

+

Therefore etc. Q. E. D.

+
+
+
+
+ + BOOK VI. +
+ + DEFINITIONS. +
+1 +

Similar rectilineal figures are such as have their angles severally equal and the sides about the equal angles proportional.

+
+
+2 +

[Reciprocally related figures. See note.]

+
+
+3 +

A straight line is said to have been cut in extreme and mean ratio when, as the whole line is to the greater segment, so is the greater to the less.

+
+
+4 +

The height of any figure is the perpendicular drawn from the vertex to the base.

+
+
+5 +

+
+
+
+ + BOOK VI. PROPOSITIONS. +
+ PROPOSITION 1. +

Triangles and parallelograms which are under the same height are to one another as their bases.

+

Let ABC, ACD be triangles and EC, CF parallelograms under the same height; I say that, as the base BC is to the base CD, so is the triangle ABC to the triangle ACD, and the parallelogram EC to the parallelogram CF.

+

+

For let BD be produced in both directions to the points H, L and let [any number of straight lines] BG, GH be made equal to the base BC, and any number of straight lines DK, KL equal to the base CD; let AG, AH, AK, AL be joined.

+

Then, since CB, BG, GH are equal to one another, the triangles ABC, AGB, AHG are also equal to one another. [I. 38] +

+

Therefore, whatever multiple the base HC is of the base BC, that multiple also is the triangle AHC of the triangle ABC.

+

For the same reason, whatever multiple the base LC is of the base CD, that multiple also is the triangle ALC of the triangle ACD; and, if the base HC is equal to the base CL, the triangle AHC is also equal to the triangle ACL, [I. 38] if the base HC is in excess of the base CL, the triangle AHC + is also in excess of the triangle ACL, and, if less, less.

+

Thus, there being four magnitudes, two bases BC, CD and two triangles ABC, ACD, equimultiples have been taken of the base BC and the triangle ABC, namely the base HC and the triangle AHC, and of the base CD and the triangle ADC other, chance, equimultiples, namely the base LC and the triangle ALC; and it has been proved that, if the base HC is in excess of the base CL, the triangle AHC + is also in excess of the triangle ALC; if equal, equal; and, if less, less.

+

Therefore, as the base BC is to the base CD, so is the triangle ABC to the triangle ACD. [V. Def. 5]

+

Next, since the parallelogram EC is double of the triangle ABC, [I. 41] and the parallelogram FC is double of the triangle ACD, while parts have the same ratio as the same multiples of them, [V. 15] therefore, as the triangle ABC is to the triangle ACD, so is the parallelogram EC to the parallelogram FC.

+

Since, then, it was proved that, as the base BC is to CD, so is the triangle ABC to the triangle ACD, and, as the triangle ABC is to the triangle ACD, so is the parallelogram EC to the parallelogram CF, therefore also, as the base BC is to the base CD, so is the parallelogram EC to the parallelogram FC. [V. 11]

+

Therefore etc. Q. E. D. + +

Under the same height. The Greek text has under the same height AC, + with a figure in which the side AC common to the two triangles is perpendicular to the base and is therefore itself the height. + But, even if the two triangles are placed contiguously so as to have a common side AC, it is quite gratuitous to require it to be perpendicular to the base. Theon, on this occasion making an improvement, altered to which are (ὅντα) under the same height, (namely) the perpendicular drawn from A to BD. + I have ventured to alter so far as to omit AC + and to draw the figure in the usual way.

+ + +

ABC, AGB, AHG. Euclid, indifferent to exact order, writes AHG, AGB, ABC. +

+ + +

Since then it was proved that, as the base BC is to CD, so is the triangle ABC to the triangle ACD. Here again words have to be supplied in translating the extremely terse Greek ἐπεὶ οὐν ὲδείχθη, ὡς μὲν ἡ βάσις ΒΓ πρὸς τὴν ΓΔ, οὔτως τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΑΓΔ τρίγωνον, literally since was proved, as the base BC to CD, so the triangle ABC to the triangle ACD. + Cf. note on V. 16, p. 165.

+

+
+
+ PROPOSITION 2. +

If a straight line be drawn parallel to one of the sides of a triangle, it will cut the sides of the triangle proportionally; and, if the sides of the triangle be cut proportionally, the line joining the points of section will be parallel to the remaining side of the triangle.

+

For let DE be drawn parallel to BC, one of the sides of the triangle ABC; I say that, as BD is to DA, so is CE to EA.

+

For let BE, CD be joined.

+

Therefore the triangle BDE is equal to

the triangle CDE; for they are on the same base DE and in the same parallels DE, BC. [I. 38]

+

And the triangle ADE is another area.

+

But equals have the same ratio to the same; [V. 7] therefore, as the triangle BDE is to the triangle ADE, so is the triangle CDE to the triangle ADE.

+

But, as the triangle BDE is to ADE, so is BD to DA; for, being under the same height, the perpendicular drawn from E to AB, they are to one another as their bases. [VI. 1]

+

For the same reason also, as the triangle CDE is to ADE, so is CE to EA.

+

Therefore also, as BD is to DA, so is CE to EA. [V. 11]

+

Again, let the sides AB, AC of the triangle ABC be cut proportionally, so that, as BD is to DA, so is CE to EA; and let DE be joined.

+

I say that DE is parallel to BC.

+

For, with the same construction, since, as BD is to DA, so is CE to EA, but, as BD is to DA, so is the triangle BDE to the triangle ADE, and, as CE is to EA, so is the triangle CDE to the triangle ADE, [VI. 1] therefore also, as the triangle BDE is to the triangle ADE, so is the triangle CDE to the triangle ADE. [V. 11] +

+

Therefore each of the triangles BDE, CDE has the same ratio to ADE.

+

Therefore the triangle BDE is equal to the triangle CDE; [V. 9] and they are on the same base DE.

+

But equal triangles which are on the same base are also in the same parallels. [I. 39]

+

Therefore DE is parallel to BC.

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 3. +

If an angle of a triangle be bisected and the straight line cutting the angle cut the base also, the segments of the base will have the same ratio as the remaining sides of the triangle; and, if the segments of the base have the same ratio as the remaining sides of the triangle, the straight line joined from the vertex to the point of section will bisect the angle of the triangle.

+

Let ABC be a triangle, and let the angle BAC be bisected by the straight line AD; I say that, as BD is to CD, so is BA to AC.

+

For let CE be drawn through C parallel to DA, and let BA +

be carried through and meet it at E.

+

Then, since the straight line AC falls upon the parallels AD, EC, the angle ACE is equal to the angle CAD. [I. 29] +

+

But the angle CAD is by hypothesis equal to the angle BAD; therefore the angle BAD is also equal to the angle ACE.

+

Again, since the straight line BAE falls upon the parallels AD, EC, the exterior angle BAD is equal to the interior angle AEC. [I. 29] +

+

But the angle ACE was also proved equal to the angle BAD; therefore the angle ACE is also equal to the angle AEC, so that the side AE is also equal to the side AC. [I. 6] +

+

And, since AD has been drawn parallel to EC, one of the sides of the triangle BCE, therefore, proportionally, as BD is to DC, so is BA to AE.

+

But AE is equal to AC; [VI. 2] therefore, as BD is to DC, so is BA to AC.

+

Again, let BA be to AC as BD to DC, and let AD be joined; I say that the angle BAC has been bisected by the straight line A.D.

+

For, with the same construction, since, as BD is to DC, so is BA to AC, and also, as BD is to DC, so is BA to AE : for AD has been drawn parallel to EC, one of the sides of the triangle BCE: [VI. 2] therefore also, as BA is to AC, so is BA to AE. [V. 11]

+

Therefore AC is equal to AE, [V. 9] so that the angle AEC is also equal to the angle ACE. [I. 5]

+

But the angle AEC is equal to the exterior angle BAD, [I. 29] and the angle ACE is equal to the alternate angle CAD; [id.] therefore the angle BAD is also equal to the angle CAD. +

+

Therefore the angle BAC has been bisected by the straight line AD.

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 4. +

In equiangular triangles the sides about the equal angles are proportional, and those are corresponding sides which subtend the equal angles.

+

Let ABC, DCE be equiangular triangles having the angle ABC equal to the angle DCE, the angle BAC to the angle CDE, and further the angle ACB to the angle CED; I say that in the triangles ABC, DCE the sides about the equal angles are proportional, and those are corresponding sides which subtend the equal angles.

+

+

For let BC be placed in a straight line with CE.

+

Then, since the angles ABC, ACB are less than two right angles, [I. 17] and the angle ACB is equal to the angle DEC, therefore the angles ABC, DEC are less than two right angles; therefore BA, ED, when produced, will meet. [I. Post. 5]

+

Let them be produced and meet at F.

+

Now, since the angle DCE is equal to the angle ABC, BF is parallel to CD. [I. 28] +

+

Again, since the angle ACB is equal to the angle DEC, AC is parallel to FE. [I. 28] +

+

Therefore FACD is a parallelogram; therefore FA is equal to DC, and AC to FD. [I. 34]

+

And, since AC has been drawn parallel to FE, one side of the triangle FBE, therefore, as BA is to AF, so is BC to CE. [VI. 2]

+

But AF is equal to CD; therefore, as BA is to CD, so is BC to CE, and alternately, as AB is to BC, so is DC to CE. [V. 16]

+

Again, since CD is parallel to BF, therefore, as BC is to CE, so is FD to DE. [VI. 2]

+

But FD is equal to AC; therefore, as BC is to CE, so is AC to DE, and alternately, as BC is to CA, so is CE to ED. [V. 16]

+

Since then it was proved that, as AB is to BC, so is DC to CE, and, as BC is to CA, so is CE to ED; therefore, ex aequali, as BA is to AC, so is CD to DE. [V. 22]

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 5. +

If two triangles have their sides proportional, the triangles will be equiangular and will have those angles equal which the corresponding sides subtend.

+

Let ABC, DEF be two triangles having their sides proportional, so that, as AB is to BC, so is DE to EF, as BC is to CA, so is EF to FD, and further, as BA is to AC, so is ED to DF; I say that the triangle ABC is equiangular with the triangle DEF, and they will have those angles equal which the corresponding sides subtend, namely the angle ABC to the angle DEF, the angle BCA to the angle EFD, and further the angle BAC to the angle EDF.

+

For on the straight line EF, and at the points E, F on it, let there be constructed the angle FEG equal to the angle ABC, and the angle EFG equal to the angle ACB; [I. 23] therefore the remaining angle at A is equal to the remaining angle at G. [I. 32] +

+

Therefore the triangle ABC is equiangular with the triangle GEF.

+

+

Therefore in the triangles ABC, GEF the sides about the equal angles are proportional, and those are corresponding sides which subtend the equal angles; [VI. 4] therefore, as AB is to BC, so is GE to EF. +

+

But, as AB is to BC, so by hypothesis is DE to EF; therefore, as DE is to EF, so is GE to EF. [V. 11] +

+

Therefore each of the straight lines DE, GE has the same ratio to EF; therefore DE is equal to GE. [V. 9] +

+

For the same reason DF is also equal to GF. +

+

Since then DE is equal to EG, and EF is common, the two sides DE, EF are equal to the two sides GE, EF; and the base DF is equal to the base FG; therefore the angle DEF is equal to the angle GEF, [I. 8] and the triangle DEF is equal to the triangle GEF, and the remaining angles are equal to the remaining angles, namely those which the equal sides subtend. [I. 4]

+

Therefore the angle DFE is also equal to the angle GFE, and the angle EDF to the angle EGF. +

+

And, since the angle FED is equal to the angle GEF, while the angle GEF is equal to the angle ABC, therefore the angle ABC is also equal to the angle DEF.

+

For the same reason the angle ACB is also equal to the angle DFE, and further, the angle at A to the angle at D; therefore the triangle ABC is equiangular with the triangle DEF. +

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 6. +

If two triangles have one angle equal to one angle and the sides about the equal angles proportional, the triangles will be equiangular and will have those angles equal which the corresponding sides subtend.

+

Let ABC, DEF be two triangles having one angle BAC equal to one angle EDF and the sides about the equal angles proportional, so that, as BA is to AC, so is ED to DF; I say that the triangle ABC is equiangular with the triangle DEF, and will have the angle ABC equal to the angle DEF, and the angle ACB to the angle DFE.

+

For on the straight line DF, and at the points D, F on it, let there be constructed the angle FDG equal to either of the angles BAC, EDF, and the angle DFG equal to the angle ACB; [I. 23] therefore the remaining angle at B is equal to the remaining angle at G. [I. 32] +

+

Therefore the triangle ABC is equiangular with the triangle DGF.

+

Therefore, proportionally, as BA is to AC, so is GD to DF. [VI. 4]

+

But, by hypothesis, as BA is to AC, so also is ED to DF; therefore also, as ED is to DF, so is GD to DF. [V. 11]

+

+

Therefore ED is equal to DG; [V. 9] and DF is common; therefore the two sides ED, DF are equal to the two sides GD, DF; and the angle EDF is equal to the angle GDF; therefore the base EF is equal to the base GF, and the triangle DEF is equal to the triangle DGF, and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend. [I. 4]

+

Therefore the angle DFG is equal to the angle DFE, and the angle DGF to the angle DEF. +

+

But the angle DFG is equal to the angle ACB; therefore the angle ACB is also equal to the angle DFE.

+

And, by hypothesis, the angle BAC is also equal to the angle EDF; therefore the remaining angle at B is also equal to the remaining angle at E; [I. 32] therefore the triangle ABC is equiangular with the triangle DEF. +

+

Therefore etc. Q. E. D.

+
+
+ + PROPOSITION 7. +

If two triangles have one angle equal to one angle, the sides about other angles proportional, and the remaining angles either both less or both not less than a right angle, the triangles will be equiangular and will have those angles equal, the sides about which are proportional.

+

Let ABC, DEF be two triangles having one angle equal to one angle, the angle BAC to the angle EDF, the sides about other angles ABC, DEF proportional, so that, as AB is to BC, so is DE to EF, and, first, each of the remaining angles at C, F less than a right angle;

I say that the triangle ABC is equiangular with the triangle DEF, the angle ABC will be equal to the angle DEF, and the remaining angle, namely the angle at C, equal to the remaining angle, the angle at F.

+

For, if the angle ABC is unequal to the angle DEF, one of them is greater.

+

Let the angle ABC be greater; and on the straight line AB, and at the point B on it, let the angle ABG be constructed equal to the angle DEF. [I. 23]

+

Then, since the angle A is equal to D, and the angle ABG to the angle DEF, therefore the remaining angle AGB is equal to the remaining angle DFE. [I. 32]

+

Therefore the triangle ABG is equiangular with the triangle DEF.

+

Therefore, as AB is to BG, so is DE to EF [VI. 4]

+

But, as DE is to EF, so by hypothesis is AB to BC; therefore AB has the same ratio to each of the straight lines BC, BG; [V. 11] therefore BC is equal to BG, [V. 9] so that the angle at C is also equal to the angle BGC. [I. 5]

+

But, by hypothesis, the angle at C is less than a right angle; therefore the angle BGC is also less than a right angle; so that the angle AGB adjacent to it is greater than a right angle. [I. 13]

+

And it was proved equal to the angle at F; therefore the angle at F is also greater than a right angle.

+

But it is by hypothesis less than a right angle : which is absurd.

+

Therefore the angle ABC is not unequal to the angle DEF; therefore it is equal to it.

+

But the angle at A is also equal to the angle at D; therefore the remaining angle at C is equal to the remaining angle at F. [I. 32]

+

Therefore the triangle ABC is equiangular with the triangle DEF.

+

But, again, let each of the angles at C, F be supposed not less than a right angle; I say again that, in this case too, the triangle ABC is equiangular with the triangle DEF.

+

+

For, with the same construction, we can prove similarly that BC is equal to BG; so that the angle at C is also equal to the angle BGC. [I. 5]

+

But the angle at C is not less than a right angle; therefore neither is the angle BGC less than a right angle.

+

Thus in the triangle BGC the two angles are not less than two right angles: which is impossible. [I. 17]

+

Therefore, once more, the angle ABC is not unequal to the angle DEF; therefore it is equal to it.

+

But the angle at A is also equal to the angle at D; therefore the remaining angle at C is equal to the remaining angle at F. [I. 32]

+

Therefore the triangle ABC is equiangular with the triangle DEF.

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 8. +

If in a right-angled triangle a perpendicular be drawn from the right angle to the base, the triangles adjoining the perpendicular are similar both to the whole and to one another.

+

Let ABC be a right-angled triangle having the angle BAC right, and let AD be drawn from A perpendicular to BC; I say that each of the triangles ABD, ADC is similar to the whole ABC and, further, they are similar to one another.

+

For, since the angle BAC is equal to the angle ADB, for each is right, and the angle at B is common to the two triangles ABC and ABD, therefore the remaining angle ACB is equal to the remaining angle BAD; [I. 32]

therefore the triangle ABC is equiangular with the triangle ABD.

+

Therefore, as BC which subtends the right angle in the triangle ABC is to BA which subtends the right angle in the triangle ABD, so is AB itself which subtends the angle at C in the triangle ABC to BD which subtends the equal angle BAD in the triangle ABD, and so also is AC to AD which subtends the angle at B common to the two triangles. [VI. 4]

+

Therefore the triangle ABC is both equiangular to the triangle ABD and has the sides about the equal angles proportional.

+

Therefore the triangle ABC is similar to the triangle ABD. [VI. Def. 1]

+

Similarly we can prove that the triangle ABC is also similar to the triangle ADC; therefore each of the triangles ABD, ADC is similar to the whole ABC.

+

I say next that the triangles ABD, ADC are also similar to one another.

+

For, since the right angle BDA is equal to the right angle ADC, and moreover the angle BAD was also proved equal to the angle at C, therefore the remaining angle at B is also equal to the remaining angle DAC; [I. 32] therefore the triangle ABD is equiangular with the triangle ADC.

+

Therefore, as BD which subtends the angle BAD in the triangle ABD is to DA which subtends the angle at C in the triangle ADC equal to the angle BAD, so is AD itself which subtends the angle at B in the triangle ABD to DC which subtends the angle DAC in the triangle ADC equal to the angle at B, and so also is BA to AC, these sides subtending the right angles; [VI. 4] therefore the triangle ABD is similar to the triangle ADC. [VI. Def. 1]

+

Therefore etc.

+ +

Porism. From this it is clear that, if in a right-angled triangle a perpendicular be drawn from the right angle to the base, the straight line so drawn is a mean proportional between the segments of the base. Q. E. D.

+ +
+
+ PROPOSITION 9. +

From a given straight line to cut off a prescribed part.

+

Let AB be the given straight line; thus it is required to cut off from AB a prescribed part.

+

Let the third part be that prescribed.

+

Let a straight line AC be drawn through from A containing with AB any angle; let a point D be taken at random on AC, and let DE, EC be made equal to AD. [I. 3]

+

+

Let BC be joined, and through D let DF be drawn parallel to it. [I. 31]

+

Then, since FD has been drawn parallel to BC, one of the sides of the triangle ABC, therefore, proportionally, as CD is to DA, so is BF to FA. [VI. 2]

+

But CD is double of DA; therefore BF is also double of FA; therefore BA is triple of AF. +

+

Therefore from the given straight line AB the prescribed third part AF has been cut off. Q. E. F. + +

any angle. The expression here and in the two following propositions is τυχοῦσα γωνία, corresponding exactly to τυχὸν σημεῖον which I have translated as a point (taken) at random +; but an angle (taken) at random + would not be so appropriate where it is a question, not of taking any angle at all, but of drawing a straight line casually so as to make any angle with another straight line.

+

+
+
+ PROPOSITION 10. +

To cut a given uncut straight line similarly to a given cut straight line.

+

Let AB be the given uncut straight line, and AC the straight line cut at the points D, E; and let them be so placed as to contain any angle; let CB be joined, and through D, E let DF, EG be drawn parallel to BC, and through D let DHK be drawn parallel to AB. [I. 31]

+

+

Therefore each of the figures FH, HB is a parallelogram; therefore DH is equal to FG and HK to GB. [I. 34]

+

Now, since the straight line HE has been drawn parallel to KC, one of the sides of the triangle DKC, therefore, proportionally, as CE is to ED, so is KH to HD. [VI. 2]

+

But KH is equal to BG, and HD to GF; therefore, as CE is to ED, so is BG to GF.

+

Again, since FD has been drawn parallel to GE, one of the sides of the triangle AGE, therefore, proportionally, as ED is to DA, so is GF to FA. [VI. 2]

+

But it was also proved that, as CE is to ED, so is BG to GF; therefore, as CE is to ED, so is BG to GF, and, as ED is to DA, so is GF to FA. +

+

Therefore the given uncut straight line AB has been cut similarly to the given cut straight line AC. Q. E. F.

+
+
+ PROPOSITION 11. +

To two given straight lines to find a third proportional.

+

Let BA, AC be the two given straight lines, and let them be placed so as to contain any angle; thus it is required to find a third proportional to BA, AC.

+

For let them be produced to the points D, E, and let BD be made equal to AC; [I. 3] let BC be joined, and through D let DE be drawn parallel to it. [I. 31]

+

+

Since, then, BC has been drawn parallel to DE, one of the sides of the triangle ADE, proportionally, as AB is to BD, so is AC to CE. [VI. 2]

+

But BD is equal to AC; therefore, as AB is to AC, so is AC to CE.

+

Therefore to two given straight lines AB, AC a third proportional to them, CE, has been found. Q. E. F. + +

to find. The Greek word, here and in the next two propositions, is προσευρεῖν, literally to find in addition. +

+

+
+
+ + PROPOSITION 12. +

To three given straight lines to find a fourth proportional.

+

Let A, B, C be the three given straight lines; thus it is required to find a fourth proportional to A, B, C.

+

+

Let two straight lines DE, DF be set out containing any angle EDF; let DG be made equal to A, GE equal to B, and further DH equal to C; let GH be joined, and let EF be drawn through E parallel to it. [I. 31]

+

Since, then, GH has been drawn parallel to EF, one of the sides of the triangle DEF, therefore, as DG is to GE, so is DH to HF. [VI. 2]

+

But DG is equal to A, GE to B, and DH to C; therefore, as A is to B, so is C to HF.

+

Therefore to the three given straight lines A, B, C a fourth proportional HF has been found. Q. E. F.

+
+
+ PROPOSITION 13. +

To two given straight lines to find a mean proportional.

+

Let AB, BC be the two given straight lines; thus it is required to find a mean proportional to AB, BC.

+

Let them be placed in a straight line, and let the semicircle ADC be described on AC;

let BD be drawn from the point B at right angles to the straight line AC, and let AD, DC be joined.

+

Since the angle ADC is an angle in a semicircle, it is right. [III. 31]

+

And, since, in the right-angled triangle ADC, DB has been drawn from the right angle perpendicular to the base, therefore DB is a mean proportional between the segments of the base, AB, BC. [VI. 8, Por.]

+

Therefore to the two given straight lines AB, BC a mean proportional DB has been found. Q. E. F.

+
+
+ PROPOSITION 14. +

In equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional; and equiangular parallelograms in which the sides about the equal angles are reciprocally proportional are equal.

+

Let AB, BC be equal and equiangular parallelograms having the angles at B equal, and let DB, BE be placed in a straight line; therefore FB, BG are also in a straight line. [I. 14] +

+

I say that, in AB, BC, the sides about the equal angles are reciprocally proportional, that is to say, that, as DB is to BE, so is GB to BF.

+

+

For let the parallelogram FE be completed.

+

Since, then, the parallelogram AB is equal to the parallelogram BC, and FE is another area, therefore, as AB is to FE, so is BC to FE. [V. 7]

+

But, as AB is to FE, so is DB to BE, [VI. 1] and, as BC is to FE, so is GB to BF. [id.] therefore also, as DB is to BE, so is GB to BF. [V. 11]

+

Therefore in the parallelograms AB, BC the sides about the equal angles are reciprocally proportional.

+

Next, let GB be to BF as DB to BE; I say that the parallelogram AB is equal to the parallelogram BC.

+

For since, as DB is to BE, so is GB to BF, while, as DB is to BE, so is the parallelogram AB to the parallelogram FE, [VI. 1] and, as GB is to BF, so is the parallelogram BC to the parallelogram FE, [VI. 1] therefore also, as AB is to FE, so is BC to FE; [V. 11] therefore the parallelogram AB is equal to the parallelogram BC. [V. 9]

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 15. +

In equal triangles which have one angle equal to one angle the sides about the equal angles are reciprocally proportional; and those triangles which have one angle equal to one angle, and in which the sides about the equal angles are reciprocally proportional, are equal.

+

Let ABC, ADE be equal triangles having one angle equal to one angle, namely the angle BAC to the angle DAE; I say that in the triangles ABC, ADE the sides about the equal angles are reciprocally proportional, that is to say, that, as CA is to AD, so is EA to AB. +

+

For let them be placed so that CA is in a straight line with AD; therefore EA is also in a straight line with AB. [I. 14]

+

Let BD be joined.

+

Since then the triangle ABC is equal to the triangle ADE, and BAD is another area, therefore, as the triangle CAB is to the triangle BAD, so is the triangle EAD to the triangle BAD. [V. 7]

+

+

But, as CAB is to BAD, so is CA to AD, [VI. 1] and, as EAD is to BAD, so is EA to AB. [id.]

+

Therefore also, as CA is to AD, so is EA to AB. [V. 11]

+

Therefore in the triangles ABC, ADE the sides about the equal angles are reciprocally proportional.

+

Next, let the sides of the triangles ABC, ADE be reciprocally proportional, that is to say, let EA be to AB as CA to AD; I say that the triangle ABC is equal to the triangle ADE.

+

For, if BD be again joined, since, as CA is to AD, so is EA to AB, while, as CA is to AD, so is the triangle ABC to the triangle BAD, and, as EA is to AB, so is the triangle EAD to the triangle BAD, [VI. 1] therefore, as the triangle ABC is to the triangle BAD, so is the triangle EAD to the triangle BAD. [V. 11]

+

Therefore each of the triangles ABC, EAD has the same ratio to BAD.

+

Therefore the triangle ABC is equal to the triangle EAD. [V. 9]

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 16. +

If four straight lines be proportional, the rectangle contained by the extremes is equal to the rectangle contained by the means; and, if the rectangle contained by the extremes be equal to the rectangle contained by the means, the four straight lines will be proportional.

+

Let the four straight lines AB, CD, E, F be proportional, so that, as AB is to CD, so is E to F; I say that the rectangle contained by AB, F is equal to the rectangle contained by CD, E.

+

+

Let AG, CH be drawn from the points A, C at right angles to the straight lines AB, CD, and let AG be made equal to F, and CH equal to E.

+

Let the parallelograms BG, DH be completed.

+

Then since, as AB is to CD, so is E to F, while E is equal to CH, and F to AG, therefore, as AB is to CD, so is CH to AG.

+

Therefore in the parallelograms BG, DH the sides about the equal angles are reciprocally proportional.

+

But those equiangular parallelograms in which the sides about the equal angles are reciprocally proportional are equal; [VI. 14] therefore the parallelogram BG is equal to the parallelogram DH.

+

And BG is the rectangle AB, F, for AG is equal to F; and DH is the rectangle CD, E, for E is equal to CH; therefore the rectangle contained by AB, F is equal to the rectangle contained by CD, E.

+

Next, let the rectangle contained by AB, F be equal to the rectangle contained by CD, E; I say that the four straight lines will be proportional, so that, as AB is to CD, so is E to F.

+

For, with the same construction, since the rectangle AB, F is equal to the rectangle CD, E, and the rectangle AB, F is BG, for AG is equal to F, and the rectangle CD, E is DH, for CH is equal to E, therefore BG is equal to DH. +

+

And they are equiangular

+

But in equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional. [VI. 14]

+

Therefore, as AB is to CD, so is CH to AG.

+

But CH is equal to E, and AG to F; therefore, as AB is to CD, so is E to F.

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 17 +

If three straight lines be proportional, the rectangle contained by the extremes is equal to the square on the mean; and, if the rectangle contained by the extremes be equal to the square on the mean, the three straight lines will be proportional.

+

Let the three straight lines A, B, C be proportional, so that, as A is to B, so is B to C; I say that the rectangle contained by A, C is equal to the square on B.

+

+

Let D be made equal to B.

+

Then, since, as A is to B, so is B to C, and B is equal to D, therefore, as A is to B, so is D to C.

+

But, if four straight lines be proportional, the rectangle contained by the extremes is equal to the rectangle contained by the means. [VI. 16]

+

Therefore the rectangle A, C is equal to the rectangle B, D.

+

But the rectangle B, D is the square on B, for B is equal to D; therefore the rectangle contained by A, C is equal to the square on B.

+

Next, let the rectangle A, C be equal to the square on B; I say that, as A is to B, so is B to C.

+

For, with the same construction, since the rectangle A, C is equal to the square on B, while the square on B is the rectangle B, D, for B is equal to D, therefore the rectangle A, C is equal to the rectangle B, D.

+

But, if the rectangle contained by the extremes be equal to that contained by the means, the four straight lines are proportional. [VI. 16]

+

Therefore, as A is to B, so is D to C.

+

But B is equal to D; therefore, as A is to B, so is B to C. +

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 18. +

On a given straight line to describe a rectilineal figure similar and similarly situated to a given rectilineal figure.

+

Let AB be the given straight line and CE the given rectilineal figure; thus it is required to describe on the straight line AB a rectilineal figure similar and similarly situated to the rectilineal figure CE.

+

+

Let DF be joined, and on the straight line AB, and at the points A, B on it, let the angle GAB be constructed equal to the angle at C, and the angle ABG equal to the angle CDF. [I. 23]

+

Therefore the remaining angle CFD is equal to the angle AGB; [I. 32] therefore the triangle FCD is equiangular with the triangle GAB. +

+

Therefore, proportionally, as FD is to GB, so is FC to GA, and CD to AB.

+

Again, on the straight line BG, and at the points B, G on it, let the angle BGH be constructed equal to the angle DFE, and the angle GBH equal to the angle FDE. [I. 23]

+

Therefore the remaining angle at E is equal to the remaining angle at H; [I. 32] therefore the triangle FDE is equiangular with the triangle GBH; therefore, proportionally, as FD is to GB, so is FE to GH, and ED to HB. [VI. 4] +

+

But it was also proved that, as FD is to GB, so is FC to GA, and CD to AB; therefore also, as FC is to AG, so is CD to AB, and FE to GH, and further ED to HB. +

+

And, since the angle CFD is equal to the angle AGB, and the angle DFE to the angle BGH, therefore the whole angle CFE is equal to the whole angle AGH.

+

For the same reason the angle CDE is also equal to the angle ABH. +

+

And the angle at C is also equal to the angle at A, and the angle at E to the angle at H. +

+

Therefore AH is equiangular with CE; and they have the sides about their equal angles proportional; therefore the rectilineal figure AH is similar to the rectilineal figure CE. [VI. Def. 1] +

+

Therefore on the given straight line AB the rectilineal figure AH has been described similar and similarly situated to the given rectilineal figure CE. Q. E. F.

+
+
+ PROPOSITION 19. +

Similar triangles are to one another in the duplicate ratio of the corresponding sides.

+

Let ABC, DEF be similar triangles having the angle at B equal to the angle at E, and such that, as AB is to BC, so is DE to EF, so that BC corresponds to EF; [V. Def. 11] I say that the triangle ABC has to the triangle DEF a ratio duplicate of that which BC has to EF.

+

+

For let a third proportional BG be taken to BC, EF, so that, as BC is to EF, so is EF to BG; [VI. 11] and let AG be joined.

+

Since then, as AB is to BC, so is DE to EF, therefore, alternately, as AB is to DE, so is BC to EF. [V. 16]

+

But, as BC is to EF, so is EF to BG; therefore also, as AB is to DE, so is EF to BG. [V. 11]

+

Therefore in the triangles ABG, DEF the sides about the equal angles are reciprocally proportional.

+

But those triangles which have one angle equal to one angle, and in which the sides about the equal angles are reciprocally proportional, are equal; [VI. 15] therefore the triangle ABG is equal to the triangle DEF.

+

Now since, as BC is to EF, so is EF to BG, and, if three straight lines be proportional, the first has to the third a ratio duplicate of that which it has to the second, [V. Def. 9] therefore BC has to BG a ratio duplicate of that which CB + has to EF.

+

But, as CB is to BG, so is the triangle ABC to the triangle ABG; [VI. 1] therefore the triangle ABC also has to the triangle ABG a ratio duplicate of that which BC has to EF.

+

But the triangle ABG is equal to the triangle DEF; therefore the triangle ABC also has to the triangle DEF a ratio duplicate of that which BC has to EF.

+

Therefore etc.

+ +

Porism. From this it is manifest that, if three straight lines be proportional, then, as the first is to the third, so is the figure described on the first to that which is similar and similarly described on the second. Q. E. D.

+ + +

and such that, as AB is to BC, so is DE to EF, literally (triangles) having the angle at B equal to the angle at E, and (having), as AB to BC, so DE to EF. +

+ +
+
+ PROPOSITION 20. +

Similar polygons are divided into similar triangles, and into triangles equal in multitude and in the same ratio as the wholes, and the polygon has to the polygon a ratio duplicate of that which the corresponding side has to the corresponding + side.

+

Let ABCDE, FGHKL be similar polygons, and let AB correspond to FG; I say that the polygons ABCDE, FGHKL are divided into similar triangles, and into triangles equal in multitude and in the same ratio as the wholes, and the polygon ABCDE has to the polygon FGHKL a ratio duplicate of that which AB has to FG.

+

Let BE, EC, GL, LH be joined.

+

+

Now, since the polygon ABCDE is similar to the polygon FGHKL, the angle BAE is equal to the angle GFL; and, as BA is to AE, so is GF to FL. [VI. Def. 1] +

+

Since then ABE, FGL are two triangles having one angle equal to one angle and the sides about the equal angles proportional, therefore the triangle ABE is equiangular with the triangle FGL; [VI. 6] so that it is also similar; [VI. 4 and Def. 1] therefore the angle ABE is equal to the angle FGL.

+

But the whole angle ABC is also equal to the whole angle FGH because of the similarity of the polygons; therefore the remaining angle EBC is equal to the angle LGH.

+

And, since, because of the similarity of the triangles ABE, FGL, as EB is to BA, so is LG to GF, and moreover also, because of the similarity of the polygons, as AB is to BC, so is FG to GH, therefore, ex aequali, as EB is to BC, so is LG to GH; [V. 22] that is, the sides about the equal angles EBC, LGH are proportional; therefore the triangle EBC is equiangular with the triangle LGH, [VI. 6] so that the triangle EBC is also similar to the triangle LGH. [VI. 4 and Def. 1] +

+

For the same reason the triangle ECD is also similar to the triangle LHK.

+

Therefore the similar polygons ABCDE, FGHKL have been divided into similar triangles, and into triangles equal in multitude.

+

I say that they are also in the same ratio as the wholes, that is, in such manner that the triangles are proportional, and ABE, EBC, ECD are antecedents, while FGL, LGH, LHK are their consequents, and that the polygon ABCDE + has to the polygon FGHKL a ratio duplicate of that which the corresponding side has to the corresponding side, that is AB to FG.

+

For let AC, FH be joined.

+

Then since, because of the similarity of the polygons, the angle ABC is equal to the angle FGH, and, as AB is to BC, so is FG to GH, the triangle ABC is equiangular with the triangle FGH; [VI. 6] therefore the angle BAC is equal to the angle GFH, and the angle BCA to the angle GHF. +

+

And, since the angle BAM is equal to the angle GFN, and the angle ABM is also equal to the angle FGN, therefore the remaining angle AMB is also equal to the remaining angle FNG; [I. 32] therefore the triangle ABM is equiangular with the triangle FGN.

+

Similarly we can prove that the triangle BMC is also equiangular with the triangle GNH.

+

Therefore, proportionally, as AM is to MB, so is FN to NG, and, as BM is to MC, so is GN to NH; so that, in addition, ex aequali, as AM is to MC, so is FN to NH. +

+

But, as AM is to MC, so is the triangle ABM to MBC, and AME to EMC; for they are to one another as their bases. [VI. 1]

+

Therefore also, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents; [V. 12] therefore, as the triangle AMB is to BMC, so is ABE to CBE.

+

But, as AMB is to BMC, so is AM to MC; therefore also, as AM is to MC, so is the triangle ABE to the triangle EBC.

+

For the same reason also, as FN is to NH, so is the triangle FGL to the triangle GLH.

+

And, as AM is to MC, so is FN to NH; therefore also, as the triangle ABE is to the triangle BEC, so is the triangle FGL to the triangle GLH; and, alternately, as the triangle ABE is to the triangle FGL, so is the triangle BEC to the triangle GLH.

+

Similarly we can prove, if BD, GK be joined, that, as the triangle BEC is to the triangle LGH, so also is the triangle ECD to the triangle LHK.

+

And since, as the triangle ABE is to the triangle FGL, so is EBC to LGH, and further ECD to LHK, therefore also, as one of the antecedents is to one of the consequents so are all the antecedents to all the consequents; [V. 12 therefore, as the triangle ABE is to the triangle FGL, so is the polygon ABCDE to the polygon FGHKL.

+

But the triangle ABE has to the triangle FGL a ratio duplicate of that which the corresponding side AB has to the corresponding side FG; for similar triangles are in the duplicate ratio of the corresponding sides. [VI. 19]

+

Therefore the polygon ABCDE also has to the polygon FGHKL a ratio duplicate of that which the corresponding side AB has to the corresponding side FG.

+

Therefore etc.

+ +

Porism. Similarly also it can be proved in the case of quadrilaterals that they are in the duplicate ratio of the corresponding sides. And it was also proved in the case of triangles; therefore also, generally, similar rectilineal figures are to one another in the duplicate ratio of the corresponding sides. Q. E. D.

+ + +

in the same ratio as the wholes. The same word ὁμόλογος is used which I have generally translated by corresponding. + But here it is followed by a dative, ὁμόλογα τοῖς ὅλοις homologous with the wholes, + instead of being used absolutely. The meaning can therefore here be nothing else but in the same ratio with + or proportional to the wholes +; and Euclid seems to recognise that he is making a special use of the word, because he explains it lower down (I. 46): the triangles are homologous to the wholes, that is, in such manner that the triangles are proportional, and ABE, EBC, ECD are antecedents, while FGL, LGH, LHK are their consequents. +

+ + +

ἑπόμενα αὐτῶν, their consequents, + is a little awkward, but may be supposed to indicate which triangles correspond to which as consequent to antecedent.

+ +
+
+ PROPOSITION 21. +

Figures which are similar to the same rectilineal figure are also similar to one another.

+

For let each of the rectilineal figures A, B be similar to C; I say that A is also similar to B.

+

For, since A is similar to C, it is equiangular with it and has the sides about the equal angles proportional. [VI. Def. 1]

+

+

Again, since B is similar to C, it is equiangular with it and has the sides about the equal angles proportional.

+

Therefore each of the figures A, B is equiangular with C and with C has the sides about the equal angles proportional; therefore A is similar to B. Q. E. D.

+
+
+ PROPOSITION 22. +

If four straight lines be proportional, the rectilineal figures similar and similarly described upon them will also be proportional; and, if the rectilineal figures similar and similarly described upon them be proportional, the straight lines will themselves also be proportional.

+

Let the four straight lines AB, CD, EF, GH be proportional, so that, as AB is to CD, so is EF to GH, and let there be described on AB, CD the similar and similarly situated rectilineal figures KAB, LCD, and on EF, GH the similar and similarly situated rectilineal figures MF, NH; I say that, as KAB is to LCD, so is MF to NH.

+

For let there be taken a third proportional O to AB, CD, and a third proportional P to EF, GH. [VI. 11]

+

Then since, as AB is to CD, so is EF to GH, and, as CD is to O, so is GH to P, therefore, ex aequali, as AB is to O, so is EF to P. [V. 22]

+

But, as AB is to O, so is KAB to LCD, [VI. 19, Por.] and, as EF is to P, so is MF to NH; therefore also, as KAB is to LCD, so is MF to NH. [V. 11]

+

+

Next, let MF be to NH as KAB is to LCD; I say also that, as AB is to CD, so is EF to GH.

+

For, if EF is not to GH as AB to CD, let EF be to QR as AB to CD, [VI. 12] and on QR let the rectilineal figure SR be described similar and similarly situated to either of the two MF, NH. [VI. 18]

+

Since then, as AB is to CD, so is EF to QR, and there have been described on AB, CD the similar and similarly situated figures KAB, LCD, and on EF, QR the similar and similarly situated figures MF, SR, therefore, as KAB is to LCD, so is MF to SR.

+

But also, by hypothesis, as KAB is to LCD, so is MF to NH; therefore also, as MF is to SR, so is MF to NH. [V. 11] +

+

Therefore MF has the same ratio to each of the figures NH, SR; therefore NH is equal to SR. [V. 9] +

+

But it is also similar and similarly situated to it; therefore GH is equal to QR. +

+

And, since, as AB is to CD, so is EF to QR, while QR is equal to GH, therefore, as AB is to CD, so is EF to GH.

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 23. +

Equiangular parallelograms have to one another the ratio compounded of the ratios of their sides.

+

Let AC, CF be equiangular parallelograms having the angle BCD equal to the angle ECG; I say that the parallelogram AC has to the parallelogram CF the ratio compounded of the ratios of the sides.

+

+

For let them be placed so that BC is in a straight line with CG; therefore DC is also in a straight line with CE. +

+

Let the parallelogram DG be completed; let a straight line K be set out, and let it be contrived that, as BC is to CG, so is K to L, and, as DC is to CE, so is L to M. [VI. 12]

+

Then the ratios of K to L and of L to M are the same as the ratios of the sides, namely of BC to CG and of DC to CE.

+

But the ratio of K to M is compounded of the ratio of K to L and of that of L to M; so that K has also to M the ratio compounded of the ratios of the sides.

+

Now since, as BC is to CG, so is the parallelogram AC to the parallelogram CH, [VI. 1] while, as BC is to CG, so is K to L, therefore also, as K is to L, so is AC to CH. [V. 11]

+

Again, since, as DC is to CE, so is the parallelogram CH to CF, [VI. 1] while, as DC is to CE, so is L to M, therefore also, as L is to M, so is the parallelogram CH to the parallelogram CF. [V. 11]

+

Since then it was proved that, as K is to L, so is the parallelogram AC to the parallelogram CH, and, as L is to M, so is the parallelogram CH to the parallelogram CF, therefore, ex aequali, as K is to M, so is AC to the parallelogram + CF.

+

But K has to M the ratio compounded of the ratios of the sides; therefore AC also has to CF the ratio compounded of the ratios of the sides. +

+

Therefore etc. Q. E. D. + +

the ratio compounded of the ratios of the sides, λόγον τὸν συγκείμενον ἐκ τῶν πλευρῶν which, meaning literally the ratio compounded of the sides, + is negligently written here and commonly for λόγον τὸν συγκείμενον ἐκ τῶν τῶν πλευρῶν (sc. λόγων).

+ + +

let it be contrived that, as BC is to CG, so is K to L. The Greek phrase is of the usual terse kind, untranslatable literally : καὶ γεγονέτω ὡς μὲν ἡ ΒΓ πρὸς τὴν ΓΗ, οὕτως ἡ Κ πρὸς τὸ Λ, the words meaning and let (there) be made, as BC to CG, so K to L, + where L is the straight line which has to be constructed.

+

+
+
+ PROPOSITION 24. +

In any parallelogram the parallelograms about the diameter are similar both to the whole and to one another.

+

Let ABCD be a parallelogram, and AC its diameter, and let EG, HK be parallelograms about AC; I say that each of the parallelograms EG, HK is similar both to the whole ABCD and to the other.

+

+

For, since EF has been drawn parallel to BC, one of the sides of the triangle ABC, proportionally, as BE is to EA, so is CF to FA. [VI. 2] +

+

Again, since FG has been drawn parallel to CD, one of the sides of the triangle ACD, proportionally, as CF is to FA, so is DG to GA. [VI. 2] +

+

But it was proved that, as CF is to FA, so also is BE to EA; therefore also, as BE is to EA, so is DG to GA, and therefore, componendo, as BA is to AE, so is DA to AG, [V. 18] and, alternately, as BA is to AD, so is EA to AG. [V. 16] +

+

Therefore in the parallelograms ABCD, EG, the sides about the common angle BAD are proportional.

+

And, since GF is parallel to DC, the angle AFG is equal to the angle DCA; and the angle DAC is common to the two triangles ADC, AGF; therefore the triangle ADC is equiangular with the triangle AGF. +

+

For the same reason the triangle ACB is also equiangular with the triangle AFE, and the whole parallelogram ABCD is equiangular with the parallelogram EG.

+

Therefore, proportionally, as AD is to DC, so is AG to GF, as DC is to CA, so is GF to FA, as AC is to CB, so is AF to FE, and further, as CB is to BA, so is FE to EA.

+

And, since it was proved that, as DC is to CA, so is GF to FA, and, as AC is to CB, so is AF to FE, therefore, ex aequali, as DC is to CB, so is GF to FE. [V. 22]

+

Therefore in the parallelograms ABCD, EG the sides about the equal angles are proportional; therefore the parallelogram ABCD is similar to the parallelogram EG. [VI. Def. 1]

+

For the same reason the parallelogram ABCD is also similar to the parallelogram KH; therefore each of the parallelograms EG, HK is similar to ABCD.

+

But figures similar to the same rectilineal figure are also similar to one another; [VI. 21] therefore the parallelogram EG is also similar to the parallelogram HK.

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 25. +

To construct one and the same figure similar to a given rectilineal figure and equal to another given rectilineal figure.

+

Let ABC be the given rectilineal figure to which the figure to be constructed must be similar, and D that to which it must be equal; thus it is required to construct one and the same figure similar to ABC and equal to D.

+

+

Let there be applied to BC the parallelogram BE equal to the triangle ABC [I. 44], and to CE the parallelogram CM equal to D in the angle FCE which is equal to the angle CBL. [I. 45]

+

Therefore BC is in a straight line with CF, and LE with EM.

+

Now let GH be taken a mean proportional to BC, CF [VI. 13], and on GH let KGH be described similar and similarly situated to ABC. [VI. 18]

+

Then, since, as BC is to GH, so is GH to CF, and, if three straight lines be proportional, as the first is to the third, so is the figure on the first to the similar and similarly situated figure described on the second, [VI. 19, Por.] therefore, as BC is to CF, so is the triangle ABC to the triangle KGH.

+

But, as BC is to CF, so also is the parallelogram BE to the parallelogram EF. [VI. 1]

+

Therefore also, as the triangle ABC is to the triangle KGH, so is the parallelogram BE to the parallelogram EF; therefore, alternately, as the triangle ABC is to the parallelogram BE, so is the triangle KGH to the parallelogram EF. [V. 16]

+

But the triangle ABC is equal to the parallelogram BE; therefore the triangle KGH is also equal to the parallelogram EF.

+

But the parallelogram EF is equal to D; therefore KGH is also equal to D.

+

And KGH is also similar to ABC.

+

Therefore one and the same figure KGH has been constructed similar to the given rectilineal figure ABC and equal to the other given figure D. Q. E. D. + +

to which the figure to be constructed must be similar, literally to which it is required to construct (one) similar, + ᾧ δεῖ ὅμοιον συστήσασθαι.

+

+
+
+ PROPOSITION 26. +

If from a parallelogram there be taken away a parallelogram similar and similarly situated to the whole and having a common angle with it, it is about the same diameter with the whole +

+

For from the parallelogram ABCD let there be taken away the parallelogram AF similar and similarly situated to ABCD, and having the angle DAB common with it;

I say that ABCD is about the same diameter with AF.

+

For suppose it is not, but, if possible, let AHC be the diameter lt of ABCD gt, let GF be produced and carried through to H, and let HK be drawn through H parallel to either of the straight lines AD, BC. [I. 31]

+

Since, then, ABCD is about the same diameter with KG, therefore, as DA is to AB, so is GA to AK. [VI. 24]

+

But also, because of the similarity of ABCD, EG, as DA is to AB, so is GA to AE; therefore also, as GA is to AK, so is GA to AE. [V. 11]

+

Therefore GA has the same ratio to each of the straight lines AK, AE.

+

Therefore AE is equal to AK [V. 9], the less to the greater : which is impossible.

+

Therefore ABCD cannot but be about the same diameter with AF; therefore the parallelogram ABCD is about the same diameter with the parallelogram AF.

+

Therefore etc. Q. E. D.

+
+
+ + PROPOSITION 27. +

Of all the parallelograms applied to the same straight line and deficient by parallelogrammic figures similar and similarly situated to that described on the half of the straight line, that parallelogram is greatest which is applied to the half of the straight line and is similar to the defect.

+

Let AB be a straight line and let it be bisected at C; let there be applied to the straight line AB the parallelogram AD deficient by the parallelogrammic figure DB described on the half of AB, that is, CB;

I say that, of all the parallelograms applied to AB and deficient by parallelogrammic figures similar and similarly situated to DB, AD is greatest.

+

For let there be applied to the straight line AB the parallelogram AF deficient by the parallelogrammic figure FB similar and similarly situated to DB; I say that AD is greater than AF.

+

For, since the parallelogram DB is similar to the parallelogram FB, they are about the same diameter. [VI. 26] +

+

Let their diameter DB be drawn, and let the figure be described.

+

Then, since CF is equal to FE, [I. 43] and FB is common, therefore the whole CH is equal to the whole KE.

+

But CH is equal to CG, since AC is also equal to CB. [I. 36]

+

Therefore GC is also equal to EK.

+

Let CF be added to each; therefore the whole AF is equal to the gnomon LMN; so that the parallelogram DB, that is, AD, is greater than the parallelogram AF.

+

Therefore etc.

+
+
+ PROPOSITION 28. +

To a given straight line to apply a parallelogram equal to a given rectilineal figure and deficient by a parallelogrammic figure similar to a given one : thus the given rectilineal figure must not be greater than the parallelogram described on the half of the straight line and similar to the defect.

+

Let AB be the given straight line, C the given rectilineal figure to which the figure to be applied to AB is required to be equal, not being greater than the parallelogram described on the half of AB and similar to the defect, and D the parallelogram to which the defect is required to be similar; thus it is required to apply to the given straight line AB a parallelogram equal to the given rectilineal figure C and deficient by a parallelogrammic figure which is similar to D.

+

Let AB be bisected at the point E, and on EB let EBFG be described similar and similarly situated to D; [VI. 18] let the parallelogram AG be completed.

+

If then AG is equal to C, that which was enjoined will have been done; for there has been applied to the given straight line AB the parallelogram AG equal to the given rectilineal figure C and deficient by a parallelogrammic figure GB which is similar to D. +

+

+

But, if not, let HE be greater than C.

+

Now HE is equal to GB; therefore GB is also greater than C. +

+

Let KLMN be constructed at once equal to the excess by which GB is greater than C and similar and similarly situated to D. [VI. 25]

+

But D is similar to GB; therefore KM is also similar to GB. [VI. 21] +

+

Let, then, KL correspond to GE, and LM to GF.

+

Now, since GB is equal to C, KM, therefore GB is greater than KM; therefore also GE is greater than KL, and GF than LM. +

+

Let GO be made equal to KL, and GP equal to LM; and let the parallelogram OGPQ be completed; therefore it is equal and similar to KM. +

+

Therefore GQ is also similar to GB; [VI. 21] therefore GQ is about the same diameter with GB. [VI. 26]

+

Let GQB be their diameter, and let the figure be described.

+

Then, since BG is equal to C, KM, and in them GQ is equal to KM, therefore the remainder, the gnomon UWV, is equal to the remainder C.

+

And, since PR is equal to OS, let QB be added to each; therefore the whole PB is equal to the whole OB.

+

But OB is equal to TE, since the side AE is also equal to the side EB; [I. 36] therefore TE is also equal to PB. +

+

Let OS be added to each; therefore the whole TS is equal to the whole, the gnomon VWU. +

+

But the gnomon VWU was proved equal to C; therefore TS is also equal to C. +

+

Therefore to the given straight line AB there has been applied the parallelogram ST equal to the given rectilineal figure C and deficient by a parallelogrammic figure QB which is similar to D. Q. E. F.

+
+
+ PROPOSITION 29. +

To a given straight line to apply a parallelogram equal to a given rectilineal figure and exceeding by a parallelogrammic figure similar to a given one.

+

Let AB be the given straight line, C the given rectilineal figure to which the figure to be applied to AB is required to be equal, and D that to which the excess is required to be similar; thus it is required to apply to the straight line AB a parallelogram equal to the rectilineal figure C and exceeding by a parallelogrammic figure similar to D.

+

+

Let AB be bisected at E; let there be described on EB the parallelogram BF similar and similarly situated to D; and let GH be constructed at once equal to the sum of BF, C and similar and similarly situated to D. [VI. 25]

+

Let KH correspond to FL and KG to FE.

+

Now, since GH is greater than FB, therefore KH is also greater than FL, and KG than FE.

+

Let FL, FE be produced, let FLM be equal to KH, and FEN to KG, and let MN be completed; therefore MN is both equal and similar to GH. +

+

But GH is similar to EL; therefore MN is also similar to EL; [VI. 21] therefore EL is about the same diameter with MN. [VI. 26]

+

Let their diameter FO be drawn, and let the figure be described.

+

Since GH is equal to EL, C, while GH is equal to MN, therefore MN is also equal to EL, C.

+

Let EL be subtracted from each; therefore the remainder, the gnomon XWV, is equal to C. +

+

Now, since AE is equal to EB, AN is also equal to NB [I. 36], that is, to LP [I. 43]. +

+

Let EO be added to each; therefore the whole AO is equal to the gnomon VWX. +

+

But the gnomon VWX is equal to C; therefore AO is also equal to C. +

+

Therefore to the given straight line AB there has been applied the parallelogram AO equal to the given rectilineal figure C and exceeding by a parallelogrammic figure QP which is similar to D, since PQ is also similar to EL [VI. 24]. Q. E. F.

+
+
+ PROPOSITION 30. +

To cut a given finite straight line in extreme and mean ratio.

+

Let AB be the given finite straight line; thus it is required to cut AB in extreme and mean ratio. +

+

On AB let the square BC be described; and let there be applied to AC the parallelogram CD equal to BC and exceeding by the figure AD similar to BC. [VI. 29]

+

Now BC is a square; therefore AD is also a square. +

+

+

And, since BC is equal to CD, let CE be subtracted from each; therefore the remainder BF is equal to the remainder AD. +

+

But it is also equiangular with it; therefore in BF, AD the sides about the equal angles are reciprocally proportional; [VI. 14] therefore, as FE is to ED, so is AE to EB. +

+

But FE is equal to AB, and ED to AE.

+

Therefore, as BA is to AE, so is AE to EB.

+

And AB is greater than AE; therefore AE is also greater than EB. +

+

Therefore the straight line AB has been cut in extreme and mean ratio at E, and the greater segment of it is AE. Q. E. F.

+
+
+ PROPOSITION 31. +

In right-angled triangles the figure on the side subtending the right angle is equal to the similar and similarly described figures on the sides containing the right angle.

+

Let ABC be a right-angled triangle having the angle BAC right; I say that the figure on BC is equal to the similar and similarly described figures on BA, AC.

+

Let AD be drawn perpendicular.

+

Then since, in the right-angled triangle ABC, AD has been drawn from the right angle at A perpendicular to the base BC, the triangles ABD, ADC adjoining the perpendicular are similar both to the whole ABC and to one another. [VI. 8]

+

+

And, since ABC is similar to ABD, therefore, as CB is to BA, so is AB to BD. [VI. Def. 1]

+

And, since three straight lines are proportional, as the first is to the third, so is the figure on the first to the similar and similarly described figure on the second. [VI. 19, Por.]

+

Therefore, as CB is to BD, so is the figure on CB to the similar and similarly described figure on BA.

+

For the same reason also, as BC is to CD, so is the figure on BC to that on CA; so that, in addition, as BC is to BD, DC, so is the figure on BC to the similar and similarly described figures on BA, AC.

+

But BC is equal to BD, DC; therefore the figure on BC is also equal to the similar and similarly described figures on BA, AC.

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 32. +

If two triangles having two sides proportional to two sides be placed together at one angle so that their corresponding sides are also parallel, the remaining sides of the triangles will be in a straight line.

+

Let ABC, DCE be two triangles having the two sides BA, AC proportional to the two sides DC, DE, so that, as AB is to AC, so is DC to DE, and AB parallel to DC, and AC to DE; I say that BC is in a straight line with CE.

+

For, since AB is parallel to DC, and the straight line AC has fallen upon them, the alternate angles BAC, ACD are equal to one another. [I. 29]

+

For the same reason the angle CDE is also equal to the angle ACD; so that the angle BAC is equal to the angle CDE.

+

+

And, since ABC, DCE are two triangles having one angle, the angle at A, equal to one angle, the angle at D, and the sides about the equal angles proportional, so that, as BA is to AC, so is CD to DE, therefore the triangle ABC is equiangular with the triangle DCE; [VI. 6] therefore the angle ABC is equal to the angle DCE. +

+

But the angle ACD was also proved equal to the angle BAC; therefore the whole angle ACE is equal to the two angles ABC, BAC. +

+

Let the angle ACB be added to each; therefore the angles ACE, ACB are equal to the angles BAC, ACB, CBA.

+

But the angles BAC, ABC, ACB are equal to two right angles; [I. 32] therefore the angles ACE, ACB are also equal to two right angles. +

+

Therefore with a straight line AC, and at the point C on it, the two straight lines BC, CE not lying on the same side make the adjacent angles ACE, ACB equal to two right angles; therefore BC is in a straight line with CE. [I. 14] +

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 33. +

In equal circles angles have the same ratio as the circumferences on which they stand, whether they stand at the centres or at the circumferences.

+

Let ABC, DEF be equal circles, and let the angles BGC, EHF be angles at their centres G, H, and the angles BAC, EDF angles at the circumferences; I say that, as the circumference BC is to the circumference EF, so is the angle BGC to the angle EHF, and the angle BAC to the angle EDF.

+

+

For let any number of consecutive circumferences CK, KL be made equal to the circumference BC, and any number of consecutive circumferences FM, MN equal to the circumference EF; and let GK, GL, HM, HN be joined.

+

Then, since the circumferences BC, CK, KL are equal to one another, the angles BGC, CGK, KGL are also equal to one another; [III. 27] therefore, whatever multiple the circumference BL is of BC, that multiple also is the angle BGL of the angle BGC.

+

For the same reason also, whatever multiple the circumference NE is of EF, that multiple also is the angle NHE of the angle EHF.

+

If then the circumference BL is equal to the circumference EN, the angle BGL is also equal to the angle EHN; [III. 27] if the circumference BL is greater than the circumference EN, the angle BGL is also greater than the angle EHN; and, if less, less.

+

There being then four magnitudes, two circumferences BC, EF, and two angles BGC, EHF, there have been taken, of the circumference BC and the angle BGC equimultiples, namely the circumference BL and the angle BGL, and of the circumference EF and the angle EHF equimultiples, namely the circumference EN and the angle EHN.

+

And it has been proved that, if the circumference BL is in excess of the circumference EN, the angle BGL is also in excess of the angle EHN; if equal, equal; and if less, less.

+

Therefore, as the circumference BC is to EF, so is the angle BGC to the angle EHF. [V. Def. 5]

+

But, as the angle BGC is to the angle EHF, so is the angle BAC to the angle EDF; for they are doubles respectively.

+

Therefore also, as the circumference BC is to the circumference EF, so is the angle BGC to the angle EHF, and the angle BAC to the angle EDF.

+

Therefore etc. Q. E. D.

+
+
+
+ +
+ BOOK VII. +
+ DEFINITIONS. +
+1 +

An unit is that by virtue of which each of the things that exist is called one.

+
+
+2 +

A number is a multitude composed of units.

+
+
+3 +

A number is a part of a number, the less of the greater, when it measures the greater;

+
+
+4 +

but parts when it does not measure it.

+
+
+5 +

The greater number is a multiple of the less when it is measured by the less.

+
+
+6 +

An even number is that which is divisible into two equal parts.

+
+
+7 +

An odd number is that which is not divisible into two equal parts, or that which differs by an unit from an even number.

+
+
+8 +

An even-times even number is that which is measured by an even number according to an even number.

+
+
+9 +

An even-times odd number is that which is measured by an even number according to an odd number.

+
+
+10 +

An odd-times odd number is that which is measured by an odd number according to an odd number.

+
+
+11 + +

A prime number is that which is measured by an unit alone.

+
+
+12 +

Numbers prime to one another are those which are measured by an unit alone as a common measure.

+
+
+13 +

A composite number is that which is measured by some number.

+
+
+14 +

Numbers composite to one another are those which are measured by some number as a common measure.

+
+
+15 +

A number is said to multiply a number when that which is multiplied is added to itself as many times as there are units in the other, and thus some number is produced.

+
+
+16 +

And, when two numbers having multiplied one another make some number, the number so produced is called plane, and its sides are the numbers which have multiplied one another.

+
+
+17 +

And, when three numbers having multiplied one another make some number, the number so produced is solid, and its sides are the numbers which have multiplied one another.

+
+
+18 +

A square number is equal multiplied by equal, or a number which is contained by two equal numbers.

+
+
+19 +

And a cube is equal multiplied by equal and again by equal, or a number which is contained by three equal numbers.

+
+
+20 +

Numbers are proportional when the first is the same multiple, or the same part, or the same parts, of the second that the third is of the fourth.

+
+
+21 +

+ Similar plane and solid numbers are those which have their sides proportional.

+
+
+22 +

A perfect number is that which is equal to its own parts.

+
+
+ +
+ BOOK VII. PROPOSITIONS. +
+ PROPOSITION I. +

+ Two unequal numbers being set out, and the less being continually subtracted in turn from the greater, if the number which is left never measures the one before it until an unit is left, the original numbers will be prime to one another. +

+

For, the less of two unequal numbers AB, CD being continually subtracted from the greater, let the number which is left never measure the one before it until an unit is left; I say that AB, CD are prime to one another, that is, that an unit alone measures AB, CD.

+

+

For, if AB, CD are not prime to one another, some number will measure them.

+

Let a number measure them, and let it be E; let CD, measuring BF, leave FA less than itself, let AF, measuring DG, leave GC less than itself, and let GC, measuring FH, leave an unit HA.

+

Since, then, E measures CD, and CD measures BF, therefore E also measures BF.

+

But it also measures the whole BA; therefore it will also measure the remainder AF.

+

But AF measures DG; therefore E also measures DG.

+

But it also measures the whole DC therefore it will also measure the remainder CG.

+

But CG measures FH; therefore E also measures FH.

+

But it also measures the whole FA; therefore it will also measure the remainder, the unit AH, though it is a number: which is impossible.

+

Therefore no number will measure the numbers AB, CD; therefore AB, CD are prime to one another. [VII. Def. 12] Q. E. D.

+
+ +
+ PROPOSITION 2. +

+ Given two numbers not prime to one another, to find their greatest common measure. +

+

Let AB, CD be the two given numbers not prime to one another.

+

Thus it is required to find the greatest common measure of AB, CD.

+

If now CD measures AB—and it also measures itself—CD is a common measure of CD, AB.

+

+

And it is manifest that it is also the greatest; for no greater number than CD will measure CD.

+

But, if CD does not measure AB, then, the less of the numbers AB, CD being continually subtracted from the greater, some number will be left which will measure the one before it.

+

For an unit will not be left; otherwise AB, CD will be prime to one another [VII. 1], which is contrary to the hypothesis.

+

Therefore some number will be left which will measure the one before it.

+

Now let CD, measuring BE, leave EA less than itself, let EA, measuring DF, leave FC less than itself, and let CF measure AE.

+

Since then, CF measures AE, and AE measures DF, therefore CF will also measure DF.

+

But it also measures itself; therefore it will also measure the whole CD.

+

But CD measures BE; therefore CF also measures BE.

+

But it also measures EA; therefore it will also measure the whole BA.

+

But it also measures CD; therefore CF measures AB, CD.

+

Therefore CF is a common measure of AB, CD.

+

I say next that it is also the greatest.

+

For, if CF is not the greatest common measure of AB, CD, some number which is greater than CF will measure the numbers AB, CD.

+

Let such a number measure them, and let it be G.

+

Now, since G measures CD, while CD measures BE, G also measures BE.

+

But it also measures the whole BA; therefore it will also measure the remainder AE.

+

But AE measures DF; therefore G will also measure DF.

+

But it also measures the whole DC; therefore it will also measure the remainder CF, that is, the greater will measure the less: which is impossible.

+

Therefore no number which is greater than CF will measure the numbers AB, CD; therefore CF is the greatest common measure of AB, CD. +

+ +

Porism. From this it is manifest that, if a number measure two numbers, it will also measure their greatest common measure.

+ +

Q. E. D.

+
+
+ PROPOSITION 3. +

+ Given three numbers not prime to one another, to find their greatest common measure. +

+

Let A, B, C be the three given numbers not prime to one another; thus it is required to find the greatest common measure of A, B, C.

+

+

For let the greatest common measure, D, of the two numbers A, B be taken; [VII. 2] then D either measures, or does not measure, C.

+

First, let it measure it.

+

But it measures A, B also; therefore D measures A, B, C; therefore D is a common measure of A, B, C.

+

I say that it is also the greatest.

+

For, if D is not the greatest common measure of A, B, C, some number which is greater than D will measure the numbers A, B, C.

+

Let such a number measure them, and let it be E.

+

Since then E measures A, B, C, it will also measure A, B; therefore it will also measure the greatest common measure of A, B. [VII. 2, Por.]

+

But the greatest common measure of A, B is D; therefore E measures D, the greater the less: which is impossible.

+

Therefore no number which is greater than D will measure the numbers A, B, C; therefore D is the greatest common measure of A, B, C. +

+

Next, let D not measure C; I say first that C, D are not prime to one another.

+

For, since A, B, C are not prime to one another, some number will measure them.

+

Now that which measures A, B, C will also measure A, B, and will measure D, the greatest common measure of A, B. [VII. 2, Por.]

+

But it measures C also; therefore some number will measure the numbers D, C; therefore D, C are not prime to one another.

+

Let then their greatest common measure E be taken. [VII. 2]

+

Then, since E measures D, and D measures A, B, therefore E also measures A, B.

+

But it measures C also; therefore E measures A, B, C; therefore E is a common measure of A, B, C.

+

I say next that it is also the greatest.

+

For, if E is not the greatest common measure of A, B, C, some number which is greater than E will measure the numbers A, B, C.

+

Let such a number measure them, and let it be F.

+

Now, since F measures A, B, C, it also measures A, B; therefore it will also measure the greatest common measure of A, B. [VII. 2, Por.]

+

But the greatest common measure of A, B is D; therefore F measures D.

+

And it measures C also; therefore F measures D, C; therefore it will also measure the greatest common measure of D, C. [VII. 2, Por.]

+

But the greatest common measure of D, C is E; therefore F measures E, the greater the less: which is impossible.

+

Therefore no number which is greater than E will measure the numbers A, B, C; therefore E is the greatest common measure of A, B, C. Q. E. D.

+
+
+ PROPOSITION 4. +

+ Any number is either a part or parts of any number, the less of the greater. +

+

Let A, BC be two numbers, and let BC be the less; I say that BC is either a part, or parts, of A.

+

For A, BC are either prime to one another or not.

+

First, let A, BC be prime to one another.

+

Then, if BC be divided into the units in it, each unit of those in BC will be some part of A; so that BC is parts of A.

+

+

Next let A, BC not be prime to one another; then BC either measures, or does not measure, A.

+

If now BC measures A, BC is a part of A.

+

But, if not, let the greatest common measure D of A, BC be taken; [VII. 2] and let BC be divided into the numbers equal to D, namely BE, EF, FC.

+

Now, since D measures A, D is a part of A.

+

But D is equal to each of the numbers BE, EF, FC; therefore each of the numbers BE, EF, FC is also a part of A; so that BC is parts of A.

+

Therefore etc. Q. E. D.

+
+ +
+ PROPOSITION 5. +

+ If a number be a part of a number, and another be the same part of another, the sum will also be the same part of the sum that the one is of the one. +

+

For let the number A be a part of BC, and another, D, the same part of another EF that A is of BC; I say that the sum of A, D is also the same part of the sum of BC, EF that A is of BC.

+

For since, whatever part A is of BC, D is also the same part of EF, therefore, as many numbers as there are in BC equal to A, so many numbers are there also in EF equal to D.

+

+

Let BC be divided into the numbers equal to A, namely BG, GC, and EF into the numbers equal to D, namely EH, HF; then the multitude of BG, GC will be equal to the multitude of EH, HF.

+

And, since BG is equal to A, and EH to D, therefore BG, EH are also equal to A, D.

+

For the same reason GC, HF are also equal to A, D.

+

Therefore, as many numbers as there are in BC equal to A, so many are there also in BC, EF equal to A, D.

+

Therefore, whatever multiple BC is of A, the same multiple also is the sum of BC, EF of the sum of A, D.

+

Therefore, whatever part A is of BC, the same part also is the sum of A, D of the sum of BC, EF. Q. E. D.

+
+ +
+ PROPOSITION 6. +

+ If a number be parts of a number, and another be the same parts of another, the sum will also be the same parts of the sum that the one is of the one. +

+

For let the number AB be parts of the number C, and another, DE, the same parts of another, F, that AB is of C; I say that the sum of AB, DE is also the same parts of the sum of C, F that AB is of C.

+

+

For since, whatever parts AB is of C, DE is also the same parts of F, therefore, as many parts of C as there are in AB, so many parts of F are there also in DE.

+

Let AB be divided into the parts of C, namely AG, GB, and DE into the parts of F, namely DH, HE; thus the multitude of AG, GB will be equal to the multitude of DH, HE.

+

And since, whatever part AG is of C, the same part is DH of F also, therefore, whatever part AG is of C, the same part also is the sum of AG, DH of the sum of C, F. [VII. 5]

+

For the same reason, whatever part GB is of C, the same part also is the sum of GB, HE of the sum of C, F.

+

Therefore, whatever parts AB is of C, the same parts also is the sum of AB, DE of the sum of C, F. Q. E. D.

+
+
+ PROPOSITION 7. +

+ If a number be that part of a number, which a number subtracted is of a number subtracted, the remainder will also be the same part of the remainder that the whole is of the whole. +

+

For let the number AB be that part of the number CD which AE subtracted is of CF subtracted; I say that the remainder EB is also the same part of the remainder FD that the whole AB is of the whole CD.

+

+

For, whatever part AE is of CF, the same part also let EB be of CG.

+

Now since, whatever part AE is of CF, the same part also is EB of CG, therefore, whatever part AE is of CF, the same part also is AB of GF. [VII. 5]

+

But, whatever part AE is of CF, the same part also, by hypothesis, is AB of CD; therefore, whatever part AB is of GF, the same part is it of CD also; therefore GF is equal to CD.

+

Let CF be subtracted from each; therefore the remainder GC is equal to the remainder FD.

+

Now since, whatever part AE is of CF, the same part also is EB of GC, while GC is equal to FD, therefore, whatever part AE is of CF, the same part also is EB of FD.

+

But, whatever part AE is of CF, the same part also is AB of CD; therefore also the remainder EB is the same part of the remainder FD that the whole AB is of the whole CD. Q. E. D.

+
+
+ PROPOSITION 8. +

+ If a number be the same parts of a number that a number subtracted is of a number subtracted, the remainder will also be the same parts of the remainder that the whole is of the whole. +

+

For let the number AB be the same parts of the number CD that AE subtracted is of CF subtracted; I say that the remainder EB is also the same parts of the remainder FD that the whole AB is of the whole CD.

+

+

For let GH be made equal to AB.

+

Therefore, whatever parts GH is of CD, the same parts also is AE of CF.

+

Let GH be divided into the parts of CD, namely GK, KH, and AE into the parts of CF, namely AL, LE; thus the multitude of GK, KH will be equal to the multitude of AL, LE.

+

Now since, whatever part GK is of CD, the same part also is AL of CF, while. CD is greater than CF, therefore GK is also greater than AL.

+

Let GM be made equal to AL.

+

Therefore, whatever part GK is of CD, the same part also is GM of CF; therefore also the remainder MK is the same part of the remainder FD that the whole GK is of the whole CD. [VII. 7]

+

Again, since, whatever part KH is of CD, the same part also is EL of CF, while CD is greater than CF, therefore HK is also greater than EL.

+

Let KN be made equal to EL.

+

Therefore, whatever part KH is of CD, the same part also is KN of CF; therefore also the remainder NH is the same part of the remainder FD that the whole KH is of the whole CD. [VII. 7]

+

But the remainder MK was also proved to be the same part of the remainder FD that the whole GK is of the whole CD; therefore also the sum of MK, NH is the same parts of DF that the whole HG is of the whole CD.

+

But the sum of MK, NH is equal to EB, and HG is equal to BA; therefore the remainder EB is the same parts of the remainder FD that the whole AB is of the whole CD. Q. E. D.

+
+
+ PROPOSITION 9. +

+ If a number be a part of a number, and another be the same part of another, alternately also, whatever part or parts the first is of the third, the same part, or the same parts, will the second also be of the fourth. +

+

For let the number A be a part of the number BC, and another, D, the same part of another, EF, that A is of BC; I say that, alternately also, whatever part or parts A is of D, the same part or parts is BC of EF also.

+

+

For since, whatever part A is of BC, the same part also is D of EF, therefore, as many numbers as there are in BC equal to A, so many also are there in EF equal to D.

+

Let BC be divided into the numbers equal to A, namely BG, GC, and EF into those equal to D, namely EH, HF; thus the multitude of BG, GC will be equal to the multitude of EH, HF.

+

Now, since the numbers BG, GC are equal to one another, and the numbers EH, HF are also equal to one another, while the multitude of BG, GC is equal to the multitude of EH, HF, therefore, whatever part or parts BG is of EH, the same part or the same parts is GC of HF also; so that, in addition, whatever part or parts BG is of EH, the same part also, or the same parts, is the sum BC of the sum EF. [VII. 5, 6]

+

But BG is equal to A, and EH to D; therefore, whatever part or parts A is of D, the same part or the same parts is BC of EF also. Q. E. D.

+
+
+ PROPOSITION 10. +

+ If a number be parts of a number, and another be the same parts of another, alternately also, whatever parts or part the first is of the third, the same parts or the same part will the second also be of the fourth. +

+

For let the number AB be parts of the number C, and another, DE, the same parts of another, F; I say that, alternately also, whatever parts or part AB is of DE, the same parts or the same part is C of F also.

+

+

For since, whatever parts AB is of C, the same parts also is DE of F, therefore, as many parts of C as there are in AB, so many parts also of F are there in DE.

+

Let AB be divided into the parts of C, namely AG, GB, and DE into the parts of F, namely DH, HE; thus the multitude of AG, GB will be equal to the multitude of DH, HE.

+

Now since, whatever part AG is of C, the same part also is DH of F, alternately also, whatever part or parts AG is of DH, the same part or the same parts is C of F also. [VII. 9]

+

For the same reason also, whatever part or parts GB is of HE, the same part or the same parts is C of F also; so that, in addition, whatever parts or part AB is of DE, the same parts also, or the same part, is C of F. [VII. 5, 6] Q. E. D.

+
+
+ PROPOSITION 11. +

+ If, as whole is to whole, so is a number subtracted to a number subtracted, the remainder will also be to the remainder as whole to whole. +

+

As the whole AB is to the whole CD, so let AE subtracted be to CF subtracted; I say that the remainder EB is also to the remainder FD as the whole AB to the whole CD.

+

Since, as AB is to CD, so is AE to CF, whatever part or parts AB is of CD, the same part or the same parts is AE of CF also; [VII. Def. 20]

+

+

Therefore also the remainder EB is the same part or parts of FD that AB is of CD. [VII. 7, 8]

+

Therefore, as EB is to FD, so is AB to CD. [VII. Def. 20] Q. E. D.

+
+
+ PROPOSITION 12. +

+ If there be as many numbers as we please in proportion, then, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents. +

+

Let A, B, C, D be as many numbers as we please in proportion, so that, as A is to B, so is C to D; I say that, as A is to B, so are A, C to B, D.

+

+

For since, as A is to B, so is C to D, whatever part or parts A is of B, the same part or parts is C of D also. [VII. Def. 20]

+

Therefore also the sum of A, C is the same part or the same parts of the sum of B, D that A is of B. [VII. 5, 6]

+

Therefore, as A is to B, so are A, C to B, D. [VII. Def. 20]

+
+ +
+ PROPOSITION 13. +

+ If four numbers be proportional, they will also be proportional alternately. +

+

Let the four numbers A, B, C, D be proportional, so that, as A is to B, so is C to D; I say that they will also be proportional alternately, so that, as A is to C, so will B be to D. +

+

For since, as A is to B, so is C to D, therefore, whatever part or parts A is of B, the same part or the same parts is C of D also. [VII. Def. 20]

+

+

Therefore, alternately, whatever part or parts A is of C, the same part or the same parts is B of D also. [VII. 10]

+

Therefore, as A is to C, so is B to D. [VII. Def. 20] Q. E. D.

+
+
+ PROPOSITION 14. +

If there be as many numbers as we please, and others equal to them in multitude, which taken two and two are in the same ratio, they will also be in the same ratio ex aequali.

+

Let there be as many numbers as we please A, B, C, and others equal to them in multitude D, E, F, which taken two and two are in the same ratio, so that, as A is to B, so is D to E, and, as B is to C, so is E to F; I say that, ex aequali, as A is to C, so also is D to F. +

+

+

For, since, as A is to B, so is D to E, therefore, alternately, as A is to D, so is B to E. [VII. 13] +

+

Again, since, as B is to C, so is E to F, therefore, alternately, as B is to E, so is C to F. [VII. 13] +

+

But, as B is to E, so is A to D; therefore also, as A is to D, so is C to F.

+

Therefore, alternately, as A is to C, so is D to F. [id.]

+
+
+ PROPOSITION 15. +

+ If an unit measure any number, and another number measure any other number the same number of times, alternately also, the unit will measure the third number the same number of times that the second measures the fourth. + +

+

For let the unit A measure any number BC, and let another number D measure any other number EF the same number of times; I say that, alternately also, the unit A measures the number D the same number of times that BC measures EF.

+

+

For, since the unit A measures the number BC the same number of times that D measures EF, therefore, as many units as there are in BC, so many numbers equal to D are there in EF also.

+

Let BC be divided into the units in it, BG, GH, HC, and EF into the numbers EK, KL, LF equal to D.

+

Thus the multitude of BG, GH, HC will be equal to the multitude of EK, KL, LF.

+

And, since the units BG, GH, HC are equal to one another, and the numbers EK, KL, LF are also equal to one another, while the multitude of the units BG, GH, HC is equal to the multitude of the numbers EK, KL, LF, therefore, as the unit BG is to the number EK, so will the unit GH be to the number KL, and the unit HC to the number LF.

+

Therefore also, as one of the antecedents is to one of the consequents, so will all the antecedents be to all the consequents; [VII. 12] therefore, as the unit BG is to the number EK, so is BC to EF.

+

But the unit BG is equal to the unit A, and the number EK to the number D.

+

Therefore, as the unit A is to the number D, so is BC to EF.

+

Therefore the unit A measures the number D the same number of times that BC measures EF. Q. E. D.

+
+ +
+ PROPOSITION 16. +

+ If two numbers by multiplying one another make certain numbers, the numbers so produced will be equal to one another. +

+

Let A, B be two numbers, and let A by multiplying B make C, and B by multiplying A make D; I say that C is equal to D.

+

+

For, since A by multiplying B has made C, therefore B measures C according to the units in A.

+

But the unit E also measures the number A according to the units in it; therefore the unit E measures A the same number of times that B measures C. +

+

Therefore, alternately, the unit E measures the number B the same number of times that A measures C. [VII. 15]

+

Again, since B by multiplying A has made D, therefore A measures D according to the units in B.

+

But the unit E also measures B according to the units in it; therefore the unit E measures the number B the same number of times that A measures D. +

+

But the unit E measured the number B the same number of times that A measures C; therefore A measures each of the numbers C, D the same number of times. +

+

Therefore C is equal to D. Q. E. D. +The Greek has οἰ γενόμενοι ἐξ αὐτῶν, the (numbers) produced from them. + By from them + Euclid means from the original numbers, + though this is not very clear even in the Greek. I think ambiguity is best avoided by leaving out the words.

+
+
+ PROPOSITION 17. +

+ If a number by multiplying two numbers make certain numbers, the numbers so produced will have the same ratio as the numbers multiplied. +

+

For let the number A by multiplying the two numbers B, C make D, E; I say that, as B is to C, so is D to E.

+

For, since A by multiplying B has made D, therefore B measures D according to the units in A.

+

+

But the unit F also measures the number A according to the units in it; therefore the unit F measures the number A the same number of times that B measures D.

+

Therefore, as the unit F is to the number A, so is B to D. [VII. Def. 20]

+

For the same reason, as the unit F is to the number A, so also is C to E; therefore also, as B is to D, so is C to E.

+

Therefore, alternately, as B is to C, so is D to E. [VII. 13] Q. E. D.

+
+ +
+ PROPOSITION 18. +

+ If two numbers by multiplying any number make certain numbers, the numbers so produced will have the same ratio as the multipliers. +

+

For let two numbers A, B by multiplying any number C make D, E; I say that, as A is to B, so is D to E.

+

+

For, since A by multiplying C has made D, therefore also C by multiplying A has made D. [VII. 16] For the same reason also C by multiplying B has made E.

+

Therefore the number C by multiplying the two numbers A, B has made D, E.

+

Therefore, as A is to B, so is D to E. [VII. 17]

+
+
+ PROPOSITION 19. +

+ If four numbers be proportional, the number produced from the first and fourth will be equal to the number produced from the second and third; and, if the number produced from the first and fourth be equal to that produced from the second and third, the four numbers will be proportional. +

+

Let A, B, C, D be four numbers in proportion, so that, as A is to B, so is C to D; and let A by multiplying D make E, and let B by multiplying C make F; I say that E is equal to F.

+

For let A by multiplying C make G.

+

Since, then, A by multiplying C has made G, and by multiplying D has made E, the number A by multiplying the two numbers C, D has made G, E.

+

Therefore, as C is to D, so is G to E. [VII. 17]

+

But, as C is to D, so is A to B; therefore also, as A is to B, so is G to E.

+

+

Again, since A by multiplying C has made G, but, further, B has also by multiplying C made F, the two numbers A, B by multiplying a certain number C have made G, F.

+

Therefore, as A is to B, so is G to F. [VII. 18]

+

But further, as A is to B, so is G to E also; therefore also, as G is to E, so is G to F.

+

Therefore G has to each of the numbers E, F the same ratio; therefore E is equal to F. [cf. V. 9] +

+

Again, let E be equal to F; I say that, as A is to B, so is C to D.

+

For, with the same construction, since E is equal to F, therefore, as G is to E, so is G to F. [cf. V. 7]

+

But, as G is to E, so is C to D, [VII. 17] and, as G is to F, so is A to B. [VII. 18] +

+

Therefore also, as A is to B, so is C to D. Q. E. D.

+
+
+ PROPOSITION 20. +

+ The least numbers of those which have the same ratio with them measure those which have the same ratio the same number of times, the greater the greater and the less the less. +

+

For let CD, EF be the least numbers of those which have the same ratio with A, B; I say that CD measures A the same number of times that EF measures B.

+

Now CD is not parts of A.

+

For, if possible, let it be so; therefore EF is also the same parts of B that CD is of A. [VII. 13 and Def. 20]

+

+

Therefore, as many parts of A as there are in CD, so many parts of B are there also in EF.

+

Let CD be divided into the parts of A, namely CG, GD, and EF into the parts of B, namely EH, HF; thus the multitude of CG, GD will be equal to the multitude of EH, HF.

+

Now, since the numbers CG, GD are equal to one another, and the numbers EH, HF are also equal to one another, while the multitude of CG, GD is equal to the multitude of EH, HF, therefore, as CG is to EH, so is GD to HF.

+

Therefore also, as one of the antecedents is to one of the consequents, so will all the antecedents be to all the consequents. [VII. 12]

+

Therefore, as CG is to EH, so is CD to EF.

+

Therefore CG, EH are in the same ratio with CD, EF, being less than they: which is impossible, for by hypothesis CD, EF are the least numbers of those which have the same ratio with them.

+

Therefore CD is not parts of A; therefore it is a part of it. [VII. 4]

+

And EF is the same part of B that CD is of A; [VII. 13 and Def. 20] therefore CD measures A the same number of times that EF measures B. Q. E. D.

+
+
+ PROPOSITION 21. +

+ Numbers prime to one another are the least of those which have the same ratio with them. +

+

Let A, B be numbers prime to one another; I say that A, B are the least of those which have the same ratio with them.

+

For, if not, there will be some numbers less than A, B which are in the same ratio with A, B.

+

+

Let them be C, D.

+

Since, then, the least numbers of those which have the same ratio measure those which have the same ratio the same number of times, the greater the greater and the less the less, that is, the antecedent the antecedent and the consequent the consequent, [VII. 20] therefore C measures A the same number of times that D measures B.

+

Now, as many times as C measures A, so many units let there be in E.

+

Therefore D also measures B according to the units in E.

+

And, since C measures A according to the units in E, therefore E also measures A according to the units in C. [VII. 16]

+

For the same reason E also measures B according to the units in D. [VII. 16]

+

Therefore E measures A, B which are prime to one another: which is impossible. [VII. Def. 12]

+

Therefore there will be no numbers less than A, B which are in the same ratio with A, B.

+

Therefore A, B are the least of those which have the same ratio with them. Q. E. D.

+
+
+ PROPOSITION 22. +

+ The least numbers of those which have the same ratio with them are prime to one another. +

+

Let A, B be the least numbers of those which have the same ratio with them; I say that A, B are prime to one another.

+

+

For, if they are not prime to one another, some number will measure them.

+

Let some number measure them, and let it be C.

+

And, as many times as C measures A, so many units let there be in D, and, as many times as C measures B, so many units let there be in E +

+

Since C measures A according to the units in D, therefore C by multiplying D has made A. [VII. Def. 15]

+

For the same reason also C by multiplying E has made B.

+

Thus the number C by multiplying the two numbers D, E has made A, B; therefore, as D is to E, so is A to B; [VII. 17] therefore D, E are in the same ratio with A, B, being less than they: which is impossible.

+

Therefore no number will measure the numbers A, B.

+

Therefore A, B are prime to one another. Q. E. D.

+
+
+ PROPOSITION 23. +

+ If two number be prime to one another, the number which measures the one of them will be prime to the remaining number. +

+

Let A, B be two numbers prime to one another, and let any number C measure A; I say that C, B are also prime to one another.

+

For, if C, B are not prime to one another, some number will measure C, B.

+

Let a number measure them, and let it be D.

+

Since D measures C, and C measures A, therefore D also measures A.

+

+

But it also measures B; therefore D measures A, B which are prime to one another: which is impossible. [VII. Def. 12]

+

Therefore no number will measure the numbers C, B.

+

Therefore C, B are prime to one another. Q. E. D.

+
+
+ PROPOSITION 24. +

+ If two numbers be prime to any number, their product also will be prime to the same. +

+

For let the two numbers A, B be prime to any number C, and let A by multiplying B make D; I say that C, D are prime to one another.

+

For, if C, D are not prime to one another, some number will measure C, D.

+

Let a number measure them, and let it be E.

+

+

Now, since C, A are prime to one another, and a certain number E measures C, therefore A, E are prime to one another. [VII. 23]

+

As many times, then, as E measures D, so many units let there be in F; therefore F also measures D according to the units in E. [VII. 16]

+

Therefore E by multiplying F has made D. [VII. Def. 15]

+

But, further, A by multiplying B has also made D; therefore the product of E, F is equal to the product of A, B.

+

But, if the product of the extremes be equal to that of the means, the four numbers are proportional; [VII. 19] therefore, as E is to A, so is B to F.

+

But A, E are prime to one another, numbers which are prime to one another are also the least of those which have the same ratio, [VII. 21] and the least numbers of those which have the same ratio with them measure those which have the same ratio the same number of times, the greater the greater and the less the less, that is, the antecedent the antecedent and the consequent the consequent; [VII. 20] therefore E measures B.

+

But it also measures C; therefore E measures B, C which are prime to one another: which is impossible. [VII. Def. 12]

+

Therefore no number will measure the numbers C, D.

+

Therefore C, D are prime to one another. Q. E. D. +ὁ ἐξ αὐτῶν γενόμενος, literally the (number) produced from them, + will henceforth be translated as their product. +

+
+
+ PROPOSITION 25. +

+ If two numbers be prime to one another, the product of one of them into itself will be prime to the remaining one. +

+

Let A, B be two numbers prime to one another, and let A by multiplying itself make C: I say that B, C are prime to one another.

+

For let D be made equal to A.

+

Since A, B are prime to one another, and A is equal to D, therefore D, B are also prime to one another.

+

+

Therefore each of the two numbers D, A is prime to B; therefore the product of D, A will also be prime to B. [VII. 24]

+

But the number which is the product of D, A is C.

+

Therefore C, B are prime to one another. Q. E. D. +The Greek, ὁ ἐκ τοῦ ἑνὸς αὐτῶν γενόμενος, literally the number produced from the one of them, + leaves multiplied into itself + to be understood.

+
+
+ PROPOSITION 26. +

+ If two numbers be prime to two numbers, both to each, their products also will be prime to one another. +

+

For let the two numbers A, B be prime to the two numbers C, D; both to each, and let A by multiplying B make E, and let C by multiplying D make F; I say that E, F are prime to one another.

+

+

For, since each of the numbers A, B is prime to C, therefore the product of A, B will also be prime to C. [VII. 24]

+

But the product of A, B is E; therefore E, C are prime to one another.

+

For the same reason E, D are also prime to one another.

+

Therefore each of the numbers C, D is prime to E.

+

Therefore the product of C, D will also be prime to E. [VII. 24]

+

But the product of C, D is F.

+

Therefore E, F are prime to one another. Q. E. D.

+
+ +
+ PROPOSITION 27. +

If two numbers be prime to one another, and each by multiplying itself make a certain number, the products will be prime to one another; and, if the original numbers by multiplying the products make certain numbers, the latter will also be prime to one another [and this is always the case with the extremes].

+

Let A, B be two numbers prime to one another, let A by multiplying itself make C, and by multiplying C make D, and let B by multiplying itself make E, and by multiplying E make F; I say that both C, E and D, F are prime to one another.

+

+

For, since A, B are prime to one another, and A by multiplying itself has made C, therefore C, B are prime to one another. [VII. 25]

+

Since then C, B are prime to one another, and B by multiplying itself has made E, therefore C, E are prime to one another. [id.]

+

Again, since A, B are prime to one another, and B by multiplying itself has made E, therefore A, E are prime to one another. [id.]

+

Since then the two numbers A, C are prime to the two numbers B, E, both to each, therefore also the product of A, C is prime to the product of B, E. [VII. 26]

+

And the product of A, C is D, and the product of B, E is F.

+

Therefore D, F are prime to one another. Q. E. D.

+
+
+ PROPOSITION 28. +

+ If two numbers be prime to one another, the sum will also be prime to each of them; and, if the sum of two numbers be prime to any one of them, the original numbers will also be prime to one another. +

+

For let two numbers AB, BC prime to one another be added; I say that the sum AC is also prime to each of the numbers AB, BC.

+

+

For, if CA, AB are not prime to one another, some number will measure CA, AB.

+

Let a number measure them, and let it be D.

+

Since then D measures CA, AB, therefore it will also measure the remainder BC.

+

But it also measures BA; therefore D measures AB, BC which are prime to one another: which is impossible. [VII. Def. 12]

+

Therefore no number will measure the numbers CA, AB; therefore CA, AB are prime to one another.

+

For the same reason AC, CB are also prime to one another.

+

Therefore CA is prime to each of the numbers AB, BC.

+

Again, let CA, AB be prime to one another; I say that AB, BC are also prime to one another.

+

For, if AB, BC are not prime to one another, some number will measure AB, BC.

+

Let a number measure them, and let it be D.

+

Now, since D measures each of the numbers AB, BC, it will also measure the whole CA.

+

But it also measures AB; therefore D measures CA, AB which are prime to one another: which is impossible. [VII. Def. 12]

+

Therefore no number will measure the numbers AB, BC.

+

Therefore AB, BC are prime to one another. Q. E. D.

+
+
+ PROPOSITION 29. +

+ Any prime number is prime to any number which it does not measure. +

+

Let A be a prime number, and let it not measure B; I say that B, A are prime to one another.

+

For, if B, A are not prime to one another, some number will measure them.

+

+

Let C measure them.

+

Since C measures B, and A does not measure B, therefore C is not the same with A.

+

Now, since C measures B, A, therefore it also measures A which is prime, though it is not the same with it: which is impossible.

+

Therefore no number will measure B, A.

+

Therefore A, B are prime to one another. Q. E. D.

+
+
+ PROPOSITION 30. +

+ If two numbers by multiplying one another make some number, and any prime number measure the product, it will also measure one of the original numbers. +

+

For let the two numbers A, B by multiplying one another make C, and let any prime number D measure C; I say that D measures one of the numbers A, B.

+

+

For let it not measure A.

+

Now D is prime; therefore A, D are prime to one another. [VII. 29]

+

And, as many times as D measures C, so many units let there be in E.

+

Since then D measures C according to the units in E, therefore D by multiplying E has made C. [VII. Def. 15]

+

Further, A by multiplying B has also made C; therefore the product of D, E is equal to the product of A, B.

+

Therefore, as D is to A, so is B to E. [VII. 19]

+

But D, A are prime to one another, primes are also least, [VII. 21] and the least measure the numbers which have the same ratio the same number of times, the greater the greater and the less the less, that is, the antecedent the antecedent and the consequent the consequent; [VII. 20] therefore D measures B.

+

Similarly we can also show that, if D do not measure B, it will measure A.

+

Therefore D measures one of the numbers A, B. Q. E. D.

+
+
+ PROPOSITION 31. +

+ Any composite number is measured by some prime number. +

+

Let A be a composite number; I say that A is measured by some prime number.

+

For, since A is composite, some number will measure it.

+

Let a number measure it, and let it be B.

+

+

Now, if B is prime, what was enjoined will have been done.

+

But if it is composite, some number will measure it.

+

Let a number measure it, and let it be C.

+

Then, since C measures B, and B measures A, therefore C also measures A.

+

And, if C is prime, what was enjoined will have been done.

+

But if it is composite, some number will measure it.

+

Thus, if the investigation be continued in this way, some prime number will be found which will measure the number before it, which will also measure A.

+

For, if it is not found, an infinite series of numbers will measure the number A, each of which is less than the other: which is impossible in numbers.

+

Therefore some prime number will be found which will measure the one before it, which will also measure A.

+

Therefore any composite number is measured by some prime number. + i.e. the implied problem of finding a prime number which measures A. + In the Greek the sentence stops here, but it is necessary to add the words the number before it, which will also measure A, + which are found a few lines further down. It is possible that the words may have fallen out of P here by a simple mistake due to ὁμοιοτέλευτον (Heiberg).

+
+
+ PROPOSITION 32. +

+ Any number either is prime or is measured by some prime number. +

+

Let A be a number; I say that A either is prime or is measured by some prime number.

+

+

If now A is prime, that which was enjoined will have been done.

+

But if it is composite, some prime number will measure it. [VII. 31]

+

Therefore any number either is prime or is measured by some prime number. Q. E. D.

+
+
+ PROPOSITION 33. +

+ Given as many numbers as we please, to find the least of those which have the same ratio with them. +

+

Let A, B, C be the given numbers, as many as we please; thus it is required to find the least of those which have the same ratio with A, B, C.

+

A, B, C are either prime to one another or not.

+

+

Now, if A, B, C are prime to one another, they are the least of those which have the same ratio with them. [VII. 21]

+

But, if not, let D the greatest common measure of A, B, C be taken, [VII. 3] and, as many times as D measures the numbers A, B, C + respectively, so many units let there be in the numbers E, F, G respectively.

+

Therefore the numbers E, F, G measure the numbers A, B, C respectively according to the units in D. [VII. 16]

+

Therefore E, F, G measure A, B, C the same number of times; therefore E, F, G are in the same ratio with A, B, C. [VII. Def. 20]

+

I say next that they are the least that are in that ratio.

+

For, if E, F, G are not the least of those which have the same ratio with A, B, C, there will be numbers less than E, F, G which are in the same ratio with A, B, C.

+

Let them be H, K, L; therefore H measures A the same number of times that the numbers K, L measure the numbers B, C respectively.

+

Now, as many times as H measures A, so many units let there be in M; therefore the numbers K, L also measure the numbers B, C respectively according to the units in M.

+

And, since H measures A according to the units in M, therefore M also measures A according to the units in H. [VII. 16]

+

For the same reason M also measures the numbers B, C according to the units in the numbers K, L respectively;

+

Therefore M measures A, B, C.

+

Now, since H measures A according to the units in M, therefore H by multiplying M has made A. [VII. Def. 15]

+

For the same reason also E by multiplying D has made A.

+

Therefore the product of E, D is equal to the product of H, M.

+

Therefore, as E is to H, so is M to D. [VII. 19]

+

But E is greater than H; therefore M is also greater than D.

+

And it measures A, B, C: which is impossible, for by hypothesis D is the greatest common measure of A, B, C.

+

Therefore there cannot be any numbers less than E, F, G which are in the same ratio with A, B, C.

+

Therefore E, F, G are the least of those which have the same ratio with A, B, C. Q. E. D. +literally (as usual) each of the numbers E, F, G measures each of the numbers A, B, C. +

+
+ +
+ PROPOSITION 34. +

+ Given two numbers, to find the least number which they measure. +

+

Let A, B be the two given numbers; thus it is required to find the least number which they measure.

+

Now A, B are either prime to one another or not.

+

+

First, let A, B be prime to one another, and let A by multiplying B make C; therefore also B by multiplying A has made C. [VII. 16]

+

Therefore A, B measure C +

+

I say next that it is also the least number they measure.

+

For, if not, A, B will measure some number which is less than C.

+

Let them measure D.

+

Then, as many times as A measures D, so many units let there be in E, and, as many times as B measures D, so many units let there be in F; therefore A by multiplying E has made D, and B by multiplying F has made D; [VII. Def. 15] therefore the product of A, E is equal to the product of B, F.

+

Therefore, as A is to B, so is F + E. [VII. 19]

+

But A, B are prime, primes are also least, [VII. 21] and the least measure the numbers which have the same ratio the same number of times, the greater the greater and the less the less; [VII. 20] therefore B measures E, as consequent consequent.

+

And, since A by multiplying B, E has made C, D, therefore, as B is to E, so is C to D. [VII. 17]

+

But B measures E; therefore C also measures D, the greater the less: which is impossible.

+

Therefore A, B do not measure any number less than C; therefore C is the least that is measured by A, B.

+

Next, let A, B not be prime to one another, and let F, E, the least numbers of those which have the same ratio with A, B, be taken; [VII. 33] therefore the product of A, E is equal to the product of B, F. [VII. 19]

+

And let A by multiplying E make C; therefore also B by multiplying F has made C; therefore A, B measure C.

+

+

I say next that it is also the least number that they measure.

+

For, if not, A, B will measure some number which is less than C.

+

Let them measure D.

+

And, as many times as A measures D, so many units let there be in G, and, as many times as B measures D, so many units let there be in H.

+

Therefore A by multiplying G has made D, and B by multiplying H has made D.

+

Therefore the product of A, G is equal to the product of B, H; therefore, as A is to B, so is H to G. [VII. 19]

+

But, as A is to B, so is F to E.

+

Therefore also, as F is to E, so is H to G.

+

But F, E are least, and the least measure the numbers which have the same ratio the same number of times, the greater the greater and the less the less; [VII. 20] therefore E measures G.

+

And, since A by multiplying E, G has made C, D, therefore, as E is to G, so is C to D. [VII. 17]

+

But E measures G; therefore C also measures D, the greater the less: which is impossible.

+

Therefore A, B will not measure any number which is less than C.

+

Therefore C is the least that is measured by A, B. Q. E. D.

+
+ +
+ PROPOSITION 35. +

+ If two numbers measure any number, the least number measured by them will also measure the same. +

+

For let the two numbers A, B measure any number CD, and let E be the least that they measure; I say that E also measures CD.

+

+

For, if E does not measure CD, let E, measuring DF, leave CF less than itself.

+

Now, since A, B measure E, and E measures DF, therefore A, B will also measure DF.

+

But they also measure the whole CD; therefore they will also measure the remainder CF which is less than E: which is impossible.

+

Therefore E cannot fail to measure CD; therefore it measures it. Q. E. D.

+
+
+ PROPOSITION 36. +

+ Given three numbers, to find the least number which they measure. +

+

Let A, B, C be the three given numbers; thus it is required to find the least number which they measure.

+

+

Let D, the least number measured by the two numbers A, B, be taken. [VII. 34]

+

Then C either measures, or does not measure, D.

+

First, let it measure it.

+

But A, B also measure D; therefore A, B, C measure D.

+

I say next that it is also the least that they measure.

+

For, if not, A, B, C will measure some number which is less than D.

+

Let them measure E.

+

Since A, B, C measure E, therefore also A, B measure E.

+

Therefore the least number measured by A, B will also measure E. [VII. 35]

+

But D is the least number measured by A, B; therefore D will measure E, the greater the less: which is impossible.

+

Therefore A, B, C will not measure any number which is less than D; therefore D is the least that A, B, C measure. +

+

Again, let C not measure D, and let E, the least number measured by C, D, be taken. [VII. 34]

+

+

Since A, B measure D, and D measures E, therefore also A, B measure E.

+

But C also measures E; therefore also A, B, C measure E.

+

I say next that it is also the least that they measure.

+

For, if not, A, B, C will measure some number which is less than E.

+

Let them measure F.

+

Since A, B, C measure F, therefore also A, B measure F; therefore the least number measured by A, B will also measure F. [VII. 35]

+

But D is the least number measured by A, B; therefore D measures F.

+

But C also measures F; therefore D, C measure F, so that the least number measured by D, C will also measure F.

+

But E is the least number measured by C, D; therefore E measures F, the greater the less: which is impossible.

+

Therefore A, B, C will not measure any number which is less than E.

+

Therefore E is the least that is measured by A, B, C. Q. E. D.

+
+
+ PROPOSITION 37. +

+ If a number be measured by any number, the number which is measured will have a part called by the same name as the measuring number. +

+

For let the number A be measured by any number B; I say that A has a part called by the same name as B.

+

+

For, as many times as B measures A, so many units let there be in C.

+

Since B measures A according to the units in C, and the unit D also measures the number C according to the units in it, therefore the unit D measures the number C the same number of times as B measures A.

+

Therefore, alternately, the unit D measures the number B the same number of times as C measures A; [VII. 15] therefore, whatever part the unit D is of the number B, the same part is C of A also.

+

But the unit D is a part of the number B called by the same name as it; therefore C is also a part of A called by the same name as B, so that A has a part C which is called by the same name as B. Q. E. D.

+
+
+ PROPOSITION 38. +

+ If a number have any part whatever, it will be measured by a number called by the same name as the part. +

+

For let the number A have any part whatever, B, and let C be a number called by the same name as the part B; I say that C measures A.

+

+

For, since B is a part of A called by the same name as C, and the unit D is also a part of C called by the same name as it, therefore, whatever part the unit D is of the number C, the same part is B of A also; therefore the unit D measures the number C the same number of times that B measures A.

+

Therefore, alternately, the unit D measures the number B the same number of times that C measures A. [VII. 15]

+

Therefore C measures A. Q. E. D.

+
+
+ PROPOSITION 39. +

+ To find the number which is the least that will have given parts. +

+

Let A, B, C be the given parts; thus it is required to find the number which is the least that will have the parts A, B, C.

+

+

Let D, E, F be numbers called by the same name as the parts A, B, C, and let G, the least number measured by D, E, F, be taken. [VII. 36]

+

Therefore G has parts called by the same name as D, E, F. [VII. 37]

+

But A, B, C are parts called by the same name as D, E, F; therefore G has the parts A, B, C.

+

I say next that it is also the least number that has.

+

For, if not, there will be some number less than G which will have the parts A, B, C.

+

Let it be H.

+

Since H has the parts A, B, C, therefore H will be measured by numbers called by the same name as the parts A, B, C. [VII. 38]

+

But D, E, F are numbers called by the same name as the parts A, B, C; therefore H is measured by D, E, F.

+

And it is less than G : which is impossible.

+

Therefore there will be no number less than G that will have the parts A, B, C. Q. E. D.

+
+
+
+
+ + BOOK VIII. +
+ PROPOSITIONS +
+ PROPOSITION 1. +

+ If there be as many numbers as we please in continued proportion, and the extremes of them be prime to one another, the numbers are the least of those which have the same ratio with them. +

+

Let there be as many numbers as we please, A, B, C, D, in continued proportion, and let the extremes of them A, D be prime to one another; I say that A, B, C, D are the least of those which have the same ratio with them.

+

+

For, if not, let E, F, G, H be less than A, B, C, D, and in the same ratio with them.

+

Now, since A, B, C, D are in the same ratio with E, F, G, H, and the multitude of the numbers A, B, C, D is equal to the multitude of the numbers E, F, G, H, therefore, ex aequali, as A is to D, so is E to H. [VII. 14] +

+

But A, D are prime, primes are also least, [VII. 21] and the least numbers measure those which have the same ratio the same number of times, the greater the greater and the less the less, that is, the antecedent the antecedent and the consequent the consequent. [VII. 20]

+

Therefore A measures E, the greater the less: which is impossible.

+

Therefore E, F, G, H which are less than A, B, C, D are not in the same ratio with them.

+

Therefore A, B, C, D are the least of those which have the same ratio with them. Q. E. D.

+
+
+ PROPOSITION 2. +

+ To find numbers in continued proportion, as many as may be prescribed, and the least that are in a given ratio. +

+

Let the ratio of A to B be the given ratio in least numbers; thus it is required to find numbers in continued proportion, as many as may be prescribed, and the least that are in the ratio of A to B.

+

+

Let four be prescribed; let A by multiplying itself make C, and by multiplying B let it make D; let B by multiplying itself make E; further, let A by multiplying C, D, E make F, G, H, and let B by multiplying E make K.

+

Now, since A by multiplying itself has made C, and by multiplying B has made D, therefore, as A is to B, so is C to D. [VII. 17]

+

Again, since A by multiplying B has made D, and B by multiplying itself has made E, therefore the numbers A, B by multiplying B have made the numbers D, E respectively.

+

Therefore, as A is to B, so is D to E. [VII. 18]

+

But, as A is to B, so is C to D; therefore also, as C is to D, so is D to E.

+

And, since A by multiplying C, D has made F, G, therefore, as C is to D, so is F to G. [VII. 17]

+

But, as C is to D, so was A to B; therefore also, as A is to B, so is F to G.

+

Again, since A by multiplying D, E has made G, H, therefore, as D is to E, so is G to H. [VII. 17]

+

But, as D is to E, so is A to B.

+

Therefore also, as A is to B, so is G to H.

+

And, since A, B by multiplying E have made H, K, therefore, as A is to B, so is H to K. [VII. 18]

+

But, as A is to B, so is F to G, and G to H.

+

Therefore also, as F is to G, so is G to H, and H to K; therefore C, D, E, and F, G, H, K are proportional in the ratio of A to B.

+

I say next that they are the least numbers that are so.

+

For, since A, B are the least of those which have the same ratio with them, and the least of those which have the same ratio are prime to one another, [VII. 22] therefore A, B are prime to one another.

+

And the numbers A, B by multiplying themselves respectively have made the numbers C, E, and by multiplying the numbers C, E respectively have made the numbers F, K; therefore C, E and F, K are prime to one another respectively. [VII. 27]

+

But, if there be as many numbers as we please in continued proportion, and the extremes of them be prime to one another, they are the least of those which have the same ratio with them. [VIII. 1]

+

Therefore C, D, E and F, G, H, K are the least of those which have the same ratio with A, B. Q. E. D.

+ +

Porism. From this it is manifest that, if three numbers in continued proportion be the least of those which have the same ratio with them, the extremes of them are squares, and, if four numbers, cubes.

+ +
+
+ PROPOSITION 3. +

If as many numbers as we please in continued proportion be the least of those which have the same ratio with them, the extremes of them are prime to one another.

+

Let as many numbers as we please, A, B, C, D, in continued proportion be the least of those which have the same ratio with them;

+ I say that the extremes of them A, D are prime to one another.

+

For let two numbers E, F, the least that are in the ratio of A, B, C, D, be taken, [VII. 33] then three others G, H, K with the same property; and others, more by one continually, [VIII. 2] until the multitude taken becomes equal to the multitude of the numbers A, B, C, D.

+

Let them be taken, and let them be L, M, N, O.

+

Now, since E, F are the least of those which have the same ratio with them, they are prime to one another. [VII. 22]

+

And, since the numbers E, F by multiplying themselves respectively have made the numbers G, K, and by multiplying the numbers G, K respectively have made the numbers L, O, [VIII. 2, Por.] therefore both G, K and L, O are prime to one another. [VII. 27]

+

And, since A, B, C, D are the least of those which have the same ratio with them, while L, M, N, O are the least that are in the same ratio with A, B, C, D, and the multitude of the numbers A, B, C, D is equal to the multitude of the numbers L, M, N, O, therefore the numbers A, B, C, D are equal to the numbers L, M, N, O respectively; therefore A is equal to L, and D to O.

+

And L, O are prime to one another.

+

Therefore A, D are also prime to one another. Q. E. D.

+
+ +
+ PROPOSITION 4. +

+ Given as many ratios as we please in least numbers, to find numbers in continued proportion which are the least in the given ratios. +

+

Let the given ratios in least numbers be that of A to B, that of C to D, and that of E to F; thus it is required to find numbers in continued proportion which are the least that are in the ratio of A to B, in the ratio of C to D, and in the ratio of E to F.

+

+

Let G, the least number measured by B, C, be taken. [VII. 34]

+

And, as many times as B measures G, so many times also let A measure H, and, as many times as C measures G, so many times also let D measure K.

+

Now E either measures or does not measure K.

+

First, let it measure it.

+

And, as many times as E measures K, so many times let F measure L also.

+

Now, since A measures H the same number of times that B measures G, therefore, as A is to B, so is H to G. [VII. Def. 20, VII. 13]

+

For the same reason also, as C is to D, so is G to K, and further, as E is to F, so is K to L; therefore H, G, K, L are continuously proportional in the ratio of A to B, in the ratio of C to D, and in the ratio of E to F.

+

I say next that they are also the least that have this property.

+

For, if H, G, K, L are not the least numbers continuously proportional in the ratios of A to B, of C to D, and of E to F, let them be N, O, M, P.

+

Then since, as A is to B, so is N to O, while A, B are least, and the least numbers measure those which have the same ratio the same number of times, the greater the greater and the less the less, that is, the antecedent the antecedent and the consequent the consequent; therefore B measures O. [VII. 20]

+

For the same reason C also measures O; therefore B, C measure O; therefore the least number measured by B, C will also measure O. [VII. 35]

+

But G is the least number measured by B, C; therefore G measures O, the greater the less: which is impossible.

+

Therefore there will be no numbers less than H, G, K, L which are continuously in the ratio of A to B, of C to D, and of E to F.

+

Next, let E not measure K.

+

+

Let M, the least number measured by E, K, be taken.

+

And, as many times as K measures M, so many times let H, G measure N, O respectively, and, as many times as E measures M, so many times let F + measure P also.

+

Since H measures N the same number of times that G measures O, therefore, as H is to G, so is N to O. [VII. 13 and Def. 20]

+

But, as H is to G, so is A to B; therefore also, as A is to B, so is N to O.

+

For the same reason also, as C is to D, so is O to M. +

+

Again, since E measures M the same number of times that F measures P, therefore, as E is to F, so is M to P; [VII. 13 and Def. 20] therefore N, O, M, P are continuously proportional in the ratios of A to B, of C to D, and of E to F.

+

I say next that they are also the least that are in the ratios A : B, C : D, E : F.

+

For, if not, there will be some numbers less than N, O, M, P continuously proportional in the ratios A : B, C : D, E : F.

+

Let them be Q, R, S, T.

+

Now since, as Q is to R, so is A to B, while A, B are least, and the least numbers measure those which have the same ratio with them the same number of times, the antecedent the antecedent and the consequent the consequent, [VII. 20] therefore B measures R.

+

For the same reason C also measures R; therefore B, C measure R.

+

Therefore the least number measured by B, C will also measure R. [VII. 35]

+

But G is the least number measured by B, C; therefore G measures R.

+

And, as G is to R, so is K to S: [VII. 13] therefore K also measures S.

+

But E also measures S; therefore E, K measure S.

+

Therefore the least number measured by E, K will also measure S. [VII. 35]

+

But M is the least number measured by E, K; therefore M measures S, the greater the less: which is impossible.

+

Therefore there will not be any numbers less than N, O, M, P continuously proportional in the ratios of A to B, of C to D, and of E to F; therefore N, O, M, P are the least numbers continuously proportional in the ratios A : B, C : D, E : F. Q. E. D. +69, 71, 99. the ratios A : B, C : D, E : F. This abbreviated expression is in the Greek ὀ ΑΒ, ΓΔ, ΕΖ λόγοι.

+
+ +
+ PROPOSITION 5. +

+ Plane numbers have to one another the ratio compounded of the ratios of their sides. +

+

Let A, B be plane numbers, and let the numbers C, D be the sides of A, and E, F of B; I say that A has to B the ratio compounded of the ratios of the sides.

+

+

For, the ratios being given which C has to E and D to F, let the least numbers G, H, K that are continuously in the ratios C : E, D : F be taken, so that, as C is to E, so is G to H, and, as D is to F, so is H to K. [VIII. 4]

+

And let D by multiplying E make L.

+

Now, since D by multiplying C has made A, and by multiplying E has made L, therefore, as C is to E, so is A to L. [VII. 17]

+

But, as C is to E, so is G to H; therefore also, as G is to H, so is A to L.

+

Again, since E by multiplying D has made L, and further by multiplying F has made B, therefore, as D is to F, so is L to B. [VII. 17]

+

But, as D is to F, so is H to K; therefore also, as H is to K, so is L to B.

+

But it was also proved that, as G is to H, so is A to L; therefore, ex aequali, as G is to K, so is A to B. [VII. 14] +

+

But G has to K the ratio compounded of the ratios of the sides; therefore A also has to B the ratio compounded of the ratios of the sides. Q. E. D. +1, 5, 29, 31. compounded of the ratios of their sides. As in VI. 23, the Greek has the less exact phrase, compounded of their sides. +

+
+
+ PROPOSITION 6. +

+ If there be as many numbers as we please in continued proportion, and the first do not measure the second, neither will any other measure any other. +

+

Let there be as many numbers as we please, A, B, C, D, E, in continued proportion, and let A not measure B; I say that neither will any other measure any other.

+

+

Now it is manifest that A, B, C, D, E do not measure one another in order; for A does not even measure B.

+

I say, then, that neither will any other measure any other.

+

For, if possible, let A measure C.

+

And, however many A, B, C are, let as many numbers F, G, H, the least of those which have the same ratio with A, B, C, be taken. [VII. 33]

+

Now, since F, G, H are in the same ratio with A, B, C, and the multitude of the numbers A, B, C is equal to the multitude of the numbers F, G, H, therefore, ex aequali, as A is to C, so is F to H. [VII. 14]

+

And since, as A is to B, so is F to G, while A does not measure B, therefore neither does F measure G; [VII. Def. 20] therefore F is not an unit, for the unit measures any number.

+

Now F, H are prime to one another. [VIII. 3]

+

And, as F is to H, so is A to C; therefore neither does A measure C.

+

Similarly we can prove that neither will any other measure any other. Q. E. D.

+
+
+ PROPOSITION 7. +

+ If there be as many numbers as we please in continued proportion, and the first measure the last, it will measure the second also. +

+

Let there be as many numbers as we please, A, B, C, D, in continued proportion; and let A measure D; I say that A also measures B.

+

+

For, if A does not measure B, neither will any other of the numbers measure any other. [VIII. 6]

+

But A measures D.

+

Therefore A also measures B. Q. E. D.

+
+ +
+ PROPOSITION 8. +

+ If between two numbers there fall numbers in continued proportion with them, then, however many numbers fall between them in continued proportion, so many will also fall in continued proportion between the numbers which have the same ratio with the original numbers. +

+

Let the numbers C, D fall between the two numbers A, B in continued proportion with them, and let E be made in the same ratio to F as A is to B; I say that, as many numbers as have fallen between A, B in continued proportion, so many will also fall between E, F in continued proportion.

+

+

For, as many as A, B, C, D are in multitude, let so many numbers G, H, K, L, the least of those which have the same ratio with A, C, D, B, be taken; [VII. 33] therefore the extremes of them G, L are prime to one another. [VIII. 3]

+

Now, since A, C, D, B are in the same ratio with G, H, K, L, and the multitude of the numbers A, C, D, B is equal to the multitude of the numbers G, H, K, L, therefore, ex aequali, as A is to B, so is G to L. [VII. 14]

+

But, as A is to B, so is E to F; therefore also, as G is to L, so is E to F.

+

But G, L are prime, primes are also least, [VII. 21] and the least numbers measure those which have the same ratio the same number of times, the greater the greater and the less the less, that is, the antecedent the antecedent and the consequent the consequent. [VII. 20]

+

Therefore G measures E the same number of times as L measures F.

+

Next, as many times as G measures E, so many times let H, K also measure M, N respectively; therefore G, H, K, L measure E, M, N, F the same number of times.

+

Therefore G, H, K, L are in the same ratio with E, M, N, F. [VII. Def. 20]

+

But G, H, K, L are in the same ratio with A, C, D, B; therefore A, C, D, B are also in the same ratio with E, M, N, F.

+

But A, C, D, B are in continued proportion; therefore E, M, N, F are also in continued proportion.

+

Therefore, as many numbers as have fallen between A, B in continued proportion with them, so many numbers have also fallen between E, F in continued proportion. Q. E. D. +1. fall. The Greek word is ἐμπίπτειν, fall in +=can be interpolated. +

+
+
+ PROPOSITION 9. +

+ If two numbers be prime to one another, and numbers fall between them in continued proportion, then, however many numbers fall between them in continued proportion, so many will also fall between each of them and an unit in continued proportion. +

+

Let A, B be two numbers prime to one another, and let C, D fall between them in continued proportion, and let the unit E be set out; I say that, as many numbers as fall between A, B in continued proportion, so many will also fall between either of the numbers A, B and the unit in continued proportion.

+

For let two numbers F, G, the least that are in the ratio of A, C, D, B, be taken, three numbers H, K, L with the same property, and others more by one continually, until their multitude is equal to the multitude of A, C, D, B. [VIII. 2]

+

+

Let them be taken, and let them be M, N, O, P.

+

It is now manifest that F by multiplying itself has made H and by multiplying H has made M, while G by multiplying itself has made L and by multiplying L has made P. [VIII. 2, Por.]

+

And, since M, N, O, P are the least of those which have the same ratio with F, G, and A, C, D, B are also the least of those which have the same ratio with F, G, [VIII. 1] while the multitude of the numbers M, N, O, P is equal to the multitude of the numbers A, C, D, B, therefore M, N, O, P are equal to A, C, D, B respectively; therefore M is equal to A, and P to B.

+

Now, since F by multiplying itself has made H, therefore F measures H according to the units in F.

+

But the unit E also measures F according to the units in it; therefore the unit E measures the number F the same number of times as F measures H.

+

Therefore, as the unit E is to the number F, so is F to H. [VII. Def. 20]

+

Again, since F by multiplying H has made M, therefore H measures M according to the units in F.

+

But the unit E also measures the number F according to the units in it; therefore the unit E measures the number F the same number of times as H measures M.

+

Therefore, as the unit E is to the number F, so is H to M.

+

But it was also proved that, as the unit E is to the number F, so is F to H; therefore also, as the unit E is to the number F, so is F to H, and H to M.

+

But M is equal to A; therefore, as the unit E is to the number F, so is F to H, and H to A.

+

For the same reason also, as the unit E is to the number G, so is G to L and L to B.

+

Therefore, as many numbers as have fallen between A, B in continued proportion, so many numbers also have fallen between each of the numbers A, B and the unit E in continued proportion. Q. E. D.

+
+
+ PROPOSITION 10. +

+ If numbers fall between each of two numbers and an unit in continued proportion, however many numbers fall between each of them and an unit in continued proportion, so many also will fall between the numbers themselves in continued proportion. + +

+

For let the numbers D, E and F, G respectively fall between the two numbers A, B and the unit C in continued proportion; I say that, as many numbers as have fallen between each of the numbers A, B and the unit C in continued proportion, so many numbers will also fall between A, B in continued proportion.

+

For let D by multiplying F make H, and let the numbers D, F by multiplying H make K, L respectively.

+

+

Now, since, as the unit C is to the number D, so is D to E, therefore the unit C measures the number D the same number of times as D measures E. [VII. Def. 20]

+

But the unit C measures the number D according to the units in D; therefore the number D also measures E according to the units in D; therefore D by multiplying itself has made E.

+

Again, since, as C is to the number D, so is E to A, therefore the unit C measures the number D the same number of times as E measures A.

+

But the unit C measures the number D according to the units in D; therefore E also measures A according to the units in D; therefore D by multiplying E has made A.

+

For the same reason also F by multiplying itself has made G, and by multiplying G has made B.

+

And, since D by multiplying itself has made E and by multiplying F has made H, therefore, as D is to F, so is E to H. [VII. 17]

+

For the same reason also, as D is to F, so is H to G. [VII. 18] +

+

Therefore also, as E is to H, so is H to G.

+

Again, since D by multiplying the numbers E, H has made A, K respectively, therefore, as E is to H, so is A to K. [VII. 17]

+

But, as E is to H, so is D to F; therefore also, as D is to F, so is A to K.

+

Again, since the numbers D, F by multiplying H have made K, L respectively, therefore, as D is to F, so is K to L. [VII. 18]

+

But, as D is to F, so is A to K; therefore also, as A is to K, so is K to L.

+

Further, since F by multiplying the numbers H, G has made L, B respectively, therefore, as H is to G, so is L to B. [VII. 17]

+

But, as H is to G, so is D to F; therefore also, as D is to F, so is L to B.

+

But it was also proved that, as D is to F, so is A to K and K to L; therefore also, as A is to K, so is K to L and L to B.

+

Therefore A, K, L, B are in continued proportion.

+

Therefore, as many numbers as fall between each of the numbers A, B and the unit C in continued proportion, so many also will fall between A, B in continued proportion. Q. E. D.

+
+
+ PROPOSITION 11. +

+ Between two square numbers there is one mean proportional number, and the square has to the square the ratio duplicate of that which the side has to the side. +

+

Let A, B be square numbers, and let C be the side of A, and D of B; I say that between A, B there is one mean proportional number, and A has to B the ratio duplicate of that which C has to D.

+

For let C by multiplying D make E.

+

+

Now, since A is a square and C is its side, therefore C by multiplying itself has made A.

+

For the same reason also D by multiplying itself has made B.

+

Since then C by multiplying the numbers C, D has made A, E respectively, therefore, as C is to D, so is A to E. [VII. 17]

+

For the same reason also, as C is to D, so is E to B. [VII. 18] +

+

Therefore also, as A is to E, so is E to B.

+

Therefore between A, B there is one mean proportional number.

+

I say next that A also has to B the ratio duplicate of that which C has to D.

+

For, since A, E, B are three numbers in proportion, therefore A has to B the ratio duplicate of that which A has to E. [V. Def. 9]

+

But, as A is to E, so is C to D.

+

Therefore A has to B the ratio duplicate of that which the side C has to D. Q. E. D.

+
+
+ PROPOSITION 12. +

+ Between two cube numbers there are two mean proportional numbers, and the cube has to the cube the ratio triplicate of that which the side has to the side. +

+

Let A, B be cube numbers, and let C be the side of A, and D of B; I say that between A, B there are two mean proportional numbers, and A has to B the ratio triplicate of that which C has to D.

+

+

For let C by multiplying itself make E, and by multiplying D let it make F; let D by multiplying itself make G, and let the numbers C, D by multiplying F make H, K respectively.

+

Now, since A is a cube, and C its side, and C by multiplying itself has made E, therefore C by multiplying itself has made E and by multiplying E has made A.

+

For the same reason also D by multiplying itself has made G and by multiplying G has made B.

+

And, since C by multiplying the numbers C, D has made E, F respectively, therefore, as C is to D, so is E to F. [VII. 17]

+

For the same reason also, as C is to D, so is F to G. [VII. 18] +

+

Again, since C by multiplying the numbers E, F has made A, H respectively, therefore, as E is to F, so is A to H. [VII. 17]

+

But, as E is to F, so is C to D.

+

Therefore also, as C is to D, so is A to H.

+

Again, since the numbers C, D by multiplying F have made H, K respectively, therefore, as C is to D, so is H to K. [VII. 18]

+

Again, since D by multiplying each of the numbers F, G has made K, B respectively, therefore, as F is to G, so is K to B. [VII. 17]

+

But, as F is to G, so is C to D; therefore also, as C is to D, so is A to H, H to K, and K to B.

+

Therefore H, K are two mean proportionals between A, B.

+

I say next that A also has to B the ratio triplicate of that which C has to D.

+

For, since A, H, K, B are four numbers in proportion, therefore A has to B the ratio triplicate of that which A has to H. [V. Def. 10]

+

But, as A is to H, so is C to D; therefore A also has to B the ratio triplicate of that which C has to D. Q. E. D.

+
+
+ PROPOSITION 13. +

+ If there be as many numbers as we please in continued proportion, and each by multiplying itself make some number, the products will be proportional; and, if the original numbers by multiplying the products make certain numbers, the latter will also be proportional. + +

+

Let there be as many numbers as we please, A, B, C, in continued proportion, so that, as A is to B, so is B to C; let A, B, C by multiplying themselves make D, E, F, and by multiplying D, E, F let them make G, H, K; I say that D, E, F and G, H, K are in continued proportion.

+

+

For let A by multiplying B make L, and let the numbers A, B by multiplying L make M. N respectively.

+

And again let B by multiplying C make O, and let the numbers B, C by multiplying O make P, Q respectively.

+

Then, in manner similar to the foregoing, we can prove that D, L, E and G, M, N, H are continuously proportional in the ratio of A to B, and further E, O, F and H, P, Q, K are continuously proportional in the ratio of B to C.

+

Now, as A is to B, so is B to C; therefore D, L, E are also in the same ratio with E, O, F, and further G, M, N, H in the same ratio with H, P, Q, K.

+

And the multitude of D, L, E is equal to the multitude of E, O, F, and that of G, M, N, H to that of H, P, Q, K; therefore, ex acquali, as D is to E, so is E to F, and, as G is to H, so is H to K. [VII. 14] Q. E. D.

+
+
+ PROPOSITION 14. +

+ If a square measure a square, the side will also measure the side; and, if the side measure the side, the square will also measure the square. +

+

Let A, B be square numbers, let C, D be their sides, and let A measure B; I say that C also measures D.

+

+

For let C by multiplying D make E; therefore A, E, B are continuously proportional in the ratio of C to D. [VIII. 11]

+

And, since A, E, B are continuously proportional, and A measures B, therefore A also measures E. [VIII. 7]

+

And, as A is to E, so is C to D; therefore also C measures D. [VII. Def. 20]

+

Again, let C measure D; I say that A also measures B.

+

For, with the same construction, we can in a similar manner prove that A, E, B are continuously proportional in the ratio of C to D.

+

And since, as C is to D, so is A to E, and C measures D, therefore A also measures E. [VII. Def. 20]

+

And A, E, B are continuously proportional; therefore A also measures B.

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 15. +

+ If a cube number measure a cube number, the side will also measure the side; and, if the side measure the side, the cube will also measure the cube. +

+

For let the cube number A measure the cube B, and let C be the side of A and D of B; I say that C measures D.

+

For let C by multiplying itself make E, and let D by multiplying itself make G; further, let C by multiplying D make F, and let C, D by multiplying F make H, K respectively.

+

+

Now it is manifest that E, F, G and A, H, K, B are continuously proportional in the ratio of C to D. [VIII. 11, 12]

+

And, since A, H, K, B are continuously proportional, and A measures B, therefore it also measures H. [VIII. 7]

+

And, as A is to H, so is C to D; therefore C also measures D. [VII. Def. 20]

+

Next, let C measure D; I say that A will also measure B.

+

For, with the same construction, we can prove in a similar manner that A, H, K, B are continuously proportional in the ratio of C to D.

+

And, since C measures D, and, as C is to D, so is A to H, therefore A also measures H, [VII. Def. 20] so that A measures B also. Q. E. D.

+
+ +
+ PROPOSITION 16. +

+ If a square number do not measure a square number, neither will the side measure the side; and, if the side do not measure the side, neither will the square measure the square. +

+

Let A, B be square numbers, and let C, D be their sides; and let A not measure B; I say that neither does C measure D.

+

For, if C measures D, A will also measure B. [VIII. 14]

+

+

But A does not measure B; therefore neither will C measure D.

+

Again, let C not measure D; I say that neither will A measure B.

+

For, if A measures B, C will also measure D. [VIII. 14]

+

But C does not measure D; therefore neither will A measure B. Q. E. D.

+
+
+ PROPOSITION 17. +

+ If a cube number do not measure a cube number, neither will the side measure the side; and, if the side do not measure the side, neither will the cube measure the cube. +

+

For let the cube number A not measure the cube number B, and let C be the side of A, and D of B; I say that C will not measure D.

+

+

For if C measures D, A will also measure B. [VIII. 15]

+

But A does not measure B; therefore neither does C measure D.

+

Again, let C not measure D; I say that neither will A measure B.

+

For, if A measures B, C will also measure D. [VIII. 15]

+

But C does not measure D; therefore neither will A measure B. Q. E. D.

+
+
+ PROPOSITION 18. +

+ Between two similar plane numbers there is one mean proportional number; and the plane number has to the plane number the ratio duplicate of that which the corresponding side has to the corresponding side. +

+

Let A, B be two similar plane numbers, and let the numbers C, D be the sides of A, and E, F of B.

+

+

Now, since similar plane numbers are those which have their sides proportional, [VII. Def. 21] therefore, as C is to D, so is E to F.

+

I say then that between A, B there is one mean proportional number, and A has to B the ratio duplicate of that which C has to E, or D to F, that is, of that which the corresponding side has to the corresponding side.

+

Now since, as C is to D, so is E to F, therefore, alternately, as C is to E, so is D to F. [VII. 13]

+

And, since A is plane, and C, D are its sides, therefore D by multiplying C has made A.

+

For the same reason also E by multiplying F has made B.

+

Now let D by multiplying E make G.

+

Then, since D by multiplying C has made A, and by multiplying E has made G, therefore, as C is to E, so is A to G. [VII. 17]

+

But, as C is to E, so is D to F; therefore also, as D is to F, so is A to G.

+

Again, since E by multiplying D has made G, and by multiplying F has made B, therefore, as D is to F, so is G to B. [VII. 17]

+

But it was also proved that, as D is to F, so is A to G; therefore also, as A is to G, so is G to B.

+

Therefore A, G, B are in continued proportion.

+

Therefore between A, B there is one mean proportional number.

+

I say next that A also has to B the ratio duplicate of that which the corresponding side has to the corresponding side, that is, of that which C has to E or D to F.

+

For, since A, G, B are in continued proportion, A has to B the ratio duplicate of that which it has to G. [V. Def. 9]

+

And, as A is to G, so is C to E, and so is D to F.

+

Therefore A also has to B the ratio duplicate of that which C has to E or D to F. Q. E. D.

+
+ +
+ PROPOSITION 19. +

+ Between two similar solid numbers there fall two mean proportional numbers; and the solid number has to the similar solid number the ratio triplicate of that which the corresponding side has to the corresponding side. +

+

Let A, B be two similar solid numbers, and let C, D, E be the sides of A, and F, G, H of B.

+

Now, since similar solid numbers are those which have their sides proportional, [VII. Def. 21] therefore, as C is to D, so is F to G, and, as D is to E, so is G to H. +

+

I say that between A, B there fall two mean proportional numbers, and A has to B the ratio triplicate of that which C has to F, D to G, and also E to H.

+

+

For let C by multiplying D make K, and let F by multiplying G make L.

+

Now, since C, D are in the same ratio with F, G, and K is the product of C, D, and L the product of F, G, K, L are similar plane numbers; [VII. Def. 21] therefore between K, L there is one mean proportional number. [VIII. 18]

+

Let it be M +

+

Therefore M is the product of D, F, as was proved in the theorem preceding this. [VIII. 18]

+

Now, since D by multiplying C has made K, and by multiplying F has made M, therefore, as C is to F, so is K to M. [VII. 17]

+

But, as K is to M, so is M to L.

+

Therefore K, M, L are continuously proportional in the ratio of C to F.

+

And since, as C is to D, so is F to G, alternately therefore, as C is to F, so is D to G. [VII. 13]

+

For the same reason also, as D is to G, so is E to H. +

+

Therefore K, M, L are continuously proportional in the ratio of C to F, in the ratio of D to G, and also in the ratio of E to H.

+

Next, let E, H by multiplying M make N, O respectively.

+

Now, since A is a solid number, and C, D, E are its sides, therefore E by multiplying the product of C, D has made A.

+

But the product of C, D is K; therefore E by multiplying K has made A.

+

For the same reason also H by multiplying L has made B. +

+

Now, since E by multiplying K has made A, and further also by multiplying M has made N, therefore, as K is to M, so is A to N. [VII. 17]

+

But, as K is to M, so is C to F, D to G, and also E to H; therefore also, as C is to F, D to G, and E to H, so is A to N.

+

Again, since E, H by multiplying M have made N, O respectively, therefore, as E is to H, so is N to O. [VII. 18]

+

But, as E is to H, so is C to F and D to G; therefore also, as C is to F, D to G, and E to H, so is A to N and N to O.

+

Again, since H by multiplying M has made O, and further also by multiplying L has made B, therefore, as M is to L, so is O to B. [VII. 17]

+

But, as M is to L, so is C to F, D to G, and E to H.

+

Therefore also, as C is to F, D to G, and E to H, so not only is O to B, but also A to N and N to O.

+

Therefore A, N, O, B are continuously proportional in the aforesaid ratios of the sides.

+

I say that A also has to B the ratio triplicate of that which the corresponding side has to the corresponding side, that is, of the ratio which the number C has to F, or D to G, and also E to H.

+

For, since A, N, O, B are four numbers in continued proportion, therefore A has to B the ratio triplicate of that which A has to N. [V. Def. 10]

+

But, as A is to N, so it was proved that C is to F, D to G, and also E to H.

+

Therefore A also has to B the ratio triplicate of that which the corresponding side has to the corresponding side, that is, of the ratio which the number C has to F, D to G, and also E to H. Q. E. D.

+
+
+ PROPOSITION 20. +

+ If one mean proportional number fall between two numbers, the numbers will be similar plane numbers. +

+

For let one mean proportional number C fall between the two numbers A, B; I say that A, B are similar plane numbers.

+

Let D, E, the least numbers of those which have the same ratio with A, C, be taken; [VII. 33] therefore D measures A the same number of times that E measures C. [VII. 20]

+

Now, as many times as D measures A, so many units let there be in F; therefore F by multiplying D has made A, so that A is plane, and D, F are its sides.

+

Again, since D, E are the least of the numbers which have the same ratio with C, B, therefore D measures C the same number of times that E measures B. [VII. 20]

+

+

As many times, then, as E measures B, so many units let there be in G; therefore E measures B according to the units in G; therefore G by multiplying E has made B.

+

Therefore B is plane, and E, G are its sides.

+

Therefore A, B are plane numbers.

+

I say next that they are also similar.

+

For, † since F by multiplying D has made A, and by multiplying E has made C, therefore, as D is to E, so is A to C, that is, C to B. [VII. 17]

+

Again, † since E by multiplying F, G has made C, B respectively, therefore, as F is to G, so is C to B. [VII. 17]

+

But, as C is to B, so is D to E; therefore also, as D is to E, so is F to G.

+

And alternately, as D is to F, so is E to G. [VII. 13]

+

Therefore A, B are similar plane numbers; for their sides are proportional. Q. E. D. + +

25. For, since F......27. C to B. The text has clearly suffered corruption here. It is not necessary to infer from other facts that, as D is to E, so is A to C; for this is part of the hypotheses (ll. 6, 7). Again, there is no explanation of the statement (l. 25) that F by multiplying E has made C. It is the statement and explanation of this latter fact which are alone wanted; after which the proof proceeds as in l. 28. We might therefore substitute for ll. 25-28 the following.

+

“For, since E measures C the same number of times that D measures A [l. 8], that is, according to the units in F [l. 10], therefore F by multiplying E has made C.

+

And, since E by multiplying F, G,” + etc. etc.

+

+
+
+ PROPOSITION 21. +

+ If two mean proportional numbers fall between two numbers, the numbers are similar solid numbers. +

+

For let two mean proportional numbers C, D fall between the two numbers A, B; I say that A, B are similar solid numbers.

+

+

For let three numbers E, F, G, the least of those which have the same ratio with A, C, D, be taken; [VII. 33 or VIII. 2] therefore the extremes of them E, G are prime to one another. [VIII. 3]

+

Now, since one mean proportional number F has fallen between E, G, therefore E, G are similar plane numbers. [VIII. 20]

+

Let, then, H, K be the sides of E, and L, M of G.

+

Therefore it is manifest from the theorem before this that E, F, G are continuously proportional in the ratio of H to L and that of K to M.

+

Now, since E, F, G are the least of the numbers which have the same ratio with A, C, D, and the multitude of the numbers E, F, G is equal to the multitude of the numbers A, C, D, therefore, ex aequali, as E is to G, so is A to D. [VII. 14]

+

But E, G are prime, primes are also least, [VII. 21] and the least measure those which have the same ratio with them the same number of times, the greater the greater and the less the less, that is, the antecedent the antecedent and the consequent the consequent; [VII. 20] therefore E measures A the same number of times that G measures D.

+

Now, as many times as E measures A, so many units let there be in N.

+

Therefore N by multiplying E has made A.

+

But E is the product of H, K; therefore N by multiplying the product of H, K has made A.

+

Therefore A is solid, and H, K, N are its sides.

+

Again, since E, F, G are the least of the numbers which have the same ratio as C, D, B, therefore E measures C the same number of times that G measures B.

+

Now, as many times as E measures C, so many units let there be in O.

+

Therefore G measures B according to the units in O; therefore O by multiplying G has made B.

+

But G is the product of L, M; therefore O by multiplying the product of L, M has made B.

+

Therefore B is solid, and L, M, O are its sides; therefore A, B are solid.

+

I say that they are also similar.

+

For since N, O by multiplying E have made A, C, therefore, as N is to O, so is A to C, that is, E to F. [VII. 18]

+

But, as E is to F, so is H to L and K to M; therefore also, as H is to L, so is K to M and N to O.

+

And H, K, N are the sides of A, and O, L, M the sides of B.

+

Therefore A, B are similar solid numbers. Q. E. D.

+
+
+ PROPOSITION 22. +

+ If three numbers be in continued proportion, and the first be square, the third will also be square. +

+

Let A, B, C be three numbers in continued proportion, and let A the first be square; I say that C the third is also square.

+

+

For, since between A, C there is one mean proportional number, B, therefore A, C are similar plane numbers. [VIII. 20]

+

But A is square; therefore C is also square. Q. E. D.

+
+
+ PROPOSITION 23. +

+ If four numbers be in continued proportion, and the first be cube, the fourth will also be cube. +

+

Let A, B, C, D be four numbers in continued proportion, and let A be cube; I say that D is also cube.

+

+

For, since between A, D there are two mean proportional numbers B, C, therefore A, D are similar solid numbers. [VIII. 21]

+

But A is cube; therefore D is also cube. Q. E. D.

+
+
+ PROPOSITION 24. +

+ If two numbers have to one another the ratio which a square number has to a square number, and the first be square, the second will also be square. +

+

For let the two numbers A, B have to one another the ratio which the square number C has to the square number D, and let A be square; I say that B is also square.

+

+

For, since C, D are square, C, D are similar plane numbers.

+

Therefore one mean proportional number falls between C, D. [VIII. 18]

+

And, as C is to D, so is A to B; therefore one mean proportional number falls between A, B also. [VIII. 8]

+

And A is square; therefore B is also square. [VIII. 22] Q. E. D.

+
+
+ PROPOSITION 25. +

+ If two numbers have to one another the ratio which a cube number has to a cube number, and the first be cube, the second will also be cube. +

+

For let the two numbers A, B have to one another the ratio which the cube number C has to the cube number D, and let A be cube; I say that B is also cube.

+

For, since C, D are cube, C, D are similar solid numbers.

+

Therefore two mean proportional numbers fall between C, D. [VIII. 19]

+

+

And, as many numbers as fall between C, D in continued proportion, so many will also fall between those which have the same ratio with them; [VIII. 8] so that two mean proportional numbers fall between A, B also.

+

Let E, F so fall.

+

Since, then, the four numbers A, E, F, B are in continued proportion, and A is cube, therefore B is also cube. [VIII. 23] Q. E. D.

+
+
+ PROPOSITION 26. +

+ Similar plane numbers have to one another the ratio which a square number has to a square number. +

+

Let A, B be similar plane numbers; I say that A has to B the ratio which a square number has to a square number.

+

+

For, since A, B are similar plane numbers, therefore one mean proportional number falls between A, B. [VIII. 18]

+

Let it so fall, and let it be C; and let D, E, F, the least numbers of those which have the same ratio with A, C, B, be taken; [VII. 33 or VIII. 2] therefore the extremes of them D, F are square. [VIII. 2, Por.]

+

And since, as D is to F, so is A to B, and D, F are square, therefore A has to B the ratio which a square number has to a square number. Q. E. D.

+
+
+ PROPOSITION 27. +

+ Similar solid numbers have to one another the ratio which a cube number has to a cube number. +

+

Let A, B be similar solid numbers; I say that A has to B the ratio which a cube number has to a cube number.

+

+

For, since A, B are similar solid numbers, therefore two mean proportional numbers fall between A, B. [VIII. 19]

+

Let C, D so fall, and let E, F, G, H, the least numbers of those which have the same ratio with A, C, D, B, and equal with them in multitude, be taken; [VII. 33 or VIII. 2] therefore the extremes of them E, H are cube. [VIII. 2, Por.]

+

And, as E is to H, so is A to B; therefore A also has to B the ratio which a cube number has to a cube number. Q. E. D.

+
+
+
+
+ + BOOK IX. +
+ PROPOSITIONS +
+ PROPOSITION 1. +

+ If two similar plane numbers by multiplying one another make some number, the product will be square. +

+

Let A, B be two similar plane numbers, and let A by multiplying B make C; I say that C is square.

+

For let A by multiplying itself make D.

+

+

Therefore D is square.

+

Since then A by multiplying itself has made D, and by multiplying B has made C, therefore, as A is to B, so is D to C. [VII. 17]

+

And, since A, B are similar plane numbers, therefore one mean proportional number falls between A, B. [VIII. 18]

+

But, if numbers fall between two numbers in continued proportion, as many as fall between them, so many also fall between those which have the same ratio; [VIII. 8] so that one mean proportional number falls between D, C also.

+

And D is square; therefore C is also square. [VIII. 22] Q. E. D.

+
+ +
+ PROPOSITION 2. +

+ If two numbers by multiplying one another make a square number, they are similar plane numbers. +

+

Let A, B be two numbers, and let A by multiplying B make the square number C; I say that A, B are similar plane numbers.

+

+

For let A by multiplying itself make D; therefore D is square.

+

Now, since A by multiplying itself has made D, and by multiplying B has made C, therefore, as A is to B, so is D to C. [VII. 17]

+

And, since D is square, and C is so also, therefore D, C are similar plane numbers.

+

Therefore one mean proportional number falls between D, C. [VIII. 18]

+

And, as D is to C, so is A to B; therefore one mean proportional number falls between A, B also. [VIII. 8]

+

But, if one mean proportional number fall between two numbers, they are similar plane numbers; [VIII. 20] therefore A, B are similar plane numbers. Q. E. D.

+
+
+ PROPOSITION 3. +

+ If a cube number by multiplying itself make some number, the product will be cube. +

+

For let the cube number A by multiplying itself make B; I say that B is cube.

+

For let C, the side of A, be taken, and let C by multiplying itself make D.

+

It is then manifest that C by multiplying D has made A.

+

+

Now, since C by multiplying itself has made D, therefore C measures D according to the units in itself.

+

But further the unit also measures C according to the units in it; therefore, as the unit is to C, so is C to D. [VII. Def. 20]

+

Again, since C by multiplying D has made A, therefore D measures A according to the units in C.

+

But the unit also measures C according to the units in it; therefore, as the unit is to C, so is D to A.

+

But, as the unit is to C, so is C to D; therefore also, as the unit is to C, so is C to D, and D to A.

+

Therefore between the unit and the number A two mean proportional numbers C, D have fallen in continued proportion.

+

Again, since A by multiplying itself has made B, therefore A measures B according to the units in itself.

+

But the unit also measures A according to the units in it; therefore, as the unit is to A, so is A to B. [VII. Def. 20]

+

But between the unit and A two mean proportional numbers have fallen; therefore two mean proportional numbers will also fall between A, B. [VIII. 8]

+

But, if two mean proportional numbers fall between two numbers, and the first be cube, the second will also be cube. [VIII. 23]

+

And A is cube; therefore B is also cube. Q. E. D.

+
+ +
+ PROPOSITION 4. +

+ If a cube number by multiplying a cube number make some number, the product will be cube. +

+

For let the cube number A by multiplying the cube number B make C; I say that C is cube.

+

+

For let A by multiplying itself make D; therefore D is cube. [IX. 3]

+

And, since A by multiplying itself has made D, and by multiplying B has made C therefore, as A is to B, so is D to C. [VII. 17]

+

And, since A, B are cube numbers, A, B are similar solid numbers.

+

Therefore two mean proportional numbers fall between A, B; [VIII. 19] so that two mean proportional numbers will fall between D, C also. [VIII. 8]

+

And D is cube; therefore C is also cube [VIII. 23] Q. E. D.

+
+
+ PROPOSITION 5. +

+ If a cube number by multiplying any number make a cube number, the multiplied number will also be cube. +

+

For let the cube number A by multiplying any number B make the cube number C; I say that B is cube.

+

+

For let A by multiplying itself make D; therefore D is cube. [IX. 3]

+

Now, since A by multiplying itself has made D, and by multiplying B has made C, therefore, as A is to B, so is D to C. [VII. 17]

+

And since D, C are cube, they are similar solid numbers.

+

Therefore two mean proportional numbers fall between D, C. [VIII. 19]

+

And, as D is to C, so is A to B; therefore two mean proportional numbers fall between A, B also. [VIII. 8]

+

And A is cube; therefore B is also cube. [VIII. 23]

+
+
+ PROPOSITION 6. +

+ If a number by multiplying itself make a cube number, it will itself also be cube. +

+

For let the number A by multiplying itself make the cube number B; I say that A is also cube.

+

+

For let A by multiplying B make C.

+

Since, then, A by multiplying itself has made B, and by multiplying B has made C, therefore C is cube.

+

And, since A by multiplying itself has made B, therefore A measures B according to the units in itself.

+

But the unit also measures A according to the units in it.

+

Therefore, as the unit is to A, so is A to B. [VII. Def. 20]

+

And, since A by multiplying B has made C, therefore B measures C according to the units in A.

+

But the unit also measures A according to the units in it.

+

Therefore, as the unit is to A, so is B to C. [VII. Def. 20]

+

But, as the unit is to A, so is A to B; therefore also, as A is to B, so is B to C.

+

And, since B, C are cube, they are similar solid numbers.

+

Therefore there are two mean proportional numbers between B, C. [VIII. 19]

+

And, as B is to C, so is A to B.

+

Therefore there are two mean proportional numbers between A, B also. [VIII. 8]

+

And B is cube; therefore A is also cube. [cf. VIII. 23] Q. E. D.

+
+
+ PROPOSITION 7. +

+ If a composite number by multiplying any number make some number, the product will be solid. +

+

For let the composite number A by multiplying any number B make C; I say that C is solid.

+

+

For, since A is composite, it will be measured by some number. [VII. Def. 13]

+

Let it be measured by D; and, as many times as D measures A, so many units let there be in E.

+

Since then D measures A according to the units in E, therefore E by multiplying D has made A. [VII. Def. 15]

+

And, since A by multiplying B has made C, and A is the product of D, E, therefore the product of D, E by multiplying B has made C.

+

Therefore C is solid, and D, E, B are its sides. Q. E. D.

+
+
+ PROPOSITION 8. +

+ If as many numbers as we please beginning from an unit be in continued proportion, the third from the unit will be square, as will also those which successively leave out one; the fourth will be cube, as will also all those which leave out two; and the seventh will be at once cube and square, as will also those which leave out five. +

+

Let there be as many numbers as we please, A, B, C, D, E, F, beginning from an unit and in continued proportion; I say that B, the third from the unit, is square, as are also all those which leave out one; C, the fourth, is cube, as are also all those which leave out two; and F, the seventh, is at once cube and square, as are also all those which leave out five.

+

+

For since, as the unit is to A, so is A to B, therefore the unit measures the number A the same number of times that A measures B. [VII. Def. 20]

+

But the unit measures the number A according to the units in it; therefore A also measures B according to the units in A.

+

Therefore A by multiplying itself has made B; therefore B is square.

+

And, since B, C, D are in continued proportion, and B is square, therefore D is also square. [VIII. 22]

+

For the same reason F is also square.

+

Similarly we can prove that all those which leave out one are square.

+

I say next that C, the fourth from the unit, is cube, as are also all those which leave out two.

+

For since, as the unit is to A, so is B to C, therefore the unit measures the number A the same number of times that B measures C.

+

But the unit measures the number A according to the units in A; therefore B also measures C according to the units in A.

+

Therefore A by multiplying B has made C.

+

Since then A by multiplying itself has made B, and by multiplying B has made C, therefore C is cube.

+

And, since C, D, E, F are in continued proportion, and C is cube, therefore F is also cube. [VIII. 23]

+

But it was also proved square; therefore the seventh from the unit is both cube and square.

+

Similarly we can prove that all the numbers which leave out five are also both cube and square. Q. E. D.

+
+
+ PROPOSITION 9. +

+ If as many numbers as we please beginning from an unit be in continued proportion, and the number after the unit be square, all the rest will also be square. And, if the number after the unit be cube, all the rest will also be cube. +

+

Let there be as many numbers as we please, A, B, C, D, E, F, beginning from an unit and in continued proportion, and let A, the number after the unit, be square; I say that all the rest will also be square.

+

+

Now it has been proved that B, the third from the unit, is square, as are also all those which leave out one; [IX. 8] I say that all the rest are also square.

+

For, since A, B, C are in continued proportion, and A is square, therefore C is also square. [VIII. 22]

+

Again, since B, C, D are in continued proportion, and B is square, D is also square. [VIII. 22]

+

Similarly we can prove that all the rest are also square.

+

Next, let A be cube; I say that all the rest are also cube.

+

Now it has been proved that C, the fourth from the unit, is cube, as also are all those which leave out two; [IX. 8] I say that all the rest are also cube.

+

For, since, as the unit is to A, so is A to B, therefore the unit measures A the same number of times as A measures B.

+

But the unit measures A according to the units in it; therefore A also measures B according to the units in itself; therefore A by multiplying itself has made B.

+

And A is cube.

+

But, if a cube number by multiplying itself make some number, the product is cube. [IX. 3]

+

Therefore B is also cube.

+

And, since the four numbers A, B, C, D are in continued proportion, and A is cube, D also is cube. [VIII. 23]

+

For the same reason E is also cube, and similarly all the rest are cube. Q. E. D.

+
+
+ PROPOSITION 10. +

+ If as many numbers as we please beginning from an unit be in continued proportion, and the number after the unit be not square, neither will any other be square except the third from the unit and all those which leave out one. And, if the number after the unit be not cube, neither will any other be cube except the fourth from the unit and all those which leave out two. +

+

Let there be as many numbers as we please, A, B, C, D, E, F, beginning from an unit and in continued proportion, and let A, the number after the unit, not be square; I say that neither will any other be square except the third from the unit ltand those which leave out onegt.

+

+

For, if possible, let C be square.

+

But B is also square; [IX. 8] [therefore B, C have to one another the ratio which a square number has to a square number].

+

And, as B is to C, so is A to B; therefore A, B have to one another the ratio which a square number has to a square number; [so that A, B are similar plane numbers]. [VIII. 26, converse]

+

And B is square; therefore A is also square: which is contrary to the hypothesis.

+

Therefore C is not square.

+

Similarly we can prove that neither is any other of the numbers square except the third from the unit and those which leave out one.

+

Next, let A not be cube.

+

I say that neither will any other be cube except the fourth from the unit and those which leave out two.

+

For, if possible, let D be cube.

+

Now C is also cube; for it is fourth from the unit. [IX. 8]

+

And, as C is to D, so is B to C; therefore B also has to C the ratio which a cube has to a cube.

+

And C is cube; therefore B is also cube. [VIII. 25]

+

And since, as the unit is to A, so is A to B, and the unit measures A according to the units in it, therefore A also measures B according to the units in itself; therefore A by multiplying itself has made the cube number B.

+

But, if a number by multiplying itself make a cube number, it is also itself cube. [IX. 6]

+

Therefore A is also cube: which is contrary to the hypothesis.

+

Therefore D is not cube.

+

Similarly we can prove that neither is any other of the numbers cube except the fourth from the unit and those which leave out two. Q. E. D.

+
+
+ PROPOSITION II. +

+ If as many numbers as we please beginning from an unit be in continued proportion, the less measures the greater according to some one of the numbers which have place among the proportional numbers. + +

+

Let there be as many numbers as we please, B, C, D, E, beginning from the unit A and in continued proportion; I say that B, the least of the numbers B, C, D, E, measures E according to some one of the numbers C, D.

+

+

For since, as the unit A is to B, so is D to E, therefore the unit A measures the number B the same number of times as D measures E; therefore, alternately, the unit A measures D the same number of times as B measures E. [VII. 15]

+

But the unit A measures D according to the units in it; therefore B also measures E according to the units in D; so that B the less measures E the greater according to some number of those which have place among the proportional numbers.—

+ +

Porism. And it is manifest that, whatever place the measuring number has, reckoned from the unit, the same place also has the number according to which it measures, reckoned from the number measured, in the direction of the number before it.—

+ +

Q. E. D.

+
+ +
+ PROPOSITION 12. +

+ If as many numbers as we please beginning from an unit be in continued proportion, by however many prime numbers the last is measured, the next to the unit will also be measured by the same. +

+

Let there be as many numbers as we please, A, B, C, D, beginning from an unit, and in continued proportion; I say that, by however many prime numbers D is measured, A will also be measured by the same.

+

+

For let D be measured by any prime number E; I say that E measures A.

+

For suppose it does not; now E is prime, and any prime number is prime to any which it does not measure; [VII. 29] therefore E, A are prime to one another.

+

And, since E measures D, let it measure it according to F, therefore E by multiplying F has made D.

+

Again, since A measures D according to the units in C, [IX. 11 and Por.] therefore A by multiplying C has made D.

+

But, further, E has also by multiplying F made D; therefore the product of A, C is equal to the product of E, F.

+

Therefore, as A is to E, so is F to C. [VII. 19]

+

But A, E are prime, primes are also least, [VII. 21] and the least measure those which have the same ratio the same number of times, the antecedent the antecedent and the consequent the consequent; [VII. 20] therefore E measures C.

+

Let it measure it according to G; therefore E by multiplying G has made C.

+

But, further, by the theorem before this, A has also by multiplying B made C. [IX. 11 and Por.]

+

Therefore the product of A, B is equal to the product of E, G.

+

Therefore, as A is to E, so is G to B. [VII. 19]

+

But A, E are prime, primes are also least, [VII. 21] and the least numbers measure those which have the same ratio with them the same number of times, the antecedent the antecedent and the consequent the consequent: [VII. 20] therefore E measures B.

+

Let it measure it according to H; therefore E by multiplying H has made B.

+

But further A has also by multiplying itself made B; [IX. 8] therefore the product of E, H is equal to the square on A.

+

Therefore, as E is to A, so is A to H. [VII. 19]

+

But A, E are prime, primes are also least, [VII. 21] and the least measure those which have the same ratio the same number of times, the antecedent the antecedent and the consequent the consequent; [VII. 20] therefore E measures A, as antecedent antecedent.

+

But, again, it also does not measure it: which is impossible.

+

Therefore E, A are not prime to one another.

+

Therefore they are composite to one another.

+

But numbers composite to one another are measured by some number. [VII. Def. 14]

+

And, since E is by hypothesis prime, and the prime is not measured by any number other than itself, therefore E measures A, E, so that E measures A.

+

[But it also measures D; therefore E measures A, D.]

+

Similarly we can prove that, by however many prime numbers D is measured, A will also be measured by the same. Q. E. D.

+
+
+ PROPOSITION 13. +

If as many numbers as we please beginning from an unit be in continued proportion, and the number after the unit be prime, the greatest will not be measured by any except those which have a place among the proportional numbers.

+

Let there be as many numbers as we please, A, B, C, D, beginning from an unit and in continued proportion, and let A, the number after the unit, be prime; I say that D, the greatest of them, will not be measured by any other number except A, B, C.

+

+

For, if possible, let it be measured by E, and let E not be the same with any of the numbers A, B, C.

+

It is then manifest that E is not prime.

+

For, if E is prime and measures D, it will also measure A [IX. 12], which is prime, though it is not the same with it: which is impossible.

+

Therefore E is not prime.

+

Therefore it is composite.

+

But any composite number is measured by some prime number; [VII. 31] therefore E is measured by some prime number.

+

I say next that it will not be measured by any other prime except A.

+

For, if E is measured by another, and E measures D, that other will also measure D; so that it will also measure A [IX. 12], which is prime, though it is not the same with it: which is impossible.

+

Therefore A measures E.

+

And, since E measures D, let it measure it according to F.

+

I say that F is not the same with any of the numbers A, B, C.

+

For, if F is the same with one of the numbers A, B, C, and measures D according to E, therefore one of the numbers A, B, C also measures D according to E.

+

But one of the numbers A, B, C measures D according to some one of the numbers A, B, C; [IX. 11] therefore E is also the same with one of the numbers A, B, C: which is contrary to the hypothesis.

+

Therefore F is not the same as any one of the numbers A, B, C.

+

Similarly we can prove that F is measured by A, by proving again that F is not prime.

+

For, if it is, and measures D, it will also measure A [IX. 12], which is prime, though it is not the same with it: which is impossible; therefore F is not prime.

+

Therefore it is composite.

+

But any composite number is measured by some prime number; [VII. 31] therefore F is measured by some prime number.

+

I say next that it will not be measured by any other prime except A.

+

For, if any other prime number measures F, and F measures D, that other will also measure D; so that it will also measure A [IX. 12], which is prime, though it is not the same with it: which is impossible.

+

Therefore A measures F.

+

And, since E measures D according to F, therefore E by multiplying F has made D.

+

But, further, A has also by multiplying C made D; [IX. 11] therefore the product of A, C is equal to the product of E, F.

+

Therefore, proportionally, as A is to E, so is F to C. [VII. 19]

+

But A measures E; therefore F also measures C.

+

Let it measure it according to G.

+

Similarly, then, we can prove that G is not the same with any of the numbers A, B, and that it is measured by A.

+

And, since F measures C according to G therefore F by multiplying G has made C.

+

But, further, A has also by multiplying B made C; [IX. 11] therefore the product of A, B is equal to the product of F, G.

+

Therefore, proportionally, as A is to F, so is G to B. [VII. 19]

+

But A measures F; therefore G also measures B.

+

Let it measure it according to H.

+

Similarly then we can prove that H is not the same with A.

+

And, since G measures B according to H, therefore G by multiplying H has made B.

+

But further A has also by multiplying itself made B; [IX. 8] therefore the product of H, G is equal to the square on A.

+

Therefore, as H is to A, so is A to G. [VII. 19]

+

But A measures G;lt therefore H also measures A, which is prime, though it is not the same with it: which is absurd.

+

Therefore D the greatest will not be measured by any other number except A, B, C. Q. E. D.

+
+
+ PROPOSITION 14. +

+ If a number be the least that is measured by prime numbers, it will not be measured by any other prime number except those originally measuring it. +

+

For let the number A be the least that is measured by the prime numbers B, C, D; I say that A will not be measured by any other prime number except B, C, D.

+

For, if possible, let it be measured by the prime number E, and let E not be the same with any one of the numbers B, C, D.

+

+

Now, since E measures A, let it measure it according to F; therefore E by multiplying F has made A. +

+

And A is measured by the prime numbers B, C, D.

+

But, if two numbers by multiplying one another make some number, and any prime number measure the product, it will also measure one of the original numbers; [VII. 30] therefore B, C, D will measure one of the numbers E, F.

+

Now they will not measure E; for E is prime and not the same with any one of the numbers B, C, D.

+

Therefore they will measure F, which is less than A: which is impossible, for A is by hypothesis the least number measured by B, C, D.

+

Therefore no prime number will measure A except B, C, D. Q. E. D.

+
+ +
+ PROPOSITION 15. +

+ If three numbers in continued proportion be the least of those which have the same ratio with them, any two whatever added together will be prime to the remaining number. +

+

Let A, B, C, three numbers in continued proportion, be the least of those which have the same ratio with them; I say that any two of the numbers A, B, C whatever added together are prime to the remaining number, namely A, B to C; B, C to A; and further A, C to B.

+

+

For let two numbers DE, EF, the least of those which have the same ratio with A, B, C, be taken. [VIII. 2]

+

It is then manifest that DE by multiplying itself has made A, and by multiplying EF has made B, and, further, EF by multiplying itself has made C. [VIII. 2]

+

Now, since DE, EF are least, they are prime to one another. [VII. 22]

+

But, if two numbers be prime to one another, their sum is also prime to each; [VII. 28] therefore DF is also prime to each of the numbers DE, EF.

+

But further DE is also prime to EF; therefore DF, DE are prime to EF.

+

But, if two numbers be prime to any number, their product is also prime to the other; [VII. 24] so that the product of FD, DE is prime to EF; hence the product of FD, DE is also prime to the square on EF. [VII. 25]

+

But the product of FD, DE is the square on DE together with the product of DE, EF; [II. 3] therefore the square on DE together with the product of DE, EF is prime to the square on EF.

+

And the square on DE is A, the product of DE, EF is B, and the square on EF is C; therefore A, B added together are prime to C.

+

Similarly we can prove that B, C added together are prime to A.

+

I say next that A, C added together are also prime to B.

+

For, since DF is prime to each of the numbers DE, EF, the square on DF is also prime to the product of DE, EF. [VII. 24, 25]

+

But the squares on DE, EF together with twice the product of DE, EF are equal to the square on DF; [II. 4] therefore the squares on DE, EF together with twice the product of DE, EF are prime to the product of DE, EF.

+

Separando, the squares on DE, EF together with once the product of DE, EF are prime to the product of DE, EF.

+

Therefore, separando again, the squares on DE, EF are prime to the product of DE, EF.

+

And the square on DE is A, the product of DE, EF is B, and the square on EF is C.

+

Therefore A, C added together are prime to B. Q. E. D.

+
+ +
+ PROPOSITION 16. +

+ If two numbers be prime to one another, the second will not be to any other number as the first is to the second. +

+

For let the two numbers A, B be prime to one another; I say that B is not to any other number as A is to B.

+

+

For, if possible, as A is to B, so let B be to C.

+

Now A, B are prime, primes are also least, [VII. 21] and the least numbers measure those which have the same ratio the same number of times, the antecedent the antecedent and the consequent the consequent; [VII. 20] therefore A measures B as antecedent antecedent.

+

But it also measures itself; therefore A measures A, B which are prime to one another: which is absurd.

+

Therefore B will not be to C, as A is to B. Q. E. D.

+
+
+ PROPOSITION 17. +

+ If there be as many numbers as we please in continued proportion, and the extremes of them be prime to one another, the last will not be to any other number as the first to the second. +

+

For let there be as many numbers as we please, A, B, C, D, in continued proportion, and let the extremes of them, A, D, be prime to one another; I say that D is not to any other number as A is to B.

+

+

For, if possible, as A is to B, so let D be to E; therefore, alternately, as A is to D, so is B to E. [VII. 13]

+

But A, D are prime, primes are also least, [VII. 21] and the least numbers measure those which have the same ratio the same number of times, the antecedent the antecedent and the consequent the consequent. [VII. 20]

+

Therefore A measures B.

+

And, as A is to B, so is B to C.

+

Therefore B also measures C; so that A also measures C.

+

And since, as B is to C, so is C to D, and B measures C, therefore C also measures D.

+

But A measured C; so that A also measures D.

+

But it also measures itself; therefore A measures A, D which are prime to one another : which is impossible.

+

Therefore D will not be to any other number as A is to B. Q. E. D.

+
+
+ PROPOSITION 18. +

+ Given two numbers, to investigate whether it is possible to find a third proportional to them. +

+

Let A, B be the given two numbers, and let it be required to investigate whether it is possible to find a third proportional to them.

+

Now A, B are either prime to one another or not.

+

And, if they are prime to one another, it has been proved that it is impossible to find a third proportional to them. [IX. 16]

+

Next, let A, B not be prime to one another, and let B by multiplying itself make C.

+

Then A either measures C or does not measure it.

+

+

First, let it measure it according to D; therefore A by multiplying D has made C.

+

But, further, B has also by multiplying itself made C; therefore the product of A, D is equal to the square on B.

+

Therefore, as A is to B, so is B to D; [VII. 19] therefore a third proportional number D has been found to A, B.

+

Next, let A not measure C; I say that it is impossible to find a third proportional number to A, B.

+

For, if possible, let D, such third proportional, have been found.

+

Therefore the product of A, D is equal to the square on B.

+

But the square on B is C; therefore the product of A, D is equal to C.

+

Hence A by multiplying D has made C; therefore A measures C according to D.

+

But, by hypothesis, it also does not measure it: which is absurd.

+

Therefore it is not possible to find a third proportional number to A, B when A does not measure C. Q. E. D.

+
+ +
+ PROPOSITION 19. +

+ Given three numbers, to investigate when it is possible to find a fourth proportional to them. +

+

Let A, B, C be the given three numbers, and let it be required to investigate when it is possible to find a fourth proportional to them.

+

+

Now either they are not in continued proportion, and the extremes of them are prime to one another; or they are in continued proportion, and the extremes of them are not prime to one another; or they are not in continued proportion, nor are the extremes of them prime to one another; or they are in continued proportion, and the extremes of them are prime to one another.

+

If then A, B, C are in continued proportion, and the extremes of them A, C are prime to one another, it has been proved that it is impossible to find a fourth proportional number to them. [IX. 17]

+

†Next, let A, B, C not be in continued proportion, the extremes being again prime to one another; I say that in this case also it is impossible to find a fourth proportional to them.

+

For, if possible, let D have been found, so that, as A is to B, so is C to D, and let it be contrived that, as B is to C, so is D to E.

+

Now, since, as A is to B, so is C to D, and, as B is to C, so is D to E, therefore, ex aequali, as A is to C, so is C to E. [VII. 14]

+

But A, C are prime, primes are also least, [VII. 21] and the least numbers measure those which have the same ratio, the antecedent the antecedent and the consequent the consequent. [VII. 20]

+

Therefore A measures C as antecedent antecedent.

+

But it also measures itself; therefore A measures A, C which are prime to one another: which is impossible.

+

Therefore it is not possible to find a fourth proportional to A, B, C.†

+

Next, let A, B, C be again in continued proportion, but let A, C not be prime to one another.

+

I say that it is possible to find a fourth proportional to them.

+

For let B by multiplying C make D; therefore A either measures D or does not measure it.

+

First, let it measure it according to E; therefore A by multiplying E has made D.

+

But, further, B has also by multiplying C made D; therefore the product of A, E is equal to the product of B, C; therefore, proportionally, as A is to B, so is C to E; [VII. 19] therefore E has been found a fourth proportional to A, B, C.

+

Next, let A not measure D; I say that it is impossible to find a fourth proportional number to A, B, C.

+

For, if possible, let E have been found; therefore the product of A, E is equal to the product of B, C. [VII. 19]

+

But the product of B, C is D; therefore the product of A, E is also equal to D.

+

Therefore A by multiplying E has made D; therefore A measures D according to E, so that A measures D.

+

But it also does not measure it: which is absurd.

+

Therefore it is not possible to find a fourth proportional number to A, B, C when A does not measure D.

+

Next, let A, B, C not be in continued proportion, nor the extremes prime to one another.

+

And let B by multiplying C make D.

+

Similarly then it can be proved that, if A measures D, it is possible to find a fourth proportional to them, but, if it does not measure it, impossible. Q. E. D.

+
+
+ PROPOSITION 20. +

+ Prime numbers are more than any assigned multitude of prime numbers. +

+

Let A, B, C be the assigned prime numbers; I say that there are more prime numbers than A, B, C.

+

+

For let the least number measured by A, B, C be taken, and let it be DE; let the unit DF be added to DE.

+

Then EF is either prime or not.

+

First, let it be prime; then the prime numbers A, B, C, EF have been found which are more than A, B, C.

+

Next, let EF not be prime; therefore it is measured by some prime number. [VII. 31]

+

Let it be measured by the prime number G.

+

I say that G is not the same with any of the numbers A, B, C.

+

For, if possible, let it be so.

+

Now A, B, C measure DE; therefore G also will measure DE.

+

But it also measures EF.

+

Therefore G, being a number, will measure the remainder, the unit DF: which is absurd.

+

Therefore G is not the same with any one of the numbers A, B, C.

+

And by hypothesis it is prime.

+

Therefore the prime numbers A, B, C, G have been found which are more than the assigned multitude of A, B, C. Q. E. D.

+
+
+ PROPOSITION 21. +

+ If as many even numbers as we please be added together, the whole is even. +

+

For let as many even numbers as we please, AB, BC, CD, DE, be added together; I say that the whole AE is even.

+

+

For, since each of the numbers AB, BC, CD, DE is even, it has a half part; [VII. Def. 6] so that the whole AE also has a half part.

+

But an even number is that which is divisible into two equal parts; [id.] therefore AE is even. Q. E. D.

+
+
+ PROPOSITION 22. +

+ If as many odd numbers as we please be added together, and their multitude be even, the whole will be even. +

+

For let as many odd numbers as we please, AB, BC, CD, DE, even in multitude, be added together; I say that the whole AE is even.

+

For, since each of the numbers AB, BC, CD, DE is odd, if an unit be subtracted from each, each of the remainders will be even; [VII. Def. 7] so that the sum of them will be even. [IX. 21]

+

+

But the multitude of the units is also even.

+

Therefore the whole AE is also even. [IX. 21] Q. E. D.

+
+
+ PROPOSITION 23. +

+ If as many odd numbers as we please be added together, and their multitude be odd, the whole will also be odd. +

+

For let as many odd numbers as we please, AB, BC, CD, the multitude of which is odd, be added together; I say that the whole AD is also odd.

+

+

Let the unit DE be subtracted from CD; therefore the remainder CE is even. [VII. Def. 7]

+

But CA is also even; [IX. 22] therefore the whole AE is also even. [IX. 21]

+

And DE is an unit.

+

Therefore AD is odd. [VII. Def. 7] Q. E. D. +3. Literally let there be as many numbers as we please, of which let the multitude be odd. + This form, natural in Greek, is awkward in English.

+
+
+ PROPOSITION 24. +

+ If from an even number an even number be subtracted, the remainder will be even. +

+

For from the even number AB let the even number BC be subtracted: I say that the remainder CA is even.

+

+

For, since AB is even, it has a half part. [VII. Def. 6]

+

For the same reason BC also has a half part; so that the remainder [CA also has a half part, and] AC is therefore even. Q. E. D.

+
+
+ PROPOSITION 25. +

+ If from an even number an odd number be subtracted, the remainder will be odd. +

+

For from the even number AB let the odd number BC be subtracted; I say that the remainder CA is odd.

+

+

For let the unit CD be subtracted from BC; therefore DB is even. [VII. Def. 7]

+

But AB is also even; therefore the remainder AD is also even. [IX. 24]

+

And CD is an unit; therefore CA is odd. [VII. Def. 7] Q. E. D.

+
+
+ PROPOSITION 26. +

+ If from an odd number an odd number be subtracted, the remainder will be even. +

+

For from the odd number AB let the odd number BC be subtracted; I say that the remainder CA is even.

+

+

For, since AB is odd, let the unit BD be subtracted; therefore the remainder AD is even. [VII. Def. 7]

+

For the same reason CD is also even; [VII. Def. 7] so that the remainder CA is also even. [IX. 24] Q. E. D.

+
+ +
+ PROPOSITION 27. +

+ If from an odd number an even number be subtracted, the remainder will be odd. +

+

For from the odd number AB let the even number BC be subtracted; I say that the remainder CA is odd.

+

+

Let the unit AD be subtracted; therefore DB is even. [VII. Def. 7]

+

But BC is also even; therefore the remainder CD is even. [ + IX. 24 + ]

+

Therefore CA is odd. [VII. Def. 7] Q. E. D.

+
+
+ PROPOSITION 28. +

+ If an odd number by multiplying an even number make some number, the product will be even. +

+

For let the odd number A by multiplying the even number B make C; I say that C is even.

+

+

For, since A by multiplying B has made C, therefore C is made up of as many numbers equal to B as there are units in A. [VII. Def. 15]

+

And B is even; therefore C is made up of even numbers.

+

But, if as many even numbers as we please be added together, the whole is even. [IX. 21]

+

Therefore C is even. Q. E. D.

+
+
+ PROPOSITION 29. +

+ If an odd number by multiplying an odd number make some number, the product will be odd. +

+

For let the odd number A by multiplying the odd number B make C; I say that C is odd.

+

+

For, since A by multiplying B has made C, therefore C is made up of as many numbers equal to B as there are units in A. [VII. Def. 15]

+

And each of the numbers A, B is odd; therefore C is made up of odd numbers the multitude of which is odd.

+

Thus C is odd. [IX. 23] Q. E. D.

+
+
+ PROPOSITION 30. +

+ If an odd number measure an even number, it will also measure the half of it. +

+

For let the odd number A measure the even number B; I say that it will also measure the half of it.

+

+

For, since A measures B, let it measure it according to C; I say that C is not odd.

+

For, if possible, let it be so.

+

Then, since A measures B according to C, therefore A by multiplying C has made B.

+

Therefore B is made up of odd numbers the multitude of which is odd.

+

Therefore B is odd: [IX. 23] which is absurd, for by hypothesis it is even.

+

Therefore C is not odd; therefore C is even.

+

Thus A measures B an even number of times.

+

For this reason then it also measures the half of it. Q. E. D.

+
+
+ PROPOSITION 31. +

+ If an odd number be prime to any number, it will also be prime to the double of it. +

+

For let the odd number A be prime to any number B, and let C be double of B; I say that A is prime to C.

+

+

For, if they are not prime to one another, some number will measure them.

+

Let a number measure them, and let it be D.

+

Now A is odd; therefore D is also odd.

+

And since D which is odd measures C, and C is even, therefore [D] will measure the half of C also. [IX. 30]

+

But B is half of C; therefore D measures B.

+

But it also measures A; therefore D measures A, B which are prime to one another: which is impossible.

+

Therefore A cannot but be prime to C.

+

Therefore A, C are prime to one another. Q. E. D.

+
+
+ PROPOSITION 32. +

+ Each of the numbers which are continually doubled beginning from a dyad is even-times even only. +

+

For let as many numbers as we please, B, C, D, have been continually doubled beginning from the dyad A; I say that B, C, D are eventimes even only.

+

+

Now that each of the numbers B, C, D is even-times even is manifest; for it is doubled from a dyad.

+

I say that it is also even-times even only.

+

For let an unit be set out.

+

Since then as many numbers as we please beginning from an unit are in continued proportion, and the number A after the unit is prime, therefore D, the greatest of the numbers A, B, C, D, will not be measured by any other number except A, B, C. [IX. 13]

+

And each of the numbers A, B, C is even; therefore D is even-times even only. [VII. Def. 8]

+

Similarly we can prove that each of the numbers B, C is even-times even only. Q. E. D.

+
+
+ PROPOSITION 33. +

+ If a number have its half odd, it is even-times odd only. +

+

For let the number A have its half odd; I say that A is even-times odd only.

+

+

Now that it is even-times odd is manifest; for the half of it, being odd, measures it an even number of times. [VII. Def. 9]

+

I say next that it is also even-times odd only.

+

For, if A is even-times even also, it will be measured by an even number according to an even number; [VII. Def. 8] so that the half of it will also be measured by an even number though it is odd: which is absurd.

+

Therefore A is even-times odd only. Q. E. D.

+
+
+ PROPOSITION 34. +

If a number neither be one of those which are continually doubled from a dyad, nor have its half odd, it is both eventimes even and even-times odd.

+

For let the number A neither be one of those doubled from a dyad, nor have its half odd; I say that A is both even-times even and even-times odd.

+

+

Now that A is even-times even is manifest; for it has not its half odd. [VII. Def. 8]

+

I say next that it is also even-times odd.

+

For, if we bisect A, then bisect its half, and do this continually, we shall come upon some odd number which will measure A according to an even number.

+

For, if not, we shall come upon a dyad, and A will be among those which are doubled from a dyad: which is contrary to the hypothesis.

+

Thus A is even-times odd.

+

But it was also proved even-times even.

+

Therefore A is both even-times even and even-times odd. Q. E. D.

+
+
+ PROPOSITION 35. +

+ If as many numbers as we please be in continued proportion, and there be subtracted from the second and the last numbers equal to the first, then, as the excess of the second is to the first, so will the excess of the last be to all those before it. +

+

Let there be as many numbers as we please in continued proportion, A, BC, D, EF, beginning from A as least, and let there be subtracted from BC and EF the numbers BG, FH, each equal to A; I say that, as GC is to A, so is EH to A, BC, D.

+

+

For let FK be made equal to BC, and FL equal to D.

+

Then, since FK is equal to BC, and of these the part FH is equal to the part BG, therefore the remainder HK is equal to the remainder GC.

+

And since, as EF is to D, so is D to BC, and BC to A, while D is equal to FL, BC to FK, and A to FH, therefore, as EF is to FL, so is LF to FK, and FK to FH.

+

Separando, as EL is to LF, so is LK to FK, and KH to FH. [VII. 11, 13]

+

Therefore also, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents; [VII. 12] therefore, as KH is to FH, so are EL, LK. KH to LF, FK, HF.

+

But KH is equal to CG, FH to A, and LF, FK, HF to D, BC, A; therefore, as CG is to A, so is EH to D, BC, A.

+

Therefore, as the excess of the second is to the first, so is the excess of the last to all those before it. Q. E. D.

+
+
+ PROPOSITION 36. +

+ If as many numbers as we please beginning from an unit be set out continuously in double proportion, until the sum of all becomes prime, and if the sum multiplied into the last make some number, the product will be perfect. +

+

For let as many numbers as we please, A, B, C, D, beginning from an unit be set out in double proportion, until the sum of all becomes prime, let E be equal to the sum, and let E by multiplying D make FG; I say that FG is perfect.

+

For, however many A, B, C, D are in multitude, let so many E, HK, L, M be taken in double proportion beginning from E; therefore, ex aequali, as A is to D, so is E to M. [VII. 14]

+

Therefore the product of E, D is equal to the product of A, M. [VII. 19]

+

And the product of E, D is FG; therefore the product of A, M is also FG.

+

Therefore A by multiplying M has made FG; therefore M measures FG according to the units in A.

+

And A is a dyad; therefore FG is double of M.

+

+

But M, L, HK, E are continuously double of each other; therefore E, HK, L, M, FG are continuously proportional in double proportion.

+

Now let there be subtracted from the second HK and the last FG the numbers HN, FO, each equal to the first E; therefore, as the excess of the second is to the first, so is the excess of the last to all those before it. [IX. 35]

+

Therefore, as NK is to E, so is OG to M, L, KH, E.

+

And NK is equal to E; therefore OG is also equal to M, L, HK, E.

+

But FO is also equal to E, and E is equal to A, B, C, D and the unit.

+

Therefore the whole FG is equal to E, HK, L, M and A, B, C, D and the unit; and it is measured by them.

+

I say also that FG will not be measured by any other number except A, B, C, D, E, HK, L, M and the unit.

+

For, if possible, let some number P measure FG, and let P not be the same with any of the numbers A, B, C, D, E, HK, L, M.

+

And, as many times as P measures FG, so many units let there be in Q; therefore Q by multiplying P has made FG.

+

But, further, E has also by multiplying D made FG; therefore, as E is to Q, so is P to D. [VII. 19]

+

And, since A, B, C, D are continuously proportional beginning from an unit, therefore D will not be measured by any other number except A, B, C. [IX. 13]

+

And, by hypothesis, P is not the same with any of the numbers A, B, C; therefore P will not measure D.

+

But, as P is to D, so is E to Q; therefore neither does E measure Q. [VII. Def. 20]

+

And E is prime; and any prime number is prime to any number which it does not measure. [VII. 29]

+

Therefore E, Q are prime to one another.

+

But primes are also least, [VII. 21] and the least numbers measure those which have the same ratio the same number of times, the antecedent the antecedent and the consequent the consequent; [VII. 20] and, as E is to Q, so is P to D; therefore E measures P the same number of times that Q measures D.

+

But D is not measured by any other number except A, B, C; therefore Q is the same with one of the numbers A, B, C.

+

Let it be the same with B.

+

And, however many B, C, D are in multitude, let so many E, HK, L be taken beginning from E.

+

Now E, HK, L are in the same ratio with B, C, D; therefore, ex aequali, as B is to D, so is E to L. [VII. 14]

+

Therefore the product of B, L is equal to the product of D, E. [VII. 19]

+

But the product of D, E is equal to the product of Q, P; therefore the product of Q, P is also equal to the product of B, L.

+

Therefore, as Q is to B, so is L to P. [VII. 19]

+

And Q is the same with B; therefore L is also the same with P; which is impossible, for by hypothesis P is not the same with any of the numbers set out.

+

Therefore no number will measure FG except A, B, C, D, E, HK, L, M and the unit.

+

And FG was proved equal to A, B, C, D, E, HK, L, M and the unit; and a perfect number is that which is equal to its own parts; [VII. Def. 22] therefore FG is perfect. Q. E. D.

+
+
+
+ +
+ + BOOK X. +
+ DEFINITIONS I. +
+1 +

Those magnitudes are said to be commensurable which are measured by the same measure, and those incommensurable which cannot have any common measure.

+
+
+2 +

Straight lines are commensurable in square when the squares on them are measured by the same area, and incommensurable in square when the squares on them cannot possibly have any area as a common measure.

+
+
+3 +

With these hypotheses, it is proved that there exist straight lines infinite in multitude which are commensurable and incommensurable respectively, some in length only, and others in square also, with an assigned straight line. Let then the assigned straight line be called rational, and those straight lines which are commensurable with it, whether in length and in square or in square only, rational, but those which are incommensurable with it irrational.

+
+
+4 +

And let the square on the assigned straight line be called rational and those areas which are commensurable with it rational, but those which are incommensurable with it irrational, and the straight lines which produce them irrational, that is, in case the areas are squares, the sides themselves, but in case they are any other rectilineal figures, the straight lines on which are described squares equal to them.

+
+
+ + +
+ PROPOSITIONS I—47. +
+ PROPOSITION 1. +

Two unequal magnitudes being set out, if from the greater there be subtracted a magnitude greater than its half, and from that which is left a magnitude greater than its half, and if this process be repeated continually, there will be left some magnitude which will be less than the lesser magnitude set out.

+

Let AB, C be two unequal magnitudes of which AB is the greater: I say that, if from AB there be subtracted a magnitude greater than its half, and from that which is left a magnitude greater than its half, and if this process be repeated continually, there will be left some magnitude which will be less than the magnitude C.

+

+

For C if multiplied will sometime be greater than AB. [cf. v. Def. 4]

+

Let it be multiplied, and let DE be a multiple of C, and greater than. AB; let DE be divided into the parts DF, FG, GE equal to C, from AB let there be subtracted BH greater than its half, and, from AH, HK greater than its half, and let this process be repeated continually until the divisions in AB are equal in multitude with the divisions in DE.

+

Let, then, AK, KH, HB be divisions which are equal in multitude with DF, FG, GE.

+

Now, since DE is greater than AB, and from DE there has been subtracted EG less than its half, and, from AB, BH greater than its half, therefore the remainder GD is greater than the remainder HA.

+

And, since GD is greater than HA, and there has been subtracted, from GD, the half GF, and, from HA, HK greater than its half, therefore the remainder DF is greater than the remainder AK.

+

But DF is equal to C; therefore C is also greater than AK.

+

Therefore AK is less than C.

+

Therefore there is left of the magnitude AB the magnitude AK which is less than the lesser magnitude set out, namely C. Q. E. D.

+

And the theorem can be similarly proved even if the parts subtracted be halves.

+
+ +
+ PROPOSITION 2. +

+ If, when the less of two unequal magnitudes is continually subtracted in turn from the greater, that which is left never measures the one before it, the magnitudes will be incommensurable. +

+

For, there being two unequal magnitudes AB, CD, and AB being the less, when the less is continually subtracted in turn from the greater, let that which is left over never measure the one before it; I say that the magnitudes AB, CD are incommensurable.

+

+

For, if they are commensurable, some magnitude will measure them.

+

Let a magnitude measure them, if possible, and let it be E; let AB, measuring FD, leave CF less than itself, let CF measuring BG, leave AG less than itself, and let this process be repeated continually, until there is left some magnitude which is less than E.

+

Suppose this done, and let there be left AG less than E.

+

Then, since E measures AB, while AB measures DF, therefore E will also measure FD.

+

But it measures the whole CD also; therefore it will also measure the remainder CF.

+

But CF measures BG; therefore E also measures BG.

+

But it measures the whole AB also; therefore it will also measure the remainder AG, the greater the less: which is impossible.

+

Therefore no magnitude will measure the magnitudes AB, CD; therefore the magnitudes AB, CD are incommensurable.

+

Therefore etc. [X. Def. 1]

+
+
+ PROPOSITION 3. +

Given two commensurable magnitudes, to find their greatest common measure.

+

Let the two given commensurable magnitudes be AB, CD of which AB is the less; thus it is required to find the greatest common measure of AB, CD.

+

Now the magnitude AB either measures CD or it does not.

+

If then it measures it—and it measures itself also—AB is a common measure of AB, CD.

+

And it is manifest that it is also the greatest; for a greater magnitude than the magnitude AB will not measure AB.

+

+

Next, let AB not measure CD.

+

Then, if the less be continually subtracted in turn from the greater, that which is left over will sometime measure the one before it, because AB, CD are not incommensurable; [cf. X. 2] let AB, measuring ED, leave EC less than itself, let EC, measuring FB, leave AF less than itself, and let AF measure CE.

+

Since, then, AF measures CE, while CE measures FB, therefore AF will also measure FB.

+

But it measures itself also; therefore AF will also measure the whole AB.

+

But AB measures DE; therefore AF will also measure ED.

+

But it measures CE also; therefore it also measures the whole CD.

+

Therefore AF is a common measure of AB, CD.

+

I say next that it is also the greatest.

+

For, if not, there will be some magnitude greater than AF which will measure AB, CD.

+

Let it be G.

+

Since then G measures AB, while AB measures ED, therefore G will also measure ED.

+

But it measures the whole CD also; therefore G will also measure the remainder CE.

+

But CE measures FB; therefore G will also measure FB.

+

But it measures the whole AB also, and it will therefore measure the remainder AF, the greater the less: which is impossible.

+

Therefore no magnitude greater than AF will measure AB, CD; therefore AF is the greatest common measure of AB, CD.

+

Therefore the greatest common measure of the two given commensurable magnitudes AB, CD has been found. Q. E. D.

+ +

Porism. From this it is manifest that, if a magnitude measure two magnitudes, it will also measure their greatest common measure.

+ +
+
+ PROPOSITION 4. +

Given three commensurable magnitudes, to find their greatest common measure.

+

Let A, B, C be the three given commensurable magnitudes; thus it is required to find the greatest common measure of A, B, C.

+

+

Let the greatest common measure of the two magnitudes A, B be taken, and let it be D; [X. 3] then D either measures C, or does not measure it.

+

First, let it measure it.

+

Since then D measures C, while it also measures A, B, therefore D is a common measure of A, B, C.

+

And it is manifest that it is also the greatest; for a greater magnitude than the magnitude D does not measure A, B.

+

Next, let D not measure C.

+

I say first that C, D are commensurable.

+

For, since A, B, C are commensurable, some magnitude will measure them, and this will of course measure A, B also; so that it will also measure the greatest common measure of A, B, namely D. [X. 3, Por.]

+

But it also measures C; so that the said magnitude will measure C, D; therefore C, D are commensurable.

+

Now let their greatest common measure be taken, and let it be E. [X. 3]

+

Since then E measures D, while D measures A, B, therefore E will also measure A, B.

+

But it measures C also; therefore E measures A, B, C; therefore E is a common measure of A, B, C.

+

I say next that it is also the greatest.

+

For, if possible, let there be some magnitude F greater than E, and let it measure A, B, C.

+

Now, since F measures A, B, C, it will also measure A, B, and will measure the greatest common measure of A, B. [X. 3, Por.]

+

But the greatest common measure of A, B is D; therefore F measures D.

+

But it measures C also; therefore F measures C, D; therefore F will also measure the greatest common measure of C, D. [X. 3, Por.]

+

But that is E; therefore F will measure E, the greater the less: which is impossible.

+

Therefore no magnitude greater than the magnitude E will measure A, B, C; therefore E is the greatest common measure of A, B, C if D do not measure C, and, if it measure it, D is itself the greatest common measure.

+

Therefore the greatest common measure of the three given commensurable magnitudes has been found.

+ +

Porism. From this it is manifest that, if a magnitude measure three magnitudes, it will also measure their greatest common measure.

+

Similarly too, with more magnitudes, the greatest common measure can be found, and the porism can be extended. Q. E. D.

+ + +
+
+ PROPOSITION 5. +

Commensurable magnitudes have to one another the ratio which a number has to a number.

+

Let A, B be commensurable magnitudes; I say that A has to B the ratio which a number has to a number.

+

For, since A, B are commensurable, some magnitude will measure them.

+

Let it measure them, and let it be C.

+

+

And, as many times as C measures A, so many units let there be in D; and, as many times as C measures B, so many units let there be in E.

+

Since then C measures A according to the units in D, while the unit also measures D according to the units in it, therefore the unit measures the number D the same number of times as the magnitude C measures A; therefore, as C is to A, so is the unit to D; [VII. Def. 20] therefore, inversely, as A is to C, so is D to the unit. [cf. V. 7, Por.]

+

Again, since C measures B according to the units in E, while the unit also measures E according to the units in it, therefore the unit measures E the same number of times as C measures B; therefore, as C is to B, so is the unit to E.

+

But it was also proved that, as A is to C, so is D to the unit; therefore, ex aequali, as A is to B, so is the number D to E. [V. 22] +

+

Therefore the commensurable magnitudes A, B have to one another the ratio which the number D has to the number E. Q. E. D.

+
+ +
+ PROPOSITION 6. +

If two magnitudes have to one another the ratio which a number has to a number, the magnitudes will be commensurable.

+

For let the two magnitudes A, B have to one another the ratio which the number D has to the number E; I say that the magnitudes A, B are commensurable.

+

+

For let A be divided into as many equal parts as there are units in D, and let C be equal to one of them; and let F be made up of as many magnitudes equal to C as there are units in E.

+

Since then there are in A as many magnitudes equal to C as there are units in D, whatever part the unit is of D, the same part is C of A also; therefore, as C is to A, so is the unit to D. [VII. Def. 20]

+

But the unit measures the number D; therefore C also measures A.

+

And since, as C is to A, so is the unit to D, therefore, inversely, as A is to C, so is the number D to the unit. [cf. V. 7, Por.]

+

Again, since there are in F as many magnitudes equal to C as there are units in E, therefore, as C is to F, so is the unit to E. [VII. Def. 20]

+

But it was also proved that, as A is to C, so is D to the unit; + therefore, ex aequali, as A is to F, so is D to E. [v. 22]

+

But, as D is to E, so is A to B; therefore also, as A is to B, so is it to F also. [V. 11]

+

Therefore A has the same ratio to each of the magnitudes B, F; therefore B is equal to F. [V. 9]

+

But C measures F; therefore it measures B also.

+

Further it measures A also; therefore C measures A, B.

+

Therefore A is commensurable with B.

+

Therefore etc.

+ +

Porism. From this it is manifest that, if there be two numbers, as D, E, and a straight line, as A, it is possible to make a straight line [F] such that the given straight line is to it as the number D is to the number E.

+

And, if a mean proportional be also taken between A, F, as B,

+

as A is to F, so will the square on A be to the square on B, that is, as the first is to the third, so is the figure on the first to that which is similar and similarly described on the second. [VI. 19, Por.]

+

But, as A is to F, so is the number D to the number E; therefore it has been contrived that, as the number D is to the number E, so also is the figure on the straight line A to the figure on the straight line B. Q. E. D.

+ + 15. But the unit measures the number D; therefore C also measures A. These words are redundant, though they are apparently found in all the MSS. +
+ +
+ PROPOSITION 7. +

Incommensurable magnitudes have not to one another the ratio which a number has to a number.

+

Let A, B be incommensurable magnitudes; I say that A has not to B the ratio which a number has to a number.

+

For, if A has to B the ratio which a number has to a number, A will be commensurable with B. [X. 6]

+

But it is not; therefore A has not to B the ratio which a number has to a number.

+

+

Therefore etc.

+
+
+ PROPOSITION 8. +

If two magnitudes have not to one another the ratio which a number has to a number, the magnitudes will be incommensurable.

+

For let the two magnitudes A, B not have to one another the ratio which a number has to a number; I say that the magnitudes A, B are incommensurable.

+

+

For, if they are commensurable, A will have to B the ratio which a number has to a number. [X. 5]

+

But it has not; therefore the magnitudes A, B are incommensurable.

+

Therefore etc.

+
+
+ PROPOSITION 9. +

The squares on straight lines commensurable in length have to one another the ratio which a square number has to a square number; and squares which have to one another the ratio which a square number has to a square number will also have their sides commensurable in length. But the squares on straight lines incommensurable in length have not to one another the ratio which a square number has to a square number; and squares which have not to one another the ratio which a square number has to a square number will not have their sides commensurable in length either.

+

For let A, B be commensurable in length; I say that the square on A has to the square on B the ratio which a square number has to a square number.

+

+

For, since A is commensurable in length with B, therefore A has to B the ratio which a number has to a number. [X. 5]

+

Let it have to it the ratio which C has to D.

+

Since then, as A is to B, so is C to D, while the ratio of the square on A to the square on B is duplicate of the ratio of A to B, for similar figures are in the duplicate ratio of their corresponding sides; [VI. 20, Por.] and the ratio of the square on C to the square on D is duplicate of the ratio of C to D, for between two square numbers there is one mean proportional number, and the square number has to the square number the ratio duplicate of that which the side has to the side; [VIII. 11] therefore also, as the square on A is to the square on B, so is the square on C to the square on D.

+

Next, as the square on A is to the square on B, so let the square on C be to the square on D; I say that A is commensurable in length with B.

+

For since, as the square on A is to the square on B, so is the square on C to the square on D, while the ratio of the square on A to the square on B is duplicate of the ratio of A to B, and the ratio of the square on C to the square on D is duplicate of the ratio of C to D, therefore also, as A is to B, so is C to D.

+

Therefore A has to B the ratio which the number C has to the number D; therefore A is commensurable in length with B. [X. 6]

+

Next, let A be incommensurable in length with B; I say that the square on A has not to the square on B the ratio which a square number has to a square number.

+

For, if the square on A has to the square on B the ratio which a square number has to a square number, A will be commensurable with B.

+

But it is not; therefore the square on A has not to the square on B the ratio which a square number has to a square number.

+

Again, let the square on A not have to the square on B the ratio which a square number has to a square number; I say that A is incommensurable in length with B.

+

For, if A is commensurable with B, the square on A will have to the square on B the ratio which a square number has to a square number.

+

But it has not; therefore A is not commensurable in length with B.

+

Therefore etc.

+ +

Porism. And it is manifest from what has been proved that straight lines commensurable in length are always commensurable in square also, but those commensurable in square are not always commensurable in length also.

+ +

[Lemma. It has been proved in the arithmetical books that similar plane numbers have to one another the ratio which a square number has to a square number, [VIII. 26] and that, if two numbers have to one another the ratio which a square number has to a square number, they are similar plane numbers. [Converse of VIII. 26]

+

And it is manifest from these propositions that numbers which are not similar plane numbers, that is, those which have not their sides proportional, have not to one another the ratio which a square number has to a square number.

+

For, if they have, they will be similar plane numbers: which is contrary to the hypothesis.

+

Therefore numbers which are not similar plane numbers have not to one another the ratio which a square number has to a square number.]

+ +
+
+ [PROPOSITION 10. +

To find two straight lines incommensurable, the one in length only, and the other in square also, with an assigned straight line.

+

Let A be the assigned straight line; thus it is required to find two straight lines incommensurable, the one in length only, and the other in square also, with A.

+

Let two numbers B, C be set out which have not to one another the ratio which a square number has to a square number, that is, which are not similar plane numbers; and let it be contrived that, as B is to C, so is the square on A to the square on D +

—for we have learnt how to do this— [X. 6, Por.] therefore the square on A is commensurable with the square on D. [X. 6]

+

And, since B has not to C the ratio which a square number has to a square number, therefore neither has the square on A to the square on D the ratio which a square number has to a square number; therefore A is incommensurable in length with D. [X. 9]

+

Let E be taken a mean proportional between A, D; therefore, as A is to D, so is the square on A to the square on E. [V. Def. 9]

+

But A is incommensurable in length with D; therefore the square on A is also incommensurable with the square on E; [X. 11] therefore A is incommensurable in square with E.

+

Therefore two straight lines D, E have been found incommensurable, D in length only, and E in square and of course in length also, with the assigned straight line A.]

+
+
+ PROPOSITION 11. +

If four magnitudes be proportional, and the first be commensurable with the second, the third will also be commensurable with the fourth; and, if the first be incommensurable with the second, the third will also be incommensurable with the fourth.

+

Let A, B, C, D be four magnitudes in proportion, so that, as A is to B, so is C to D, and let A be commensurable with B;

I say that C will also be commensurable with D.

+

For, since A is commensurable with B, therefore A has to B the ratio which a number has to a number. [X. 5]

+

And, as A is to B, so is C to D; therefore C also has to D the ratio which a number has to a number; therefore C is commensurable with D. [X. 6]

+

Next, let A be incommensurable with B; I say that C will also be incommensurable with D.

+

For, since A is incommensurable with B, therefore A has not to B the ratio which a number has to a number. [X. 7]

+

And, as A is to B, so is C to D; therefore neither has C to D the ratio which a number has to a number; therefore C is incommensurable with D. [X. 8]

+

Therefore etc.

+
+
+ PROPOSITION 12. +

Magnitudes commensurable with the same magnitude are commensurable with one another also.

+

For let each of the magnitudes A, B be commensurable with C; I say that A is also commensurable with B.

+

+

For, since A is commensurable with C, therefore A has to C the ratio which a number has to a number. [X. 5]

+

Let it have the ratio which D has to E.

+

Again, since C is commensurable with B, therefore C has to B the ratio which a number has to a number. [X. 5]

+

Let it have the ratio which F has to G.

+

And, given any number of ratios we please, namely the ratio which D has to E and that which F has to G, let the numbers H, K, L be taken continuously in the given ratios; [cf. VIII. 4] so that, as D is to E, so is H to K, and, as F is to G, so is K to L. +

+

Since, then, as A is to C, so is D to E, while, as D is to E, so is H to K, therefore also, as A is to C, so is H to K. [V. 11]

+

Again, since, as C is to B, so is F to G, while, as F is to G, so is K to L, therefore also, as C is to B, so is K to L. [V. 11]

+

But also, as A is to C, so is H to K; therefore, ex aequali, as A is to B, so is H to L. [V. 22]

+

Therefore A has to B the ratio which a number has to a number; therefore A is commensurable with B. [X. 6]

+

Therefore etc. Q. E. D.

+
+ +
+ PROPOSITION 13. +

If two magnitudes be commensurable, and the one of them be incommensurable with any magnitude, the remaining one will also be incommensurable with the same.

+

Let A, B be two commensurable magnitudes, and let one of them, A, be incommensurable with any other magnitude C; I say that the remaining one, B, will also be incommensurable with C.

+

+

For, if B is commensurable with C, while A is also commensurable with B, A is also commensurable with C. [X. 12]

+

But it is also incommensurable with it: which is impossible.

+

Therefore B is not commensurable with C; therefore it is incommensurable with it.

+

Therefore etc.

+ +

Lemma. Given two unequal straight lines, to find by what square the square on the greater is greater than the square on the less.

+

Let AB, C be the given two unequal straight lines, and let AB be the greater of them; thus it is required to find by what square the square on AB is greater than the square on C.

+

+

Let the semicircle ADB be described on AB, and let AD be fitted into it equal to C; [IV. 1] let DB be joined.

+

It is then manifest that the angle ADB is right, [III. 31] and that the square on AB is greater than the square on AD, that is, C, by the square on DB. [I. 47]

+

Similarly also, if two straight lines be given, the straight line the square on which is equal to the sum of the squares on them is found in this manner.

+

Let AD, DB be the given two straight lines, and let it be required to find the straight line the square on which is equal to the sum of the squares on them.

+

Let them be placed so as to contain a right angle, that formed by AD, DB; and let AB be joined.

+

It is again manifest that the straight line the square on which is equal to the sum of the squares on AD, DB is AB. [I. 47] Q. E. D.

+ +
+
+ PROPOSITION 14. +

If four straight lines be proportional, and the square on the first be greater than the square on the second by the square on a straight line commensurable with the first, the square on the third will also be greater than the square on the fourth by + the square on a straight line commensurable with the third.

+

And, if the square on the first be greater than the square on the second by the square on a straight line incommensurable with the first, the square on the third will also be greater than the square on the fourth by the square on a straight line in- + commensurable with the third.

+

Let A, B, C, D be four straight lines in proportion, so that, as A is to B, so is C to D; and let the square on A be greater than the square on B by the square on E, and let the square on C be greater than the square on D by the square on F; I say that, if A is commensurable with E, C is also commensurable with F, and, if A is incommensurable with E, C is also incommensurable with F.

+

+

For since, as A is to B, so is C to D, therefore also, as the square on A is to the square on B, so is the square on C to the square on D. [VI. 22]

+

But the squares on E, B are equal to the square on A, and the squares on D, F are equal to the square on C.

+

Therefore, as the squares on E, B are to the square on B, so are the squares on D, F to the square on D; therefore, separando, as the square on E is to the square on B, so is the square on F to the square on D; [V. 17] therefore also, as E is to B, so is F to D; [VI. 22] therefore, inversely, as B is to E, so is D to F.

+

But, as A is to B, so also is C to D; therefore, ex aequali, as A is to E, so is C to F. [V. 22]

+

Therefore, if A is commensurable with E, C is also commensurable with F, and, if A is incommensurable with E, C is also incommensurable with F. [X. 11]

+

Therefore etc.3, 5, 8, 10. Euclid speaks of the square on the first (third) being greater than the square on the second (fourth) by the square on a straight line commensurable (incommensurable) with itself (ἑαυτῇ), + and similarly in all like phrases throughout the Book. For clearness' sake I substitute the first, + the third, + or whatever it may be, for itself + in these cases.

+
+ +
+ PROPOSITION 15. +

If two commensurable magnitudes be added together, the whole will also be commensurable with each of them; and, if the whole be commensurable with one of them, the original magnitudes will also be commensurable.

+

For let the two commensurable magnitudes AB, BC be added together; I say that the whole AC is also commensurable with each of the magnitudes AB, BC.

+

+

For, since AB, BC are commensurable, some magnitude will measure them.

+

Let it measure them, and let it be D.

+

Since then D measures AB, BC, it will also measure the whole AC.

+

But it measures AB, BC also; therefore D measures AB, BC, AC; therefore AC is commensurable with each of the magnitudes AB, BC. [X. Def. 1]

+

Next, let AC be commensurable with AB; I say that AB, BC are also commensurable.

+

For, since AC, AB are commensurable, some magnitude will measure them.

+

Let it measure them, and let it be D.

+

Since then D measures CA, AB, it will also measure the remainder BC.

+

But it measures AB also; therefore D will measure AB, BC; therefore AB, BC are commensurable. [X. Def. 1]

+

Therefore etc.

+
+ +
+ PROPOSITION 16. +

If two incommensurable magnitudes be added together, the whole will also be incommensurable with each of them; and, if the whole be incommensurable with one of them, the original magnitudes will also be incommensurable.

+

For let the two incommensurable magnitudes AB, BC be added together; I say that the whole AC is also incommensurable with each of the magnitudes AB, BC.

+

For, if CA, AB are not incommensurable, some magnitude will measure them.

+

Let it measure them, if possible, and let it be D.

+

Since then D measures CA, AB, therefore it will also measure the remainder BC.

+

+

But it measures AB also; therefore D measures AB, BC.

+

Therefore AB, BC are commensurable; but they were also, by hypothesis, incommensurable: which is impossible.

+

Therefore no magnitude will measure CA, AB; therefore CA, AB are incommensurable. [X. Def. 1]

+

Similarly we can prove that AC, CB are also incommensurable.

+

Therefore AC is incommensurable with each of the magnitudes AB, BC.

+

Next, let AC be incommensurable with one of the magnitudes AB, BC.

+

First, let it be incommensurable with AB; I say that AB, BC are also incommensurable.

+

For, if they are commensurable, some magnitude will measure them.

+

Let it measure them, and let it be D.

+

Since then D measures AB, BC. therefore it will also measure the whole AC.

+

But it measures AB also; therefore D measures CA, AB.

+

Therefore CA, AB are commensurable; but they were also, by hypothesis, incommensurable: which is impossible.

+

Therefore no magnitude will measure AB, BC; therefore AB, BC are incommensurable. [X. Def. 1]

+

Therefore etc.

+ +

Lemma. + If to any straight line there be applied a parallelogram deficient by a square figure, the applied parallelogram is equal to the rectangle contained by the segments of the straight line resulting from the application. +

+

For let there be applied to the straight line AB the parallelogram AD deficient by the square figure DB; I say that AD is equal to the rectangle contained by AC, CB.

+

+

This is indeed at once manifest; for, since DB is a square, DC is equal to CB; and AD is the rectangle AC, CD, that is, the rectangle AC, CB.

+

Therefore etc.

+ +
+
+ PROPOSITION 17. +

+ If there be two unequal straight lines, and to the greater there be applied a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, and if it divide it into parts which are commensurable in length, then the square on the greater will be greater than the square on the less by the square on a straight line commensurable with the greater. +

+

And, if the square on the greater be greater than the square on the less by the square on a straight line commensurable with + the greater, and if there be applied to the greater a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, it will divide it into parts which are commensurable in length.

+

Let A, BC be two unequal straight lines, of which BC is the greater, and let there be applied to BC a parallelogram equal to the fourth part of the square on the less, A, that is, equal to the square on the half of A, and deficient by a square figure. Let this be the rectangle BD, DC, [cf. Lemma]

and let BD be commensurable in length with DC; I say that the square on BC is greater than the square on A by the square on a straight line commensurable with BC.

+

For let BC be bisected at the point E, and let EF be made equal to DE.

+

Therefore the remainder DC is equal to BF.

+

And, since the straight line BC has been cut into equal parts at E, and into unequal parts at D, therefore the rectangle contained by BD, DC, together with the square on ED, is equal to the square on EC; [II. 5]

+

And the same is true of their quadruples; therefore four times the rectangle BD, DC, together with four times the square on DE, is equal to four times the square on EC.

+

But the square on A is equal to four times the rectangle BD, DC; and the square on DF is equal to four times the square on DE, for DF is double of DE.

+

And the square on BC is equal to four times the square on EC, for again BC is double of CE.

+

Therefore the squares on A, DF are equal to the square on BC, so that the square on BC is greater than the square on A by the square on DF.

+

It is to be proved that BC is also commensurable with DF.

+

Since BD is commensurable in length with DC, therefore BC is also commensurable in length with CD. [X. 15]

+

But CD is commensurable in length with CD, BF, for CD is equal to BF. [X. 6]

+

Therefore BC is also commensurable in length with BF, CD, [X. 12] so that BC is also commensurable in length with the remainder FD; [X. 15] therefore the square on BC is greater than the square on A by the square on a straight line commensurable with BC.

+

Next, let the square on BC be greater than the square on A by the square on a straight line commensurable with BC, let a parallelogram be applied to BC equal to the fourth part of the square on A and deficient by a square figure, and let it be the rectangle BD, DC.

+

It is to be proved that BD is commensurable in length with DC.

+

With the same construction, we can prove similarly that the square on BC is greater than the square on A by the square on FD.

+

But the square on BC is greater than the square on A by the square on a straight line commensurable with BC.

+

Therefore BC is commensurable in length with FD, so that BC is also commensurable in length with the remainder, the sum of BF, DC. [X. 15]

+

But the sum of BF, DC is commensurable with DC, [X. 6] so that BC is also commensurable in length with CD; [X. 12] and therefore, separando, BD is commensurable in length with DC. [X. 15]

+

Therefore etc. +45. After saying literally that the square on BC is greater than the square on A by the square on DF, + Euclid adds the equivalent expression with δύναται in its technical sense, ἡ ΒΓ ἄρα τῆς Α μεῖζον δύναται τῇ ΔΖ. As this is untranslatable in English except by a paraphrase in practically the same words as have preceded, I have not attempted to reproduce it.

+
+
+ PROPOSITION 18. +

+ If there be two unequal straight lines, and to the greater there be applied a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, and if it divide it into parts which are incommensurable, the square on the greater will be greater than the square on the less by the square on a straight line incommensurable with the greater. +

+

And, if the square on the greater be greater than the square on the less by the square on a straight line incommensurable with the greater, and if there be applied to the greater a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, it divides it into parts which are incommensurable.

+

Let A, BC be two unequal straight lines, of which BC is the greater, and to BC let there be applied a parallelogram equal to the fourth part of the square on the less, A, and deficient by a square figure. Let this be the rectangle BD, DC, [cf. Lemma before X. 17]

and let BD be incommensurable in length with DC; I say that the square on BC is greater than the square on A by the square on a straight line incommensurable with BC.

+

For, with the same construction as before, we can prove similarly that the square on BC is greater than the square on A by the square on FD.

+

It is to be proved that BC is incommensurable in length with DF.

+

Since BD is incommensurable in length with DC, therefore BC is also incommensurable in length with CD. [X. 16]

+

But DC is commensurable with the sum of BF, DC; [X. 6] therefore BC is also incommensurable with the sum of BF, DC; [X. 13] so that BC is also incommensurable in length with the remainder FD. [X. 16]

+

And the square on BC is greater than the square on A by the square on FD; therefore the square on BC is greater than the square on A by the square on a straight line incommensurable with BC.

+

Again, let the square on BC be greater than the square on A by the square on a straight line incommensurable with BC, and let there be applied to BC a parallelogram equal to the fourth part of the square on A and deficient by a square figure. Let this be the rectangle BD, DC.

+

It is to be proved that BD is incommensurable in length with DC.

+

For, with the same construction, we can prove similarly that the square on BC is greater than the square on A by the square on FD.

+

But the square on BC is greater than the square on A by the square on a straight line incommensurable with BC; therefore BC is incommensurable in length with FD. so that BC is also incommensurable with the remainder, the sum of BF, DC. [X. 16]

+

But the sum of BF, DC is commensurable in length with DC; [X. 6] therefore BC is also incommensurable in length with DC, [X. 13] so that, separando, BD is also incommensurable in length with DC. [X. 16]

+

Therefore etc. +

+ +

[Lemma. Since it has been proved that straight lines commensurable in length are always commensurable in square also, while those commensurable in square are not always commensurable in length also, but can of course be either commensurable or incommensurable in length, it is manifest that, if any straight line be commensurable in length with a given rational straight line, it is called rational and commensurable with the other not only in length but in square also, since straight lines commensurable in length are always commensurable in square also.

+

But, if any straight line be commensurable in square with a given rational straight line, then, if it is also commensurable in length with it, it is called in this case also rational and commensurable with it both in length and in square; but, if again any straight line, being commensurable in square with a given rational straight line, be incommensurable in length with it, it is called in this case also rational but commensurable in square only.]

+ +
+
+ PROPOSITION 19. +

The rectangle contained by rational straight lines commensurable in length is rational.

+

For let the rectangle AC be contained by the rational straight lines AB, BC commensurable in length; I say that AC is rational.

+

For on AB let the square AD be described; therefore AD is rational. [X. Def. 4]

+

+

And, since AB is commensurable in length with BC, while AB is equal to BD, therefore BD is commensurable in length with BC.

+

And, as BD is to BC, so is DA to AC. [VI. 1]

+

Therefore DA is commensurable with AC. [X. 11]

+

But DA is rational; therefore AC is also rational. [X. Def. 4]

+

Therefore etc.

+
+
+ + PROPOSITION 20. +

If a rational area be applied to a rational straight line, it produces as breadth a straight line rational and commensurable in length with the straight line to which it is applied.

+

For let the rational area AC be applied to AB, a straight line once more rational in any of the aforesaid ways, producing BC as breadth; I say that BC is rational and commensurable in length with BA. For on AB let the square AD be described; therefore AD is rational. [X. Def. 4]

+

+

But AC is also rational; therefore DA is commensurable with AC.

+

And, as DA is to AC, so is DB to BC. [VI. 1]

+

Therefore DB is also commensurable with BC; [X. 11] and DB is equal to BA; therefore AB is also commensurable with BC.

+

But AB is rational; therefore BC is also rational and commensurable in length with AB.

+

Therefore etc.

+
+
+ PROPOSITION 21. +

+ The rectangle contained by rational straight lines commensurable in square only is irrational, and the side of the square equal to it is irrational. Let the latter be called + medial. +

+

For let the rectangle AC be contained by the rational straight lines AB, BC commensurable in square only; I say that AC is irrational, and the side of the square equal to it is irrational; and let the latter be called medial. +

+

For on AB let the square AD be described; therefore AD is rational. [X. Def. 4]

+

+

And, since AB is incommensurable in length with BC, for by hypothesis they are commensurable in square only, while AB is equal to BD, therefore DB is also incommensurable in length with BC.

+

And, as DB is to BC, so is AD to AC; [VI. 1] therefore DA is incommensurable with AC. [X. 11]

+

But DA is rational; therefore AC is irrational, so that the side of the square equal to AC is also irrational. [X. Def. 4]

+

And let the latter be called medial. Q. E. D.

+ +

Lemma. If there be two straight lines, then, as the first is to the second, so is the square on the first to the rectangle contained by the two straight lines.

+

Let FE, EG be two straight lines.

+

+

I say that, as FE is to EG, so is the square on FE to the rectangle FE, EG.

+

For on FE let the square DF be described, and let GD be completed.

+

Since then, as FE is to EG, so is FD to DG, [VI. 1] and FD is the square on FE, and DG the rectangle DE, EG, that is, the rectangle FE, EG, therefore, as FE is to EG, so is the square on FE to the rectangle FE, EG.

+

Similarly also, as the rectangle GE, EF is to the square on EF, that is, as GD is to FD, so is GE to EF. Q. E. D.

+ +
+
+ PROPOSITION 22. +

The square on a medial straight line, if applied to a rational straight line, produces as breadth a straight line rational and incommensurable in length with that to which it is applied.

+

Let A be medial and CB rational, and let a rectangular area BD equal to the square on A be applied to BC, producing CD as breadth; I say that CD is rational and incommensurable in length with CB.

+

For, since A is medial, the square on it is equal to a rectangular area contained by rational straight lines commensurable in square only. [X. 21]

+

+

Let the square on it be equal to GF.

+

But the square on it is also equal to BD; therefore BD is equal to GF.

+

But it is also equiangular with it; and in equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional; [VI. 14] therefore, proportionally, as BC is to EG, so is EF to CD.

+

Therefore also, as the square on BC is to the square on EG, so is the square on EF to the square on CD. [VI. 22]

+

But the square on CB is commensurable with the square on EG, for each of these straight lines is rational; therefore the square on EF is also commensurable with the square on CD. [X. 11]

+

But the square on EF is rational; therefore the square on CD is also rational; [X. Def. 4] therefore CD is rational.

+

And, since EF is incommensurable in length with EG, for they are commensurable in square only, and, as EF is to EG, so is the square on EF to the rectangle FE, EG, [Lemma] therefore the square on EF is incommensurable with the rectangle FE, EG. [X. 11]

+

But the square on CD is commensurable with the square on EF, for the straight lines are rational in square; and the rectangle DC, CB is commensurable with the rectangle FE, EG, for they are equal to the square on A; therefore the square on CD is also incommensurable with the rectangle DC, CB. [X. 13]

+

But, as the square on CD is to the rectangle DC, CB, so is DC to CB; [Lemma] therefore DC is incommensurable in length with CB. [X. 11]

+

Therefore CD is rational and incommensurable in length with CB. Q. E. D.

+
+
+ PROPOSITION 23. +

A straight line commensurable with a medial straight line is medial.

+

Let A be medial, and let B be commensurable with A; I say that B is also medial.

+

For let a rational straight line CD be set out,

and to CD let the rectangular area CE equal to the square on A be applied, producing ED as breadth; therefore ED is rational and incommensurable in length with CD. [X. 22]

+

And let the rectangular area CF equal to the square on B be applied to CD, producing DF as breadth.

+

Since then A is commensurable with B, the square on A is also commensurable with the square on B.

+

But EC is equal to the square on A, and CF is equal to the square on B; therefore EC is commensurable with CF.

+

And, as EC is to CF, so is ED to DF; [VI. 1] therefore ED is commensurable in length with DF. [X. 11]

+

But ED is rational and incommensurable in length with DC; therefore DF is also rational [X. Def. 3] and incommensurable in length with DC. [X. 13]

+

Therefore CD, DF are rational and commensurable in square only.

+

But the straight line the square on which is equal to the rectangle contained by rational straight lines commensurable in square only is medial; [X. 21] therefore the side of the square equal to the rectangle CD, DF is medial.

+

And B is the side of the square equal to the rectangle CD, DF; therefore B is medial.

+ +

Porism. From this it is manifest that an area commensurable with a medial area is medial.

+

[And in the same way as was explained in the case of rationals [Lemma following X. 18] it follows, as regards medials, that a straight line commensurable in length with a medial straight line is called medial and commensurable with it not only in length but in square also, since, in general, straight lines commensurable in length are always commensurable in square also.

+

But, if any straight line be commensurable in square with a medial straight line, then, if it is also commensurable in length with it, the straight lines are called, in this case too, medial and commensurable in length and in square, but, if in square only, they are called medial straight lines commensurable in square only.]

+ +
+
+ PROPOSITION 24. +

The rectangle contained by medial straight lines commensurable in length is medial.

+

For let the rectangle AC be contained by the medial straight lines AB, BC which are commensurable in length; I say that AC is medial.

+

For on AB let the square AD be described; therefore AD is medial.

+

+

And, since AB is commensurable in length with BC, while AB is equal to BD, therefore DB is also commensurable in length with BC; so that DA is also commensurable with AC. [VI. 1, X. 11]

+

But DA is medial; therefore AC is also medial. [X. 23, Por.] Q. E. D.

+
+
+ PROPOSITION 25. +

The rectangle contained by medial straight lines commensurable in square only is either rational or medial.

+

For let the rectangle AC be contained by the medial straight lines AB, BC which are commensurable in square only; I say that AC is either rational or medial.

+

+

For on AB, BC let the squares AD, BE be described; therefore each of the squares AD, BE is medial.

+

Let a rational straight line FG be set out, to FG let there be applied the rectangular parallelogram GH equal to AD, producing FH as breadth, to HM let there be applied the rectangular parallelogram MK equal to AC, producing HK as breadth, and further to KN let there be similarly applied NL equal to BE, producing KL as breadth; therefore FH, HK, KL are in a straight line.

+

Since then each of the squares AD, BE is medial, and AD is equal to GH, and BE to NL, therefore each of the rectangles GH, NL is also medial.

+

And they are applied to the rational straight line FG; therefore each of the straight lines FH, KL is rational and incommensurable in length with FG. [X. 22]

+

And, since AD is commensurable with BE, therefore GH is also commensurable with NL.

+

And, as GH is to NL, so is FH to KL; [VI. 1] therefore FH is commensurable in length with KL. [X. 11]

+

Therefore FH, KL are rational straight lines commensurable in length; therefore the rectangle FH, KL is rational. [X. 19]

+

And, since DB is equal to BA, and OB to BC, therefore, as DB is to BC, so is AB to BO.

+

But, as DB is to BC, so is DA to AC, [VI. 1] and, as AB is to BO, so is AC to CO; [id.] therefore, as DA is to AC, so is AC to CO.

+

But AD is equal to GH, AC to MK and CO to NL; therefore, as GH is to MK, so is MK to NL; therefore also, as FH is to HK, so is HK to KL; [VI. 1, V. 11] therefore the rectangle FH, KL is equal to the square on HK. [VI. 17]

+

But the rectangle FH, KL is rational; therefore the square on HK is also rational.

+

Therefore HK is rational.

+

And, if it is commensurable in length with FG, HN is rational; [X. 19] but, if it is incommensurable in length with FG, KH, HM are rational straight lines commensurable in square only, and therefore HN is medial. [X. 21]

+

Therefore HN is either rational or medial.

+

But HN is equal to AC; therefore AC is either rational or medial.

+

Therefore etc.

+
+
+ PROPOSITION 26. +

4 medial area does not exceed a medial area by a rational area.

+

For, if possible, let the medial area AB exceed the medial area AC by the rational area DB,

and let a rational straight line EF be set out; to EF let there be applied the rectangular parallelogram FH equal to AB, producing EH as breadth, and let the rectangle FG equal to AC be subtracted; therefore the remainder BD is equal to the remainder KH.

+

But DB is rational; therefore KH is also rational.

+

Since, then, each of the rectangles AB, AC is medial, and AB is equal to FH, and AC to FG, therefore each of the rectangles FH, FG is also medial.

+

And they are applied to the rational straight line EF; therefore each of the straight lines HE, EG is rational and incommensurable in length with EF. [X. 22]

+

And, since [DB is rational and is equal to KH, therefore] KH is [also] rational; and it is applied to the rational straight line EF; therefore GH is rational and commensurable in length with EF. [X. 20]

+

But EG is also rational, and is incommensurable in length with EF; therefore EG is incommensurable in length with GH. [X. 13]

+

And, as EG is to GH, so is the square on EG to the rectangle EG, GH; therefore the square on EG is incommensurable with the rectangle EG, GH. [X. 11]

+

But the squares on EG, GH are commensurable with the square on EG, for both are rational; and twice the rectangle EG, GH is commensurable with the rectangle EG, GH, for it is double of it; [X. 6] therefore the squares on EG, GH are incommensurable with twice the rectangle EG, GH; [X. 13] therefore also the sum of the squares on EG, GH and twice the rectangle EG, GH, that is, the square on EH [II. 4], is incommensurable with the squares on EG, GH. [X. 16]

+

But the squares on EG, GH are rational; therefore the square on EH is irrational. [X. Def. 4]

+

Therefore EH is irrational.

+

But it is also rational: which is impossible.

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 27. +

To find medial straight lines commensurable in square only which contain a rational rectangle.

+

Let two rational straight lines A, B commensurable in square only be set out; let C be taken a mean proportional between A, B, [VI. 13] and let it be contrived that,

+ as A is to B, so is C to D. [VI. 12] +

+

Then, since A, B are rational and commensurable in square only, the rectangle A, B, that is, the square on C [VI.17], is medial. [X. 21]

+

Therefore C is medial. [X. 21]

+

And since, as A is to B, so is C to D, and A, B are commensurable in square only, therefore C, D are also commensurable in square only. [X. 11]

+

And C is medial; therefore D is also medial. [X. 23, addition]

+

Therefore C, D are medial and commensurable in square only.

+

I say that they also contain a rational rectangle.

+

For since, as A is to B, so is C to D, therefore, alternately, as A is to C, so is B to D. [V. 16]

+

But, as A is to C, so is C to B; therefore also, as C is to B, so is B to D; therefore the rectangle C, D is equal to the square on B.

+

But the square on B is rational; therefore the rectangle C, D is also rational.

+

Therefore medial straight lines commensurable in square only have been found which contain a rational rectangle. Q. E. D.

+
+
+ PROPOSITION 28. +

To find medial straight lines commensurable in square only which contain a medial rectangle.

+

Let the rational straight lines A, B, C commensurable in square only be set out; let D be taken a mean proportional between A, B, [VI. 13] and let it be contrived that, as B is to C, so is D to E. [VI. 12] +

+

+

Since A, B are rational straight lines commensurable in square only, therefore the rectangle A, B, that is, the square on D [VI. 17], is medial. [X. 21]

+

Therefore D is medial. [X. 21]

+

And since B, C are commensurable in square only, and, as B is to C, so is D to E, therefore D, E are also commensurable in square only. [X. 11]

+

But D is medial; therefore E is also medial. [X. 23, addition]

+

Therefore D, E are medial straight lines commensurable in square only.

+

I say next that they also contain a medial rectangle.

+

For since, as B is to C, so is D to E, therefore, alternately, as B is to D, so is C to E. [V. 16]

+

But, as B is to D, so is D to A; therefore also, as D is to A, so is C to E; therefore the rectangle A, C is equal to the rectangle D, E. [VI. 16]

+

But the rectangle A, C is medial; [X. 21] therefore the rectangle D, E is also medial.

+

Therefore medial straight lines commensurable in square only have been found which contain a medial rectangle. Q. E. D. +

+ +

LEMMA I. To find two square numbers such that their sum is also square.

+

Let two numbers AB, BC be set out, and let them be either both even or both odd.

+

+

Then since, whether an even number is subtracted from an even number, or an odd number from an odd number, the remainder is even, [IX. 24, 26] therefore the remainder AC is even.

+

Let AC be bisected at D.

+

Let AB, BC also be either similar plane numbers, or square numbers, which are themselves also similar plane numbers.

+

Now the product of AB, BC together with the square on CD is equal to the square on BD. [II. 6]

+

And the product of AB, BC is square, inasmuch as it was proved that, if two similar plane numbers by multiplying one another make some number the product is square. [IX. 1]

+

Therefore two square numbers, the product of AB, BC, and the square on CD, have been found which, when added together, make the square on BD.

+

And it is manifest that two square numbers, the square on BD and the square on CD, have again been found such that their difference, the product of AB, BC, is a square, whenever AB, BC are similar plane numbers.

+

But when they are not similar plane numbers, two square numbers, the square on BD and the square on DC, have been found such that their difference, the product of AB, BC, is not square. Q. E. D.

+ +

LEMMA 2. To find two square numbers such that their sum is not square.

+

For let the product of AB, BC, as we said, be square, and CA even, and let CA be bisected by D.

+

+

It is then manifest that the square product of AB, BC together with the square on CD is equal to the square on BD. [See Lemma 1]

+

Let the unit DE be subtracted; therefore the product of AB, BC together with the square on CE is less than the square on BD.

+

I say then that the square product of AB, BC together with the square on CE will not be square.

+

For, if it is square, it is either equal to the square on BE, or less than the square on BE, but cannot any more be greater, lest the unit be divided.

+

First, if possible, let the product of AB, BC together with the square on CE be equal to the square on BE, and let GA be double of the unit DE.

+

Since then the whole AC is double of the whole CD, and in them AG is double of DE, therefore the remainder GC is also double of the remainder EC; therefore GC is bisected by E.

+

Therefore the product of GB, BC together with the square on CE is equal to the square on BE. [II. 6]

+

But the product of AB, BC together with the square on CE is also, by hypothesis, equal to the square on BE; therefore the product of GB, BC together with the square on CE is equal to the product of AB, BC together with the square on CE.

+

And, if the common square on CE be subtracted, it follows that AB is equal to GB: which is absurd.

+

Therefore the product of AB, BC together with the square on CE is not equal to the square on BE.

+

I say next that neither is it less than the square on BE.

+

For, if possible, let it be equal to the square on BF, and let HA be double of DF.

+

Now it will again follow that HC is double of CF; so that CH has also been bisected at F, and for this reason the product of HB, BC together with the square on FC is equal to the square on BF. [II. 6]

+

But, by hypothesis, the product of AB, BC together with the square on CE is also equal to the square on BF.

+

Thus the product of HB, BC together with the square on CF will also be equal to the product of AB, BC together with the square on CE: which is absurd.

+

Therefore the product of AB, BC together with the square on CE is not less than the square on BE.

+

And it was proved that neither is it equal to the square on BE.

+

Therefore the product of AB, BC together with the square on CE is not square. Q. E. D.

+ +
+
+ PROPOSITION 29. +

To find two rational straight lines commensurable in square only and such that the square on the greater is greater than the square on the less by the square on a straight line commensurable in length with the greater.

+

For let there be set out any rational straight line AB, and two square numbers CD, DE such that their difference CE is not square; [Lemma 1] let there be described on AB the semicircle AFB, and let it be contrived that, as DC is to CE, so is the square on BA to the square on AF. [X. 6, Por.] +

+

Let FB be joined.

+

Since, as the square on BA is to the square on AF, so is DC to CE, therefore the square on BA has to the square on AF the ratio which the number DC has to the number CE; therefore the square on BA is commensurable with the square on AF. [X. 6]

+

+

But the square on AB is rational; [X. Def. 4] therefore the square on AF is also rational; [id.] therefore AF is also rational.

+

And, since DC has not to CE the ratio which a square number has to a square number, neither has the square on BA to the square on AF the ratio which a square number has to a square number; therefore AB is incommensurable in length with AF. [X. 9]

+

Therefore BA, AF are rational straight lines commensurable in square only.

+

And since, as DC is to CE, so is the square on BA to the square on AF, therefore, convertendo, as CD is to DE, so is the square on AB to the square on BF. [V. 19, Por., III. 31, I. 47]

+

But CD has to DE the ratio which a square number has to a square number: therefore also the square on AB has to the square on BF the ratio which a square number has to a square number; therefore AB is commensurable in length with BF. [X. 9]

+

And the square on AB is equal to the squares on AF, FB; therefore the square on AB is greater than the square on AF by the square on BF commensurable with AB.

+

Therefore there have been found two rational straight lines BA, AF commensurable in square only and such that the square on the greater AB is greater than the square on the less AF by the square on BF commensurable in length with AB.

+
+
+ PROPOSITION 30. +

To find two rational straight lines commensurable in square only and such that the square on the greater is greater is greater than the square on the less by the square on a straight line incommensurable in length with the greater.

+

Let there be set out a rational straight line AB, and two square numbers CE, ED such that their sum CD is not square; [Lemma 2]

let there be described on AB the semicircle AFB, let it be contrived that, as DC is to CE, so is the square on BA to the square on AF, [X. 6, Por.] and let FB be joined.

+

Then, in a similar manner to the preceding, we can prove that BA, AF are rational straight lines commensurable in square only.

+

And since, as DC is to CE, so is the square on BA to the square on AF, therefore, convertendo, as CD is to DE, so is the square on AB to the square on BF. [V. 19, Por., III. 31, I. 47]

+

But CD has not to DE the ratio which a square number has to a square number; therefore neither has the square on AB to the square on BF the ratio which a square number has to a square number; therefore AB is incommensurable in length with BF. [X. 9]

+

And the square on AB is greater than the square on AF by the square on FB incommensurable with AB.

+

Therefore AB, AF are rational straight lines commensurable in square only, and the square on AB is greater than the square on AF by the square on FB incommensurable in length with AB. Q. E. D.

+
+
+ PROPOSITION 31. +

+ To find two medial straight lines commensurable in square only, containing a rational rectangle, and such that the square on the greater is greater than the square on the less by the square on a straight line commensurable in length with the greater. +

+

Let there be set out two rational straight lines A, B commensurable in square only and such that the square on A, being the greater, is greater than the square on B the less by the square on a straight line commensurable in length with A. [X. 29]

+

+

And let the square on C be equal to the rectangle A, B.

+

Now the rectangle A, B is medial; [X. 21] therefore the square on C is also medial; therefore C is also medial. [X. 21]

+

Let the rectangle C, D be equal to the square on B.

+

Now the square on B is rational; therefore the rectangle C, D is also rational.

+

And since, as A is to B, so is the rectangle A, B to the square on B, while the square on C is equal to the rectangle A, B, and the rectangle C, D is equal to the square on B, therefore, as A is to B, so is the square on C to the rectangle C, D.

+

But, as the square on C is to the rectangle C, D, so is C to D; therefore also, as A is to B, so is C to D.

+

But A is commensurable with B in square only; therefore C is also commensurable with D in square only. [X. 11]

+

And C is medial; therefore D is also medial. [X. 23, addition]

+

And since, as A is to B, so is C to D, and the square on A is greater than the square on B by the square on a straight line commensurable with A, therefore also the square on C is greater than the square on D by the square on a straight line commensurable with C. [X. 14]

+

Therefore two medial straight lines C, D, commensurable in square only and containing a rational rectangle, have been found, and the square on C is greater than the square on D by the square on a straight line commensurable in length with C.

+

Similarly also it can be proved that the square on C exceeds the square on D by the square on a straight line incommensurable with C, when the square on A is greater than the square on B by the square on a straight line incommensurable with A. [X. 30]

+
+
+ PROPOSITION 32. +

+ To find two medial straight lines commensurable in square only, containing a medial rectangle, and such that the square on the greater is greater than the square on the less by the square on a straight line commensurable with the greater. + +

+

Let there be set out three rational straight lines A, B, C commensurable in square only, and such that the square on A is greater than the square on C by the square on a straight line commensurable with A, [X. 29] and let the square on D be equal to the rectangle A, B.

+

+

Therefore the square on D is medial; therefore D is also medial. [X. 21]

+

Let the rectangle D, E be equal to the rectangle B, C.

+

Then since, as the rectangle A, B is to the rectangle B, C, so is A to C; while the square on D is equal to the rectangle A, B, and the rectangle D, E is equal to the rectangle B, C, therefore, as A is to C, so is the square on D to the rectangle D, E.

+

But, as the square on D is to the rectangle D, E, so is D to E; therefore also, as A is to C, so is D to E.

+

But A is commensurable with C in square only; therefore D is also commensurable with E in square only. [X. 11]

+

But D is medial; therefore E is also medial. [X. 23, addition]

+

And, since, as A is to C, so is D to E, while the square on A is greater than the square on C by the square on a straight line commensurable with A, therefore also the square on D will be greater than the square on E by the square on a straight line commensurable with D.[X. 14]

+

I say next that the rectangle D, E is also medial.

+

For, since the rectangle B, C is equal to the rectangle D, E, while the rectangle B, C is medial, [X. 21] therefore the rectangle D, E is also medial.

+

Therefore two medial straight lines D, E, commensurable in square only, and containing a medial rectangle, have been found such that the square on the greater is greater than the square on the less by the square on a straight line commensurable with the greater.

+ +

Similarly again it can be proved that the square on D is greater than the square on E by the square on a straight line incommensurable with D, when the square on A is greater than the square on C by the square on a straight line incommensurable with A. [X. 30]

+ + +

Lemma. Let ABC be a right-angled triangle having the angle A right, and let the perpendicular AD be drawn; I say that the rectangle CB, BD is equal to the square on BA, the rectangle BC, CD equal to the square on CA, the rectangle BD, DC equal to the square on AD, and, further, the rectangle BC, AD equal to the rectangle BA, AC.

+

+

And first that the rectangle CB, BD is equal to the square on BA.

+

For, since in a right-angled triangle AD has been drawn from the right angle perpendicular to the base, therefore the triangles ABD, ADC are similar both to the whole ABC and to one another. [VI. 8]

+

And since the triangle ABC is similar to the triangle ABD, therefore, as CB is to BA, so is BA to BD; [VI. 4] therefore the rectangle CB, BD is equal to the square on AB. [VI. 17]

+

For the same reason the rectangle BC, CD is also equal to the square on AC.

+

And since, if in a right-angled triangle a perpendicular be drawn from the right angle to the base, the perpendicular so drawn is a mean proportional between the segments of the base, [VI. 8, Por.] therefore, as BD is to DA, so is AD to DC; therefore the rectangle BD, DC is equal to the square on AD. [VI. 17]

+

I say that the rectangle BC, AD is also equal to the rectangle BA, AC.

+

For since, as we said, ABC is similar to ABD, therefore, as BC is to CA, so is BA to AD. [VI. 4]

+

Therefore the rectangle BC, AD is equal to the rectangle BA, AC. [VI. 16] Q. E. D.

+ +
+
+ PROPOSITION 33. +

To find two straight lines incommensurable in square which make the sum of the squares on them rational but the rectangle contained by them medial.

+

Let there be set out two rational straight lines AB, BC commensurable in square only and such that the square on the greater AB is greater than the square on the less BC by the square on a straight line incommensurable with AB, [X. 30]

+ let BC be bisected at D, let there be applied to AB a parallelogram equal to the square on either of the straight lines BD, DC and deficient by a square figure, and let it be the rectangle AE, EB; [VI. 28] let the semicircle AFB be described on AB, let EF be drawn at right angles to AB, and let AF, FB be joined.

+

Then, since AB, BC are unequal straight lines, and the square on AB is greater than the square on BC by the square on a straight line incommensurable with AB, while there has been applied to AB a parallelogram equal to the fourth part of the square on BC, that is, to the square on half of it, and deficient by a square figure, making the rectangle AE, EB, therefore AE is incommensurable with EB. [X. 18]

+

And, as AE is to EB, so is the rectangle BA, AE to the rectangle AB, BE, while the rectangle BA, AE is equal to the square on AF, and the rectangle AB, BE to the square on BF; therefore the square on AF is incommensurable with the square on FB; therefore AF, FB are incommensurable in square.

+

And, since AB is rational, therefore the square on AB is also rational; so that the sum of the squares on AF, FB is also rational. [I. 47]

+

And since, again, the rectangle AE, EB is equal to the square on EF, and, by hypothesis, the rectangle AE, EB is also equal to the square on BD, therefore FE is equal to BD; therefore BC is double of FE, so that the rectangle AB, BC is also commensurable with the rectangle AB, EF.

+

But the rectangle AB, BC is medial; [X. 21] therefore the rectangle AB, EF is also medial. [X. 23, Por.]

+

But the rectangle AB, EF is equal to the rectangle AF, FB; [Lemma] therefore the rectangle AF, FB is also medial.

+

But it was also proved that the sum of the squares on these straight lines is rational.

+

Therefore two straight lines AF, FB incommensurable in square have been found which make the sum of the squares on them rational, but the rectangle contained by them medial. Q. E. D.

+
+
+ PROPOSITION 34. +

+ To find two straight lines incommensurable in square which make the sum of the squares on them medial but the rectangle contained by them rational. +

+

Let there be set out two medial straight lines AB, BC, commensurable in square only, such that the rectangle which they contain is rational, and the square on AB is greater than the square on BC by the square on a straight line incommensurable with AB; [X. 31, ad fin.]

let the semicircle ADB be described on AB, let BC be bisected at E, let there be applied to AB a parallelogram equal to the square on BE and deficient by a square figure, namely the rectangle AF, FB; [VI. 28] therefore AF is incommensurable in length with FB. [X. 18]

+

Let FD be drawn from F at right angles to AB, and let AD, DB be joined.

+

Since AF is incommensurable in length with FB, therefore the rectangle BA, AF is also incommensurable with the rectangle AB, BF. [X. 11]

+

But the rectangle BA, AF is equal to the square on AD, and the rectangle AB, BF to the square on DB; therefore the square on AD is also incommensurable with the square on DB.

+

And, since the square on AB is medial, therefore the sum of the squares on AD, DB is also medial. [III. 31, I. 47]

+

And, since BC is double of DF, therefore the rectangle AB, BC is also double of the rectangle AB, FD.

+

But the rectangle AB, BC is rational; therefore the rectangle AB, FD is also rational. [X. 6]

+

But the rectangle AB, FD is equal to the rectangle AD, DB; [Lemma] so that the rectangle AD, DB is also rational.

+

Therefore two straight lines AD, DB incommensurable in square have been found which make the sum of the squares on them medial, but the rectangle contained by them rational. Q. E. D.

+
+
+ PROPOSITION 35. +

To find two straight lines incommensurable in square which make the sum of the squares on them medial and the rectangle contained by them medial and moreover incommensurable with the sum of the squares on them.

+

Let there be set out two medial straight lines AB, BC commensurable in square only, containing a medial rectangle, and such that the square on AB is greater than the square on BC by the square on a straight line incommensurable with AB; [X. 32 +, ad fin.] let the semicircle ADB be described on AB, and let the rest of the construction be as above.

+

+

Then, since AF is incommensurable in length with FB, [X. 18 +] AD is also incommensurable in square with DB. [X. 11 +]

+

And, since the square on AB is medial, therefore the sum of the squares on AD, DB is also medial. [III. 31 +, I. 47 +]

+

And, since the rectangle AF, FB is equal to the square on each of the straight lines BE, DF, therefore BE is equal to DF; therefore BC is double of FD, so that the rectangle AB, BC is also double of the rectangle AB, FD.

+

But the rectangle AB, BC is medial; therefore the rectangle AB, FD is also medial. [X. 32, Por.]

+

And it is equal to the rectangle AD, DB; [Lemma after X. 32 +] therefore the rectangle AD, DB is also medial.

+

And, since AB is incommensurable in length with BC, while CB is commensurable with BE, therefore AB is also incommensurable in length with BE, [X. 13 +] so that the square on AB is also incommensurable with the rectangle AB, BE. [X. 11 +]

+

But the squares on AD, DB are equal to the square on AB, [I. 47 +] and the rectangle AB, FD, that is, the rectangle AD, DB, is equal to the rectangle AB, BE; therefore the sum of the squares on AD, DB is incommensurable with the rectangle AD, DB.

+

Therefore two straight lines AD, DB incommensurable in square have been found which make the sum of the squares on them medial and the rectangle contained by them medial and moreover incommensurable with the sum of the squares on them. Q. E. D.

+
+
+ PROPOSITION 36. +

If two rational straight lines commensurable in square only be added together, the whole is irrational; and let it be called + binomial.

+

For let two rational straight lines AB, BC commensurable in square only be added together;

I say that the whole AC is irrational.

+

For, since AB is incommensurable in length with BC— for they are commensurable in square only— and, as AB is to BC, so is the rectangle AB, BC to the square on BC, therefore the rectangle AB, BC is incommensurable with the square on BC. [X. 11 +]

+

But twice the rectangle AB, BC is commensurable with the rectangle AB, BC [X. 6 +], and the squares on AB, BC are commensurable with the square on BC—for AB, BC are rational straight lines commensurable in square only— [X. 15 +] therefore twice the rectangle AB, BC is incommensurable with the squares on AB, BC. [X. 13 +]

+

And, componendo, twice the rectangle AB, BC together with the squares on AB, BC, that is, the square on AC [II. 4 +], is incommensurable with the sum of the squares on AB, BC. [X. 16 +]

+

But the sum of the squares on AB, BC is rational; therefore the square on AC is irrational, so that AC is also irrational. [X. Def. 4 +]

+

And let it be called binomial.

+
+
+ PROPOSITION 37. +

If two medial straight lines commensurable in square only and containing a rational rectangle be added together, the whole is irrational; and let it be called + a first bimedial + straight line.

+

For let two medial straight lines AB, BC commensurable in square only and containing a rational rectangle be added together;

I say that the whole AC is irrational.

+

For, since AB is incommensurable in length with BC, therefore the squares on AB, BC are also incommensurable with twice the rectangle AB, BC; [cf. X. 36, ll. 9-20] +and, componendo, the squares on AB, BC together with twice the rectangle AB, BC, that is, the square on AC [II. 4], is incommensurable with the rectangle AB, BC. [X. 16 +]

+

But the rectangle AB, BC is rational, for, by hypothesis, AB, BC are straight lines containing a rational rectangle; therefore the square on AC is irrational; therefore AC is irrational. [X. Def. 4 +]

+

And let it be called a first bimedial straight line. Q. E. D.

+
+
+ PROPOSITION 38. +

If two medial straight lines commensurable in square only and containing a medial rectangle be added together, the whole is irrational; and let it be called a + second bimedial + straight line.

+

For let two medial straight lines AB, BC commensurable in square only and containing a medial rectangle be added together;

I say that AC is irrational.

+

For let a rational straight line DE be set out, and let the parallelogram DF equal to the square on AC be applied to DE, producing DG as breadth. [I. 44 +]

+

Then, since the square on AC is equal to the squares on AB, BC and twice the rectangle AB, BC, [II. 4 +] let EH, equal to the squares on AB, BC, be applied to DE; therefore the remainder HF is equal to twice the rectangle AB, BC.

+

And, since each of the straight lines AB, BC is medial, therefore the squares on AB, BC are also medial.

+

But, by hypothesis, twice the rectangle AB, BC is also medial.

+

And EH is equal to the squares on AB, BC, while FH is equal to twice the rectangle AB, BC; therefore each of the rectangle EH, HF is medial.

+

And they are applied to the rational straight line DE; therefore each of the straight lines DH, HG is rational and incommensurable in length with DE. [X. 22 +]

+

Since then AB is incommensurable in length with BC, and, as AB is to BC, so is the square on AB to the rectangle AB, BC, therefore the square on AB is incommensurable with the rectangle AB, BC. [X. 11 +]

+

But the sum of the squares on AB, BC is commensurable with the square on AB, [X. 15 +] and twice the rectangle AB, BC is commensurable with the rectangle AB, BC. [X. 6 +]

+

Therefore the sum of the squares on AB, BC is incommensurable with twice the rectangle AB, BC. [X. 13 +]

+

But EH is equal to the squares on AB, BC, and HF is equal to twice the rectangle AB, BC.

+

Therefore EH is incommensurable with HF, so that DH is also incommensurable in length with HG. [VI. 1 +, X. 11 +]

+

Therefore DH, HG are rational straight lines commensurable in square only; so that DG is irrational. [X. 36 +]

+

But DE is rational; and the rectangle contained by an irrational and a rational straight line is irrational; [cf. X. 20 +] therefore the area DF is irrational, and the side of the square equal to it is irrational. [X. Def. 4 +]

+

But AC is the side of the square equal to DF; therefore AC is irrational.

+

And let it be called a second bimedial straight line. Q. E. D.

+
+
+ PROPOSITION 39. +

If two straight lines incommensurable in square which make the sum of the squares on them rational, but the rectangle contained by them medial, be added together, the whole straight line is irrational : and let it be called + major.

+

For let two straight lines AB, BC incommensurable in square, and fulfilling the given conditions [X. 33 +], be added together; I say that AC is irrational.

+

+

For, since the rectangle AB, BC is medial, twice the rectangle AB, BC is also medial. [X. 6 + and 23, Por.]

+

But the sum of the squares on AB, BC is rational; therefore twice the rectangle AB, BC is incommensurable with the sum of the squares on AB, BC, so that the squares on AB, BC together with twice the rectangle AB, BC that is, the square on AC, is also incommensurable with the sum of the squares on AB, BC; [X. 16 +] therefore the square on AC is irrational, so that AC is also irrational. [X. Def. 4 +]

+

And let it be called major. Q. E. D.

+
+
+ PROPOSITION 40. +

If two straight lines incommensurable in square which make the sum of the squares on them medial, but the rectangle contained by them rational, be added together, the whole straight line is irrational; and let it be called the + side of a rational plus a medial area.

+

For let two straight lines AB, BC incommensurable in square, and fulfilling the given conditions [X. 34 +], be added together; I say that AC is irrational.

+

+

For, since the sum of the squares on AB, BC is medial, while twice the rectangle AB, BC is rational, therefore the sum of the squares on AB, BC is incommensurable with twice the rectangle AB, BC; so that the square on AC is also incommensurable with twice the rectangle AB, BC. [X. 16 +]

+

But twice the rectangle AB, BC is rational; therefore the square on AC is irrational.

+

Therefore AC is irrational. [X. Def. 4 +]

+

And let it be called the side of a rational plus a medial area. Q. E. D.

+
+
+ PROPOSITION 41. +

If two straight lines incommensurable in square which make the sum of the squares on them medial, and the rectangle contained by them medial and also incommensurable with the sum of the squares on them, be added together, the whole straight line is irrational; and let it be called the + side of the sum of two medial areas.

+

For let two straight lines AB, BC incommensurable in square and satisfying the given conditions [X. 35 +] be added together; I say that AC is irrational.

+

+

Let a rational straight line DE be set out, and let there be applied to DE the rectangle DF equal to the squares on AB, BC, and the rectangle GH equal to twice the rectangle AB, BC; therefore the whole DH is equal to the square on AC. [II. 4 +]

+

Now, since the sum of the squares on AB, BC is medial, and is equal to DF, therefore DF is also medial.

+

And it is applied to the rational straight line DE; therefore DG is rational and incommensurable in length with DE. [X. 22 +]

+

For the same reason GK is also rational and incommensurable in length with GF, that is, DE.

+

And, since the squares on AB, BC are incommensurable with twice the rectangle AB, BC, DF is incommensurable with GH; so that DG is also incommensurable with GK. [VI. 1 +, X. 11 +]

+

And they are rational; therefore DG, GK are rational straight lines commensurable in square only; therefore DK is irrational and what is called binomial. [X. 36 +]

+

But DE is rational; therefore DH is irrational, and the side of the square which is equal to it is irrational. [X. Def. 4 +]

+

But AC is the side of the square equal to HD; therefore AC is irrational.

+

And let it be called the side of the sum of two medial areas. Q. E. D. +

+ +

Lemma. And that the aforesaid irrational straight lines are divided only in one way into the straight lines of which they are the sum and which produce the types in question, we will now prove after premising the following lemma.

+

Let the straight line AB be set out, let the whole be cut into unequal parts at each of the points C, D,

and let AC be supposed greater than DB; I say that the squares on AC, CB are greater than the squares on AD, DB.

+

For let AB be bisected at E.

+

Then, since AC is greater than DB, let DC be subtracted from each; therefore the remainder AD is greater than the remainder CB.

+

But AE is equal to EB; therefore DE is less than EC; therefore the points C, D are not equidistant from the point of bisection.

+

And, since the rectangle AC, CB together with the square on EC is equal to the square on EB, [II. 5 +] and, further, the rectangle AD, DB together with the square on DE is equal to the square on EB, [id.] therefore the rectangle AC, CB together with the square on EC is equal to the rectangle AD, DB together with the square on DE.

+

And of these the square on DE is less than the square on EC; therefore the remainder, the rectangle AC, CB, is also less than the rectangle AD, DB, so that twice the rectangle AC, CB is also less than twice the rectangle AD, DB.

+

Therefore also the remainder, the sum of the squares on AC, CB, is greater than the sum of the squares on AD, DB. Q. E. D. + +

3. and which produce the types in question. The Greek is ποιουσῶν τὰ προκείμενα εἴδη, and I have taken εἴδη to mean types (of irrational straight lines), + though the expression might perhaps mean satisfying the conditions in question. +

+

+ +
+
+ PROPOSITION 42. + A binomial straight line is divided into its terms at one point only. +

Let AB be a binomial straight line divided into its terms at C; therefore AC, CB are rational straight lines commensurable in square only. [X. 36 +]

+

+

I say that AB is not divided at another point into two rational straight lines commensurable in square only.

+

For, if possible, let it be divided at D also, so that AD, DB are also rational straight lines commensurable in square only.

+

It is then manifest that AC is not the same with DB.

+

For, if possible, let it be so.

+

Then AD will also be the same as CB, and, as AC is to CB, so will BD be to DA; thus AB will be divided at D also in the same way as by the division at C: which is contrary to the hypothesis.

+

Therefore AC is not the same with DB.

+

For this reason also the points C, D are not equidistant from the point of bisection.

+

Therefore that by which the squares on AC, CB differ from the squares on AD, DB is also that by which twice the rectangle AD, DB differs from twice the rectangle AC, CB, because both the squares on AC, CB together with twice the rectangle AC, CB, and the squares on AD, DB together with twice the rectangle AD, DB, are equal to the square on AB. [II. 4 +]

+

But the squares on AC, CB differ from the squares on AD, DB by a rational area, for both are rational; therefore twice the rectangle AD, DB also differs from twice the rectangle AC, CB by a rational area, though they are medial [X. 21 +]: which is absurd, for a medial area does not exceed a medial by a rational area. [x. 26 +]

+

Therefore a binomial straight line is not divided at different points; therefore it is divided at one point only. Q. E. D.

+
+
+ PROPOSITION 43. +

A first bimedial straight line is divided at one point only.

+

Let AB be a first bimedial straight line divided at C, so that AC, CB are medial straight lines commensurable in square only and containing a rational rectangle;

I say that AB is not so divided at another point.

+

For, if possible, let it be divided at D also, so that AD, DB are also medial straight lines commensurable in square only and containing a rational rectangle.

+

Since, then, that by which twice the rectangle AD, DB differs from twice the rectangle AC, CB is that by which the squares on AC, CB differ from the squares on AD, DB, while twice the rectangle AD, DB differs from twice the rectangle AC, CB by a rational area—for both are rational— therefore the squares on AC, CB also differ from the squares on AD, DB by a rational area, though they are medial: which is absurd. [x. 26 +]

+

Therefore a first bimedial straight line is not divided into its terms at different points; therefore it is so divided at one point only.

+
+
+ PROPOSITION 44. +

A second bimedial straight line is divided at one point only.

+

Let AB be a second bimedial straight line divided at C, so that AC, CB are medial straight lines commensurable in square only and containing a medial rectangle; [X. 38 +] it is then manifest that C is not at the point of bisection, because the segments are not commensurable in length.

+

I say that AB is not so divided at another point.

+

+

For, if possible, let it be divided at D also, so that AC is not the same with DB, but AC is supposed greater; it is then clear that the squares on AD, DB are also, as we proved above [Lemma], less than the squares on AC, CB; and suppose that AD, DB are medial straight lines commensurable in square only and containing a medial rectangle.

+

Now let a rational straight line EF be set out, let there be applied to EF the rectangular parallelogram EK equal to the square on AB, and let EG equal to the squares on AC, CB be subtracted; therefore the remainder HK is equal to twice the rectangle AC, CB. [II. 4 +]

+

Again, let there be subtracted EL, equal to the squares on AD, DB, which were proved less than the squares on AC, CB [Lemma +]; therefore the remainder MK is also equal to twice the rectangle AD, DB.

+

Now, since the squares on AC, CB are medial, therefore EG is medial.

+

And it is applied to the rational straight line EF; therefore EH is rational and incommensurable in length with EF. [X. 22 +]

+

For the same reason HN is also rational and incommensurable in length with EF.

+

And, since AC, CB are medial straight lines commensurable in square only, therefore AC is incommensurable in length with CB.

+

But, as AC is to CB, so is the square on AC to the rectangle AC, CB; therefore the square on AC is incommensurable with the rectangle AC, CB. [X. 11 +]

+

But the squares on AC, CB are commensurable with the square on AC; for AC, CB are commensurable in square. [x. 15 +]

+

And twice the rectangle AC, CB is commensurable with the rectangle AC, CB. [X. 6 +]

+

Therefore the squares on AC, CB are also incommensurable with twice the rectangle AC, CB. [X. 13 +]

+

But EG is equal to the squares on AC, CB, and HK is equal to twice the rectangle AC, CB; therefore EG is incommensurable with HK, so that EH is also incommensurable in length with HN. [VI. 1 +, X. 11 +]

+

And they are rational; therefore EH, HN are rational straight lines commensurable in square only.

+

But, if two rational straight lines commensurable in square only be added together, the whole is the irrational which is called binomial. [X. 36 +]

+

Therefore EN is a binomial straight line divided at H.

+

In the same way EM, MN will also be proved to be rational straight lines commensurable in square only; and EN will be a binomial straight line divided at different points, H and M.

+

And EH is not the same with MN.

+

For the squares on AC, CB are greater than the squares on AD, DB.

+

But the squares on AD, DB are greater than twice the rectangle AD, DB; therefore also the squares on AC, CB, that is, EG, are much greater than twice the rectangle AD, DB, that is, MK, so that EH is also greater than MN.

+

Therefore EH is not the same with MN. Q. E. D.

+
+ +
+ PROPOSITION 45. +

A major straight line is divided at one and the same point only.

+

Let AB be a major straight line divided at C, so that AC, CB are incommensurable in square and make the sum of the squares on AC, CB rational, but the rectangle AC, CB medial; [X. 39 +]

I say that AB is not so divided at another point.

+

For, if possible, let it be divided at D also, so that AD, DB are also incommensurable in square and make the sum of the squares on AD, DB rational, but the rectangle contained by them medial.

+

Then, since that by which the squares on AC, CB differ from the squares on AD, DB is also that by which twice the rectangle AD, DB differs from twice the rectangle AC, CB, while the squares on AC, CB exceed the squares on AD, DB by a rational area—for both are rational— therefore twice the rectangle AD, DB also exceeds twice the rectangle AC, CB by a rational area, though they are medial: which is impossible. [X. 26 +]

+

Therefore a major straight line is not divided at different points; therefore it is only divided at one and the same point. Q. E. D.

+
+ +
+ PROPOSITION 46. +

The side of a rational plus a medial area is divided at one point only.

+

Let AB be the side of a rational plus a medial area divided at C, so that AC, CB are incommensurable in square and make the sum of the squares on AC, CB medial, but twice the rectangle AC, CB rational; [X. 40 +] I say that AB is not so divided at another point.

+

+

For, if possible, let it be divided at D also, so that AD, DB are also incommensurable in square and make the sum of the squares on AD, DB medial, but twice the rectangle AD, DB rational.

+

Since then that by which twice the rectangle AC, CB differs from twice the rectangle AD, DB is also that by which the squares on AD, DB differ from the squares on AC, CB, while twice the rectangle AC, CB exceeds twice the rectangle AD, DB by a rational area, therefore the squares on AD, DB also exceed the squares on AC, CB by a rational area, though they are medial: which is impossible. [X. 26 +]

+

Therefore the side of a rational plus a medial area is not divided at different points; therefore it is divided at one point only. Q. E. D.

+
+
+ PROPOSITION 47. +

The side of the sum of two medial areas is divided at one point only.

+

Let AB be divided at C, so that AC, CB are incommensurable in square and make the sum of the squares on AC, CB medial, and the rectangle AC, CB medial and also incommensurable with the sum of the squares on them; I say that AB is not divided at another point so as to fulfil the given conditions.

+

+

For, if possible, let it be divided at D, so that again AC is of course not the same as BD, but AC is supposed greater; let a rational straight line EF be set out, and let there be applied to EF the rectangle EG equal to the squares on AC, CB, and the rectangle HK equal to twice the rectangle AC, CB; therefore the whole EK is equal to the square on AB. [II. 4 +]

+

Again, let EL, equal to the squares on AD, DB, be applied to EF; therefore the remainder, twice the rectangle AD, DB, is equal to the remainder MK.

+

And since, by hypothesis, the sum of the squares on AC, CB is medial, therefore EG is also medial.

+

And it is applied to the rational straight line EF; therefore HE is rational and incommensurable in length with EF. [X. 22 +]

+

For the same reason HN is also rational and incommensurable in length with EF.

+

And, since the sum of the squares on AC, CB is incommensurable with twice the rectangle AC, CB, therefore EG is also incommensurable with GN, so that EH is also incommensurable with HN. [VI. 1 +, X. 11 +]

+

And they are rational; therefore EH, HN are rational straight lines commensurable in square only; therefore EN is a binomial straight line divided at H. [X. 36 +]

+

Similarly we can prove that it is also divided at M. +

+

And EH is not the same with MN; therefore a binomial has been divided at different points: which is absurd. [X. 42 +]

+

Therefore a side of the sum of two medial areas is not divided at different points; therefore it is divided at one point only.

+
+
+
+ DEFINITIONS II. +
+1 +

Given a rational straight line and a binomial, divided into its terms, such that the square on the greater term is greater than the square on the lesser by the square on a straight line commensurable in length with the greater, then, if the greater term be commensurable in length with the rational straight line set out, let the whole be called a first binomial straight line;

+
+
+2 +

but if the lesser term be commensurable in length with the rational straight line set out, let the whole be called a second binomial;

+
+ +
+3 +

and if neither of the terms be commensurable in length with the rational straight line set out, let the whole be called a third binomial.

+
+
+4 +

Again, if the square on the greater term be greater than the square on the lesser by the square on a straight line incommensurable in length with the greater, then, if the greater term be commensurable in length with the rational straight line set out, let the whole be called a fourth binomial;

+
+
+5 +

if the lesser, a fifth binomial;

+
+
+6 +

and if neither, a sixth binomial.

+
+
+
+ PROPOSITIONS 48—84. +
+ PROPOSITION 48. +

To find the first binomial straight line.

+

Let two numbers AC, CB be set out such that the sum of them AB has to BC the ratio which a square number has to a square number, but has not to CA the ratio which a square number has to a square number; [Lemma I after X. 28]

let any rational straight line D be set out, and let EF be commensurable in length with D.

+

Therefore EF is also rational.

+

Let it be contrived that, as the number BA is to AC, so is the square on EF to the square on FG. [X. 6, Por.]

+

But AB has to AC the ratio which a number has to a number; therefore the square on EF also has to the square on FG the ratio which a number has to a number, so that the square on EF is commensurable with the square on FG. [X. 6]

+

And EF is rational; therefore FG is also rational.

+

And, since BA has not to AC the ratio which a square number has to a square number. neither, therefore, has the square on EF to the square on FG the ratio which a square number has to a square number; therefore EF is incommensurable in length with FG. [X. 9]

+

Therefore EF, FG are rational straight lines commensurable in square only; therefore EG is binomial. [X. 36]

+

I say that it is also a first binomial straight line.

+

For since, as the number BA is to AC, so is the square on EF to the square on FG, while BA is greater than AC, therefore the square on EF is also greater than the square on FG.

+

Let then the squares on FG, H be equal to the square on EF.

+

Now since, as BA is to AC, so is the square on EF to the square on FG, therefore, convertendo, as AB is to BC, so is the square on EF to the square on H. [V. 19, Por.]

+

But AB has to BC the ratio which a square number has to a square number; therefore the square on EF also has to the square on H the ratio which a square number has to a square number.

+

Therefore EF is commensurable in length with H; [X. 9] therefore the square on EF is greater than the square on FG by the square on a straight line commensurable with EF.

+

And EF, FG are rational, and EF is commensurable in length with D.

+

Therefore EF is a first binomial straight line. Q. E. D.

+
+
+ PROPOSITION 49. +

To find the second binomial straight line.

+

Let two numbers AC, CB be set out such that the sum of them AB has to BC the ratio which a square number has to a square number, but has not to AC the ratio which a square number has to a square number; let a rational straight line D be set out, and let EF be commensurable in length with D; therefore EF is rational.

+

+

Let it be contrived then that, as the number CA is to AB, so also is the square on EF to the square on FG; [X. 6, Por.] therefore the square on EF is commensurable with the square on FG. [X. 6]

+

Therefore FG is also rational.

+

Now, since the number CA has not to AB the ratio which a square number has to a square number, neither has the square on EF to the square on FG the ratio which a square number has to a square number.

+

Therefore EF is incommensurable in length with FG; [X. 9] therefore EF, FG are rational straight lines commensurable in square only; therefore EG is binomial. [X. 36]

+

It is next to be proved that it is also a second binomial straight line.

+

For since, inversely, as the number BA is to AC, so is the square on GF to the square on FE, while BA is greater than AC, therefore the square on GF is greater than the square on FE.

+

Let the squares on EF, H be equal to the square on GF; therefore, convertendo, as AB is to BC, so is the square on FG to the square on H. [V. 19, Por.]

+

But AB has to BC the ratio which a square number has to a square number; therefore the square on FG also has to the square on H the ratio which a square number has to a square number.

+

Therefore FG is commensurable in length with H; [X. 9] so that the square on FG is greater than the square on FE by the square on a straight line commensurable with FG.

+

And FG, FE are rational straight lines commensurable in square only, and EF, the lesser term, is commensurable in length with the rational straight line D set out.

+

Therefore EG is a second binomial straight line. Q. E. D.

+
+
+ PROPOSITION 50. +

To find the third binomial straight line.

+

Let two numbers AC, CB be set out such that the sum of them AB has to BC the ratio which a square number has to a square number, but has not to AC the ratio which a square number has to a square number.

+

+

Let any other number D, not square, be set out also, and let it not have to either of the numbers BA. AC the ratio which a square number has to a square number.

+

Let any rational straight line E be set out, and let it be contrived that, as D is to AB, so is the square on E to the square on FG; [X. 6, Por.] therefore the square on E is commensurable with the square on FG. [X. 6]

+

And E is rational; therefore FG is also rational.

+

And, since D has not to AB the ratio which a square number has to a square number, neither has the square on E to the square on FG the ratio which a square number has to a square number; therefore E is incommensurable in length with FG. [X. 9]

+

Next let it be contrived that, as the number BA is to AC, so is the square on FG to the square on GH; [X. 6, Por.] therefore the square on FG is commensurable with the square on GH. [X. 6]

+

But FG is rational; therefore GH is also rational.

+

And, since BA has not to AC the ratio which a square number has to a square number, neither has the square on FG to the square on HG the ratio which a square number has to a square number; therefore FG is incommensurable in length with GH. [X. 9]

+

Therefore FG, GH are rational straight lines commensurable in square only; therefore FH is binomial. [X. 36]

+

I say next that it is also a third binomial straight line.

+

For since, as D is to AB, so is the square on E to the square on FG, and, as BA is to AC, so is the square on FG to the square on GH, therefore, ex aequali, as D is to AC, so is the square on E to the square on GH. [V. 22]

+

But D has not to AC the ratio which a square number has to a square number; therefore neither has the square on E to the square on GH the ratio which a square number has to a square number; therefore E is incommensurable in length with GH. [X. 9]

+

And since, as BA is to AC, so is the square on FG to the square on GH, therefore the square on FG is greater than the square on GH.

+

Let then the squares on GH, K be equal to the square on FG; therefore, convertendo, as AB is to BC, so is the square on FG to the square on K. [V. 19, Por.]

+

But AB has to BC the ratio which a square number has to a square number; therefore the square on FG also has to the square on K the ratio which a square number has to a square number; therefore FG is commensurable in length with K. [X. 9]

+

Therefore the square on FG is greater than the square on GH by the square on a straight line commensurable with FG.

+

And FG, GH are rational straight lines commensurable in square only, and neither of them is commensurable in length with E.

+

Therefore FH is a third binomial straight line. Q. E. D.

+
+
+ PROPOSITION 51. +

To find the fourth binomial straight line.

+

Let two numbers AC, CB be set out such that AB neither has to BC, nor yet to AC, the ratio which a square number has to a square number.

+

Let a rational straight line D be set out, and let EF be commensurable in length with D; therefore EF is also rational.

+

Let it be contrived that, as the number BA is to AC, so is the square on EF to the square on FG; [X. 6, Por.] therefore the square on EF is commensurable with the square on FG; [X. 6] therefore FG is also rational.

+

+

Now, since BA has not to AC the ratio which a square number has to a square number, neither has the square on EF to the square on FG the ratio which a square number has to a square number; therefore EF is incommensurable in length with FG. [X. 9]

+

Therefore EF, FG are rational straight lines commensurable in square only; so that EG is binomial.

+

I say next that it is also a fourth binomial straight line.

+

For since, as BA is to AC, so is the square on EF to the square on FG, therefore the square on EF is greater than the square on FG.

+

Let then the squares on FG, H be equal to the square on EF; therefore, convertendo, as the number AB is to BC, so is the square on EF to the square on H. [V. 19, Por.]

+

But AB has not to BC the ratio which a square number has to a square number; therefore neither has the square on EF to the square on H the ratio which a square number has to a square number.

+

Therefore EF is incommensurable in length with H; [X. 9] therefore the square on EF is greater than the square on GF by the square on a straight line incommensurable with EF.

+

And EF, FG are rational straight lines commensurable in square only, and EF is commensurable in length with D.

+

Therefore EG is a fourth binomial straight line. Q. E. D.

+
+
+ PROPOSITION 52. +

To find the fifth binomial straight line.

+

Let two numbers AC, CB be set out such that AB has not to either of them the ratio which a square number has to a square number; let any rational straight line D be set out, and let EF be commensurable with D; therefore EF is rational.

+

Let it be contrived that, as CA is to AB, so is the square on EF to the square on FG. [X. 6, Por.]

+

But CA has not to AB the ratio which a square number has to a square number; therefore neither has the square on EF to the square on FG the ratio which a square number has to a square number.

+

Therefore EF, FG are rational straight lines commensurable in square only; [X. 9] therefore EG is binomial. [X. 36]

+

+

I say next that it is also a fifth binomial straight line.

+

For since, as CA is to AB, so is the square on EF to the square on FG, inversely, as BA is to AC, so is the square on FG to the square on FE; therefore the square on GF is greater than the square on FE.

+

Let then the squares on EF, H be equal to the square on GF; therefore, convertendo, as the number AB is to BC, so is the square on GF to the square on H. [V. 19, Por.]

+

But AB has not to BC the ratio which a square number has to a square number; therefore neither has the square on FG to the square on H the ratio which a square number has to a square number.

+

Therefore FG is incommensurable in length with H; [X. 9] so that the square on FG is greater than the square on FE by the square on a straight line incommensurable with FG.

+

And GF, FE are rational straight lines commensurable in square only, and the lesser term EF is commensurable in length with the rational straight line D set out.

+

Therefore EG is a fifth binomial straight line. Q. E. D.

+
+
+ PROPOSITION 53. +

+ To find the sixth binomial straight line. +

+

Let two numbers AC, CB be set out such that AB has not to either of them the ratio which a square number has to a square number; and let there also be another number D which is not square and which has not to either of the numbers BA, AC the ratio which a square number has to a square number.

+

+

Let any rational straight line E be set out, and let it be contrived that, as D is to AB, so is the square on E to the square on FG; [X. 6, Por.] therefore the square on E is commensurable with the square on FG. [X. 6]

+

And E is rational; therefore FG is also rational.

+

Now, since D has not to AB the ratio which a square number has to a square number, neither has the square on E to the square on FG the ratio which a square number has to a square number; therefore E is incommensurable in length with FG. [X. 9]

+

Again, let it be contrived that, as BA is to AC, so is the square on FG to the square on GH. [X. 6, Por.]

+

Therefore the square on FG is commensurable with the square on HG. [X. 6]

+

Therefore the square on HG is rational; therefore HG is rational.

+

And, since BA has not to AC the ratio which a square number has to a square number, neither has the square on FG to the square on GH the ratio which a square number has to a square number; therefore FG is incommensurable in length with GH. [X. 9]

+

Therefore FG, GH are rational straight lines commensurable in square only; therefore FH is binomial. [X. 36]

+

It is next to be proved that it is also a sixth binomial straight line.

+

For since, as D is to AB, so is the square on E to the square on FG, and also, as BA is to AC, so is the square on FG to the square on GH, therefore, ex aequali, as D is to AC, so is the square on E to the square on GH. [V. 22]

+

But D has not to AC the ratio which a square number has to a square number; therefore neither has the square on E to the square on GH the ratio which a square number has to a square number; therefore E is incommensurable in length with GH. [X. 9]

+

But it was also proved incommensurable with FG; therefore each of the straight lines FG, GH is incommensurable in length with E.

+

And, since, as BA is to AC, so is the square on FG to the square on GH, therefore the square on FG is greater than the square on GH.

+

Let then the squares on GH, K be equal to the square on FG; therefore, convertendo, as AB is to BC, so is the square on FG to the square on K. [V. 19, Por.]

+

But AB has not to BC the ratio which a square number has to a square number; so that neither has the square on FG to the square on K the ratio which a square number has to a square number.

+

Therefore FG is incommensurable in length with K; [X. 9] therefore the square on FG is greater than the square on GH by the square on a straight line incommensurable with FG.

+

And FG, GH are rational straight lines commensurable in square only, and neither of them is commensurable in length with the rational straight line E set out.

+

Therefore FH is a sixth binomial straight line. Q. E. D. +

+ +

Lemma. Let there be two squares AB, BC, and let them be placed so that DB is in a straight line with BE; therefore FB is also in a straight line with BG.

+

Let the parallelogram AC be completed; I say that AC is a square, that DG is a mean proportional between AB, BC, and further that DC is a mean proportional between AC, CB.

+

+

For, since DB is equal to BF, and BE to BG, therefore the whole DE is equal to the whole FG.

+

But DE is equal to each of the straight lines AH, KC, and FG is equal to each of the straight lines AK, HC; [I. 34] therefore each of the straight lines AH, KC is also equal to each of the straight lines AK, HC.

+

Therefore the parallelogram AC is equilateral.

+

And it is also rectangular; therefore AC is a square.

+

And since, as FB is to BG, so is DB to BE, while, as FB is to BG, so is AB to DG, and, as DB is to BE, so is DG to BC, [VI. 1] therefore also, as AB is to DG, so is DG to BC. [V. 11]

+

Therefore DG is a mean proportional between AB, BC.

+

I say next that DC is also a mean proportional between AC, CB.

+

For since, as AD is to DK, so is KG to GC— for they are equal respectively— and, componendo, as AK is to KD, so is KC to CG, [V. 18] while, as AK is to KD, so is AC to CD, and, as KC is to CG, so is DC to CB, [VI. 1] therefore also, as AC is to DC, so is DC to BC. [V. 11]

+

Therefore DC is a mean proportional between AC, CB. Being what it was proposed to prove.

+ +
+
+ PROPOSITION 54. +

If an area be contained by a rational straight line and the first binomial, the side + of the area is the irrational straight line which is called binomial.

+

For let the area AC be contained by the rational straight line AB and the first binomial AD; I say that the side + of the area AC is the irrational straight line which is called binomial.

+

For, since AD is a first binomial straight line, let it be divided into its terms at E, and let AE be the greater term.

+

It is then manifest that AE, ED are rational straight lines commensurable in square only, the square on AE is greater than the square on ED by the square on a straight line commensurable with AE, and AE is commensurable in length with the rational straight line AB set out. [X. Deff. II. 1]

+

Let ED be bisected at the point F.

+

+

Then, since the square on AE is greater than the square on ED by the square on a straight line commensurable with AE, therefore, if there be applied to the greater AE a parallelogram equal to the fourth part of the square on the less, that is, to the square on EF, and deficient by a square figure, it divides it into commensurable parts. [X. 17]

+

Let then the rectangle AG, GE equal to the square on EF be applied to AE; therefore AG is commensurable in length with EG.

+

Let GH, EK, FL be drawn from G, E, F parallel to either of the straight lines AB, CD; let the square SN be constructed equal to the parallelogram AH, and the square NQ equal to GK, [II. 14] and let them be placed so that MN is in a straight line with NO; therefore RN is also in a straight line with NP.

+

And let the parallelogram SQ be completed; therefore SQ is a square. [Lemma]

+

Now, since the rectangle AG, GE is equal to the square on EF, therefore, as AG is to EF, so is FE to EG; [VI. 17] therefore also, as AH is to EL, so is EL to KG; [VI. 1] therefore EL is a mean proportional between AH, GK.

+

But AH is equal to SN, and GK to NQ; therefore EL is a mean proportional between SN, NQ.

+

But MR is also a mean proportional between the same SN, NQ; [Lemma] therefore EL is equal to MR, so that it is also equal to PO.

+

But AH, GK are also equal to SN, NQ; therefore the whole AC is equal to the whole SQ, that is, to the square on MO; therefore MO is the side + of AC.

+

I say next that MO is binomial.

+

For, since AG is commensurable with GE, therefore AE is also commensurable with each of the straight lines AG, GE. [X. 15]

+

But AE is also, by hypothesis, commensurable with AB; therefore AG, GE are also commensurable with AB. [X. 12]

+

And AB is rational; therefore each of the straight lines AG, GE is also rational; therefore each of the rectangles AH, GK is rational, [X. 19] and AH is commensurable with GK.

+

But AH is equal to SN, and GK to NQ; therefore SN, NQ, that is, the squares on MN, NO, are rational and commensurable.

+

And, since AE is incommensurable in length with ED, while AE is commensurable with AG, and DE is commensurable with EF, therefore AG is also incommensurable with EF, [X. 13] so that AH is also incommensurable with EL. [VI. 1, X. 11]

+

But AH is equal to SN, and EL to MR; therefore SN is also incommensurable with MR.

+

But, as SN is to MR, so is PN to NR; [VI. 1] therefore PN is incommensurable with NR. [X. 11]

+

But PN is equal to MN, and NR to NO; therefore MN is incommensurable with NO.

+

And the square on MN is commensurable with the square on NO, and each is rational; therefore MN, NO are rational straight lines commensurable in square only.

+

Therefore MO is binomial [X. 36] and the side + of AC. Q. E. D. +2. side. + I use the word side + in the sense explained in the note on X. Def. 4 (P. 13 above), i.e. as short for side of a square equal to. + The Greek is ἡ τὸ χωίον δυναμένη.

+
+
+ PROPOSITION 55. +

+ If an area be contained by a rational straight line and the second binomial, the side + of the area is the irrational straight line which is called a first bimedial. +

+

For let the area ABCD be contained by the rational straight line AB and the second binomial AD; I say that the side + of the area AC is a first bimedial straight line.

+

For, since AD is a second binomial straight line, let it be divided into its terms at E, so that AE is the greater term; therefore AE, ED are rational straight lines commensurable in square only, the square on AE is greater than the square on ED by the square on a straight line commensurable with AE, and the lesser term ED is commensurable in length with AB. [X. Deff. II. 2]

+

Let ED be bisected at F, and let there be applied to AE the rectangle AG, GE equal to the square on EF and deficient by a square figure; therefore AG is commensurable in length with GE. [X. 17]

+

Through G, E, F let GH, EK, FL be drawn parallel to AB, CD, let the square SN be constructed equal to the parallelogram AH, and the square NQ equal to GK, and let them be placed so that MN is in a straight line with NO; therefore RN is also in a straight line with NP.

+

+

Let the square SQ be completed.

+

It is then manifest from what was proved before that MR is a mean proportional between SN, NQ and is equal to EL, and that MO is the side + of the area AC.

+

It is now to be proved that MO is a first bimedial straight line.

+

Since AE is incommensurable in length with ED, while ED is commensurable with AB, therefore AE is incommensurable with AB. [X. 13]

+

And, since AG is commensurable with EG, AE is also commensurable with each of the straight lines AG, GE. [X. 15]

+

But AE is incommensurable in length with AB; therefore AG, GE are also incommensurable with AB. [X. 13]

+

Therefore BA, AG and BA, GE are pairs of rational straight lines commensurable in square only; so that each of the rectangles AH, GK is medial. [X. 21]

+

Hence each of the squares SN, NQ is medial.

+

Therefore MN, NO are also medial.

+

And, since AG is commensurable in length with GE, AH is also commensurable with GK, [VI. 1. X. 11] that is, SN is commensurable with NQ, that is, the square on MN with the square on NO.

+

And, since AE is incommensurable in length with ED, while AE is commensurable with AG, and ED is commensurable with EF, therefore AG is incommensurable with EF; [X. 13] so that AH is also incommensurable with EL, that is, SN is incommensurable with MR, that is, PN with NR, [VI. 1, X. 11] that is, MN is incommensurable in length with NO.

+

But MN, NO were proved to be both medial and commensurable in square; therefore MN, NO are medial straight lines commensurable in square only.

+

I say next that they also contain a rational rectangle.

+

For, since DE is, by hypothesis, commensurable with each of the straight lines AB, EF, therefore EF is also commensurable with EK. [X. 12]

+

And each of them is rational; therefore EL, that is, MR is rational, [X. 19] and MR is the rectangle MN, NO.

+

But, if two medial straight lines commensurable in square only and containing a rational rectangle be added together, the whole is irrational and is called a first bimedial straight line. [X. 37]

+

Therefore MO is a first bimedial straight line. Q. E. D. +39. Therefore BA, AG and BA, GE are pairs of rational straight lines commensurable in square only. The text has Therefore BA, AG, GE are rational straight lines commensurable in square only, + which I have altered because it would naturally convey the impression that any two of the three straight lines are commensurable in square only, whereas AG, GE are commensurable in length (I. 18), and it is only the other two pairs which are commensurable in square only.

+
+ +
+ PROPOSITION 56. +

+ If an area be contained by a rational straight line and the third binomial, the side + of the area is the irrational straight line called a second bimedial. +

+

For let the area ABCD be contained by the rational straight line AB and the third binomial AD divided into its terms at E, of which terms AE is the greater; I say that the side + of the area AC is the irrational straight line called a second bimedial.

+

For let the same construction be made as before.

+

+

Now, since AD is a third binomial straight line, therefore AE, ED are rational straight lines commensurable in square only, the square on AE is greater than the square on ED by the square on a straight line commensurable with AE, and neither of the terms AE, ED is commensurable in length with AB. [X. Deff. II. 3]

+

Then, in manner similar to the foregoing, we shall prove that MO is the side + of the area AC, and MN, NO are medial straight lines commensurable in square only; so that MO is bimedial.

+

It is next to be proved that it is also a second bimedial straight line.

+

Since DE is incommensurable in length with AB, that is, with EK, and DE is commensurable with EF, therefore EF is incommensurable in length with EK. [X. 13]

+

And they are rational; therefore FE, EK are rational straight lines commensurable in square only.

+

Therefore EL, that is, MR, is medial. [X. 21]

+

And it is contained by MN, NO; therefore the rectangle MN, NO is medial.

+

Therefore MO is a second bimedial straight line. [X. 38] Q. E. D.

+
+
+ PROPOSITION 57. +

+ If an area be contained by a rational straight line and the fourth binomial, the side + of the area is the irrational straight line called major. +

+

For let the area AC be contained by the rational straight line AB and the fourth binomial AD divided into its terms at E, of which terms let AE be the greater; I say that the side + of the area AC is the irrational straight line called major.

+

For, since AD is a fourth binomial straight line, therefore AE, ED are rational straight lines commensurable in square only, the square on AE is greater than the square on ED by the square on a straight line incommensurable with AE, and AE is commensurable in length with AB. [X. Deff. II. 4]

+

Let DE be bisected at F, and let there be applied to AE a parallelogram, the rectangle AG, GE, equal to the square on EF; therefore AG is incommensurable in length with GE. [X. 18]

+

Let GH, EK, FL be drawn parallel to AB, and let the rest of the construction be as before; it is then manifest that MO is the side + of the area AC.

+

+

It is next to be proved that MO is the irrational straight line called major.

+

Since AG is incommensurable with EG, AH is also incommensurable with GK, that is, SN with NQ; [VI. 1, X. 11] therefore MN, NO are incommensurable in square.

+

And, since AE is commensurable with AB, AK is rational; [X. 19] and it is equal to the squares on MN, NO; therefore the sum of the squares on MN, NO is also rational.

+

And, since DE is incommensurable in length with AB, that is, with EK, while DE is commensurable with EF, therefore EF is incommensurable in length with EK. [X. 13]

+

Therefore EK, EF are rational straight lines commensurable in square only; therefore LE, that is, MR, is medial. [X. 21]

+

And it is contained by MN, NO; therefore the rectangle MN, NO is medial.

+

And the [sum] of the squares on MN, NO is rational, and MN, NO are incommensurable in square.

+

But, if two straight lines incommensurable in square and making the sum of the squares on them rational, but the rectangle contained by them medial, be added together, the whole is irrational and is called major. [X. 39]

+

Therefore MO is the irrational straight line called major and is the side + of the area AC. Q. E. D.

+
+ +
+ PROPOSITION 58. +

+ If an area be contained by a rational straight line and the fifth binomial, the side + of the area is the irrational straight line called the side of a rational plus a medial area. +

+

For let the area AC be contained by the rational straight line AB and the fifth binomial AD divided into its terms at E, so that AE is the greater term; I say that the side + of the area AC is the irrational straight line called the side of a rational plus a medial area.

+

For let the same construction be made as before shown; it is then manifest that MO is the side + of the area AC. +

+

+

It is then to be proved that MO is the side of a rational plus a medial area.

+

For, since AG is incommensurable with GE, [X. 18] therefore AH is also commensurable with HE, [VI. 1, X. 11] that is, the square on MN with the square on NO; therefore MN, NO are incommensurable in square.

+

And, since AD is a fifth binomial straight line, and ED the lesser segment, therefore ED is commensurable in length with AB. [X. Deff. II. 5]

+

But AE is incommensurable with ED; therefore AB is also incommensurable in length with AE. [X. 13]

+

Therefore AK, that is, the sum of the squares on MN, NO, is medial. [X. 21]

+

And, since DE is commensurable in length with AB, that is, with EK, while DE is commensurable with EF, therefore EF is also commensurable with EK. [X. 12]

+

And EK is rational; therefore EL, that is, MR, that is, the rectangle MN, NO, is also rational. [X. 19]

+

Therefore MN, NO are straight lines incommensurable in square which make the sum of the squares on them medial, but the rectangle contained by them rational.

+

Therefore MO is the side of a rational plus a medial area [X. 40] and is the side + of the area AC. Q. E. D.

+
+ +
+ PROPOSITION 59. +

+ If an area be contained by a rational straight line and the sixth binomial, the side + of the area is the irrational straight line called the side of the sum of two medial areas. +

+

For let the area ABCD be contained by the rational straight line AB and the sixth binomial AD, divided into its terms at E, so that AE is the greater term; I say that the side + of AC is the side of the sum of two medial areas.

+

Let the same construction be made as before shown.

+

+

It is then manifest that MO is the side + of AC, and that MN is incommensurable in square with NO. +

+

Now, since EA is incommensurable in length with AB, therefore EA, AB are rational straight lines commensurable in square only; therefore AK, that is, the sum of the squares on MN, NO, is medial. [X. 21]

+

Again, since ED is incommensurable in length with AB, therefore FE is also incommensurable with EK; [X. 13] therefore FE, EK are rational straight lines commensurable in square only; therefore EL, that is, MR, that is, the rectangle MN, NO, is medial. [X. 21]

+

And, since AE is incommensurable with EF, AK is also incommensurable with EL. [VI. 1, X. 11]

+

But AK is the sum of the squares on MN, NO, and EL is the rectangle MN, NO; therefore the sum of the squares on MN, NO is incommensurable with the rectangle MN, NO. +

+

And each of them is medial, and MN, NO are incommensurable in square.

+

Therefore MO is the side of the sum of two medial areas [X. 41], and is the side + of AC. Q. E. D. +

+ +

[LEMMA. If a straight line be cut into unequal parts, the squares on the unequal parts are greater than twice the rectangle contained by the unequal parts.

+

+

Let AB be a straight line, and let it be cut into unequal parts at C, and let AC be the greater; I say that the squares on AC, CB are greater than twice the rectangle AC, CB.

+

For let AB be bisected at D.

+

Since then a straight line has been cut into equal parts at D, and into unequal parts at C, therefore the rectangle AC, CB together with the square on CD is equal to the square on AD, [II. 5] so that the rectangle AC, CB is less than double of the square on AD.

+

But the squares on AC, CB are double of the squares on AD, DC; [II. 9] therefore the squares on AC, CB are greater than twice the rectangle AC, CB. Q. E. D.]

+ +
+
+ PROPOSITION 60. +

The square on the binomial straight line applied to a rational straight line produces as breadth the first binomial.

+

Let AB be a binomial straight line divided into its terms at C, so that AC is the greater term; let a rational straight line DE be set out, and let DEFG equal to the square on AB be applied to DE producing DG as its breadth; I say that DG is a first binomial straight line.

+

+

For let there be applied to DE the rectangle DH equal to the square on AC, and KL equal to the square on BC; therefore the remainder, twice the rectangle AC, CB, is equal to MF.

+

Let MG be bisected at N, and let NO be drawn parallel [to ML or GF].

+

Therefore each of the rectangles MO, NF is equal to once the rectangle AC, CB.

+

Now, since AB is a binomial divided into its terms at C, therefore AC, CB are rational straight lines commensurable in square only; [X. 36] therefore the squares on AC, CB are rational and commensurable with one another, so that the sum of the squares on AC, CB is also rational. [X. 15]

+

And it is equal to DL; therefore DL is rational.

+

And it is applied to the rational straight line DE; therefore DM is rational and commensurable in length with DE. [X. 20]

+

Again, since AC, CB are rational straight lines commensurable in square only, therefore twice the rectangle AC, CB, that is MF, is medial. [X. 21]

+

And it is applied to the rational straight line ML; therefore MG is also rational and incommensurable in length with ML, that is, DE. [X. 22]

+

But MD is also rational and is commensurable in length with DE; therefore DM is incommensurable in length with MG. [X. 13]

+

And they are rational; therefore DM, MG are rational straight lines commensurable in square only; therefore DG is binomial. [X. 36]

+

It is next to be proved that it is also a first binomial straight line.

+

Since the rectangle AC, CB is a mean proportional between the squares on AC, CB, [cf. Lemma after X. 53] therefore MO is also a mean proportional between DH, KL.

+

Therefore, as DH is to MO, so is MO to KL, that is, as DK is to MN, so is MN to MK; [VI. 1] therefore the rectangle DK, KM is equal to the square on MN. [VI. 17]

+

And, since the square on AC is commensurable with the square on CB, DH is also commensurable with KL, so that DK is also commensurable with KM. [VI. 1, X. 11]

+

And, since the squares on AC, CB are greater than twice the rectangle AC, CB, [Lemma] therefore DL is also greater than MF, so that DM is also greater than MG. [VI. 1]

+

And the rectangle DK, KM is equal to the square on MN, that is, to the fourth part of the square on MG, and DK is commensurable with KM.

+

But, if there be two unequal straight lines, and to the greater there be applied a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, and if it divide it into commensurable parts, the square on the greater is greater than the square on the less by the square on a straight line commensurable with the greater; [X. 17] therefore the square on DM is greater than the square on MG by the square on a straight line commensurable with DM.

+

And DM, MG are rational, and DM, which is the greater term, is commensurable in length with the rational straight line DE set out.

+

Therefore DG is a first binomial straight line. [X. Deff. II. 1] Q. E. D.

+
+
+ PROPOSITION 61. +

The square on the first bimedial straight line applied to a rational straight line produces as breadth the second binomial.

+

Let AB be a first bimedial straight line divided into its medials at C, of which medials AC is the greater; let a rational straight line DE be set out, and let there be applied to DE the parallelogram DF equal to the square on AB, producing DG as its breadth; I say that DG is a second binominal straight line.

+

+

For let the same construction as before be made.

+

Then, since AB is a first bimedial divided at C, therefore AC, CB are medial straight lines commensurable in square only, and containing a rational rectangle, [X. 37] so that the squares on AC, CB are also medial. [X. 21]

+

Therefore DL is medial. [X. 15 and 23, Por.]

+

And it has been applied to the rational straight line DE; therefore MD is rational and incommensurable in length with DE. [X. 22]

+

Again, since twice the rectangle AC, CB is rational, MF is also rational.

+

And it is applied to the rational straight line ML; therefore MG is also rational and commensurable in length with ML, that is, DE; [X. 20] therefore DM is incommensurable in length with MG. [X. 13]

+

And they are rational; therefore DM, MG are rational straight lines commensurable in square only; therefore DG is binomial. [X. 36]

+

It is next to be proved that it is also a second binomial straight line.

+

For, since the squares on AC, CB are greater than twice the rectangle AC, CB, therefore DL is also greater than MF, so that DM is also greater than MG. [VI. 1]

+

And, since the square on AC is commensurable with the square on CB, DH is also commensurable with KL, so that DK is also commensurable with KM. [VI. 1, X. 11]

+

And the rectangle DK, KM is equal to the square on MN; therefore the square on DM is greater than the square on MG by the square on a straight line commensurable with DM. [X. 17]

+

And MG is commensurable is length with DE.

+

Therefore DG is a second binomial straight line. [X. Deff. II. 2]

+
+
+ PROPOSITION 62. +

The square on the second bimedial straight line applied to a rational straight line produces as breadth the third binomial.

+

Let AB be a second bimedial straight line divided into its medials at C, so that AC is the greater segment; let DE be any rational straight line, and to DE let there be applied the parallelogram DF equal to the square on AB and producing DG as its breadth;

I say that DG is a third binomial straight line.

+

Let the same construction be made as before shown.

+

Then, since AB is a second bimedial divided at C, therefore AC, CB are medial straight lines commensurable in square only and containing a medial rectangle, [X. 38] so that the sum of the squares on AC, CB is also medial. [X. 15 and 23 Por.]

+

And it is equal to DL; therefore DL is also medial.

+

And it is applied to the rational straight line DE; therefore MD is also rational and incommensurable in length with DE. [X. 22]

+

For the same reason, MG is also rational and incommensurable in length with ML, that is, with DE; therefore each of the straight lines DM, MG is rational and incommensurable in length with DE.

+

And, since AC is incommensurable in length with CB, and, as AC is to CB, so is the square on AC to the rectangle AC, CB, therefore the square on AC is also incommensurable with the rectangle AC, CB. [X. 11]

+

Hence the sum of the squares on AC, CB is incommensurable with twice the rectangle AC, CB, [X. 12, 13] that is, DL is incommensurable with MF, so that DM is also incommensurable with MG. [VI. 1, X. 11]

+

And they are rational; therefore DG is binomial. [X. 36]

+

It is to be proved that it is also a third binomial straight line.

+

In manner similar to the foregoing we may conclude that DM is greater than MG, and that DK is commensurable with KM.

+

And the rectangle DK, KM is equal to the square on MN; therefore the square on DM is greater than the square on MG by the square on a straight line commensurable with DM.

+

And neither of the straight lines DM, MG is commensurable in length with DE.

+

Therefore DG is a third binomial straight line. [X. Deff. II. 3] Q. E. D.

+
+
+ PROPOSITION 63. +

The square on the major straight line applied to a rational straight line produces as breadth the fourth binomial.

+

Let AB be a major straight line divided at C, so that AC is greater than CB; let DE be a rational straight line, and to DE let there be applied the parallelogram DF equal to the square on AB and producing DG as its breadth; I say that DG is a fourth binomial straight line.

+

+

Let the same construction be made as before shown.

+

Then, since AB is a major straight line divided at C, AC, CB are straight lines incommensurable in square which make the sum of the squares on them rational, but the rectangle contained by them medial. [X. 39]

+

Since then the sum of the squares on AC, CB is rational, therefore DL is rational; therefore DM is also rational and commensurable in length with DE. [X. 20]

+

Again, since twice the rectangle AC, CB, that is, MF, is medial, and it is applied to the rational straight line ML, therefore MG is also rational and incommensurable in length with DE; [X. 22] therefore DM is also incommensurable in length with MG. [X. 13]

+

Therefore DM, MG are rational straight lines commensurable in square only; therefore DG is binomial. [X. 36]

+

It is to be proved that it is also a fourth binomial straight line.

+

In manner similar to the foregoing we can prove that DM is greater than MG, and that the rectangle DK, KM is equal to the square on MN.

+

Since then the square on AC is incommensurable with the square on CB, therefore DH is also incommensurable with KL, so that DK is also incommensurable with KM. [VI. 1, X. 11]

+

But, if there be two unequal straight lines, and to the greater there be applied a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, and if it divide it into incommensurable parts, then the square on the greater will be greater than the square on the less by the square on a straight line incommensurable in length with the greater; [X. 18] therefore the square on DM is greater than the square on MG by the square on a straight line incommensurable with DM.

+

And DM, MG are rational straight lines commensurable in square only, and DM is commensurable with the rational straight line DE set out.

+

Therefore DG is a fourth binomial straight line. [X. Deff. II. 4] Q. E. D.

+
+ +
+ PROPOSITION 64. +

The square on the side of a rational plus a medial area applied to a rational straight line produces as breadth the fifth binomial.

+

Let AB be the side of a rational plus a medial area, divided into its straight lines at C, so that AC is the greater; let a rational straight line DE be set out, and let there be applied to DE the parallelogram DF equal to the square on AB, producing DG as its breadth; I say that DG is a fifth binomial straight line.

+

+

Let the same construction as before be made.

+

Since then AB is the side of a rational plus a medial area, divided at C, therefore AC, CB are straight lines incommensurable in square which make the sum of the squares on them medial, but the rectangle contained by them rational. [X. 40]

+

Since then the sum of the squares on AC, CB is medial, therefore DL is medial, so that DM is rational and incommensurable in length with DE. [X. 22]

+

Again, since twice the rectangle AC, CB, that is MF, is rational, therefore MG is rational and commensurable with DE. [X. 20]

+

Therefore DM is incommensurable with MG; [X. 13] therefore DM, MG are rational straight lines commensurable in square only; therefore DG is binomial. [X. 36]

+

I say next that it is also a fifth binomial straight line.

+

For it can be proved similarly that the rectangle DK, KM is equal to the square on MN, and that DK is incommensurable in length with KM; therefore the square on DM is greater than the square on MG by the square on a straight line incommensurable with DM. [X. 18]

+

And DM, MG are commensurable in square only, and the less, MG, is commensurable in length with DE.

+

Therefore DG is a fifth binomial. Q. E. D.

+
+
+ PROPOSITION 65. +

The square on the side of the sum of two medial areas applied to a rational straight line produces as breadth the sixth binomial.

+

Let AB be the side of the sum of two medial areas, divided at C, let DE be a rational straight line, and let there be applied to DE the parallelogram DF equal to the square on AB, producing DG as its breadth; I say that DG is a sixth binomial straight line.

+

For let the same construction be made as before.

+

Then, since AB is the side of the sum of two medial areas, divided at C, therefore AC, CB are straight lines incommensurable in square which make the sum of the squares on them medial, the rectangle contained by them medial, and moreover the sum of the squares on them incommensurable with the rectangle contained by them, [X. 41]

so that, in accordance with what was before proved, each of the rectangles DL, MF is medial.

+

And they are applied to the rational straight line DE; therefore each of the straight lines DM, MG is rational and incommensurable in length with DE. [X. 22]

+

And, since the sum of the squares on AC, CB is incommensurable with twice the rectangle AC, CB, therefore DL is incommensurable with MF.

+

Therefore DM is also incommensurable with MG; [VI. 1, X. 11] therefore DM, MG are rational straight lines commensurable in square only; therefore DG is binomial. [X. 36]

+

I say next that it is also a sixth binomial straight line.

+

Similarly again we can prove that the rectangle DK, KM is equal to the square on MN, and that DK is incommensurable in length with KM; and, for the same reason, the square on DM is greater than the square on MG by the square on a straight line incommensurable in length with DM.

+

And neither of the straight lines DM, MG is commensurable in length with the rational straight line DE set out.

+

Therefore DG is a sixth binomial straight line. Q. E. D.

+
+
+ PROPOSITION 66. +

A straight line commensurable in length with a binomial straight line is itself also binomial and the same in order.

+

Let AB be binomial, and let CD be commensurable in length with AB;

I say that CD is binomial and the same in order with AB.

+

For, since AB is binomial, let it be divided into its terms at E, and let AE be the greater term; therefore AE, EB are rational straight lines commensurable in square only. [X. 36]

+

Let it be contrived that, as AB is to CD, so is AE to CF; [VI. 12] therefore also the remainder EB is to the remainder FD as AB is to CD. [V. 19]

+

But AB is commensurable in length with CD; therefore AE is also commensurable with CF, and EB with FD. [X. 11]

+

And AE, EB are rational; therefore CF, FD are also rational.

+

And, as AE is to CF, so is EB to FD. [V. 11]

+

Therefore, alternately, as AE is to EB, so is CF to FD. [V. 16]

+

But AE, EB are commensurable in square only; therefore CF, FD are also commensurable in square only. [X. 11]

+

And they are rational; therefore CD is binomial. [X. 36]

+

I say next that it is the same in order with AB.

+

For the square on AE is greater than the square on EB either by the square on a straight line commensurable with AE or by the square on a straight line incommensurable with it.

+

If then the square on AE is greater than the square on EB by the square on a straight line commensurable with AE, the square on CF will also be greater than the square on FD by the square on a straight line commensurable with CF. [X. 14]

+

And, if AE is commensurable with the rational straight line set out, CF will also be commensurable with it, [X. 12] and for this reason each of the straight lines AB, CD is a first binomial, that is, the same in order. [X. Deff. II. 1]

+

But, if EB is commensurable with the rational straight line set out, FD is also commensurable with it, [X. 12] and for this reason again CD will be the same in order with AB, for each of them will be a second binomial. [X. Deff. II. 2]

+

But, if neither of the straight lines AE, EB is commensurable with the rational straight line set out, neither of the straight lines CF, FD will be commensurable with it, [X. 13] and each of the straight lines AB, CD is a third binomial. [X. Deff. II. 3]

+

But, if the square on AE is greater than the square on EB by the square on a straight line incommensurable with AE, the square on CF is also greater than the square on FD by the square on a straight line incommensurable with CF. [X. 14]

+

And, if AE is commensurable with the rational straight line set out, CF is also commensurable with it, and each of the straight lines AB, CD is a fourth binomial. [X. Deff. II. 4]

+

But, if EB is so commensurable, so is FD also, and each of the straight lines AB, CD will be a fifth binomial. [X. Deff. II. 5]

+

But, if neither of the straight lines AE, EB is so commensurable, neither of the straight lines CF, FD is commensurable with the rational straight line set out, and each of the straight lines AB, CD will be a sixth binomial. [X. Deff. II. 6]

+

Hence a straight line commensurable in length with a binomial straight line is binomial and the same in order. Q. E. D.

+
+
+ PROPOSITION 67. +

A straight line commensurable in length with a bimedial straight line is itself also bimedial and the same in order.

+

Let AB be bimedial, and let CD be commensurable in length with AB; I say that CD is bimedial and the same in order with AB.

+

+

For, since AB is bimedial, let it be divided into its medials at E; therefore AE, EB are medial straight lines commensurable in square only. [X. 37, 38]

+

And let it be contrived that, as AB is to CD, so is AE to CF; therefore also the remainder EB is to the remainder FD as AB is to CD. [V. 19]

+

But AB is commensurable in length with CD; therefore AE, EB are also commensurable with CF, FD respectively. [X. 11]

+

But AE, EB are medial; therefore CF, FD are also medial. [X. 23]

+

And since, as AE is to EB, so is CF to FD, [V. 11] and AE, EB are commensurable in square only, CF, FD are also commensurable in square only. [X. 11]

+

But they were also proved medial; therefore CD is bimedial.

+

I say next that it is also the same in order with AB.

+

For since, as AE is to EB, so is CF to FD, therefore also, as the square on AE is to the rectangle AE, EB, so is the square on CF to the rectangle CF, FD; therefore, alternately, as the square on AE is to the square on CF, so is the rectangle AE, EB to the rectangle CF, FD. [V. 16]

+

But the square on AE is commensurable with the square on CF; therefore the rectangle AE, EB is also commensurable with the rectangle CF, FD.

+

If therefore the rectangle AE, EB is rational, the rectangle CF, FD is also rational, [and for this reason CD is a first bimedial]; [X. 37] but if medial, medial, [X. 23, Por.] and each of the straight lines AB, CD is a second bimedial. [X. 38]

+

And for this reason CD will be the same in order with AB. Q. E. D.

+
+ +
+ PROPOSITION 68. +

A straight line commensurable with a major straight line is itself also major.

+

Let AB be major, and let CD be commensurable with AB; I say that CD is major.

+

Let AB be divided at E; therefore AE, EB are straight lines incommensurable in square which make the sum of the squares on them rational, but the rectangle contained by them medial. [X. 39]

+

+

Let the same construction be made as before.

+

Then since, as AB is to CD, so is AE to CF, and EB to FD, therefore also, as AE is to CF, so is EB to FD. [V. 11]

+

But AB is commensurable with CD; therefore AE, EB are also commensurable with CF, FD respectively. [X. 11]

+

And since, as AE is to CF, so is EB to FD, alternately also, as AE is to EB, so is CF to FD; [V. 16] therefore also, componendo, as AB is to BE, so is CD to DF; [V. 18] therefore also, as the square on AB is to the square on BE, so is the square on CD to the square on DF. [VI. 20]

+

Similarly we can prove that, as the square on AB is to the square on AE, so also is the square on CD to the square on CF.

+

Therefore also, as the square on AB is to the squares on AE, EB, so is the square on CD to the squares on CF, FD; therefore also, alternately, as the square on AB is to the square on CD, so are the squares on AE, EB to the squares on CF, FD. [V. 16]

+

But the square on AB is commensurable with the square on CD; therefore the squares on AE, EB are also commensurable with the squares on CF, FD.

+

And the squares on AE, EB together are rational; therefore the squares on CF, FD together are rational.

+

Similarly also twice the rectangle AE, EB is commensurable with twice the rectangle CF, FD.

+

And twice the rectangle AE, EB is medial; therefore twice the rectangle CF, FD is also medial. [X. 23, Por.]

+

Therefore CF, FD are straight lines incommensurable in square which make, at the same time, the sum of the squares on them rational, but the rectangle contained by them medial; therefore the whole CD is the irrational straight line called major. [X. 39]

+

Therefore a straight line commensurable with the major straight line is major. Q. E. D.

+
+
+ PROPOSITION 69. +

A straight line commensurable with the side of a rational plus a medial area is itself also the side of a rational plus a medial area.

+

Let AB be the side of a rational plus a medial area, and let CD be commensurable with AB; it is to be proved that CD is also the side of a rational plus a medial area.

+

Let AB be divided into its straight lines at E; therefore AE, EB are straight lines incommensurable in square which make the sum of the squares on them medial, but the rectangle contained by them rational. [X. 40]

+

+

Let the same construction be made as before.

+

We can then prove similarly that CF, FD are incommensurable in square, and the sum of the squares on AE, EB is commensurable with the sum of the squares on CF, FD, and the rectangle AE, EB with the rectangle CF, FD; so that the sum of the squares on CF, FD is also medial, and the rectangle CF, FD rational.

+

Therefore CD is the side of a rational plus a medial area. Q. E. D.

+
+ +
+ PROPOSITION 70. +

A straight line commensurable with the side of the sum of two medial areas is the side of the sum of two medial areas.

+

Let AB be the side of the sum of two medial areas, and CD commensurable with AB; it is to be proved that CD is also the side of the sum of two medial areas.

+

For, since AB is the side of the sum of two medial areas,

let it be divided into its straight lines at E; therefore AE, EB are straight lines incommensurable in square which make the sum of the squares on them medial, the rectangle contained by them medial, and furthermore the sum of the squares on AE, EB incommensurable with the rectangle AE, EB. [X. 41]

+

Let the same construction be made as before.

+

We can then prove similarly that CF, FD are also incommensurable in square, the sum of the squares on AE, EB is commensurable with the sum of the squares on CF, FD, and the rectangle AE, EB with the rectangle CF, FD; so that the sum of the squares on CF, FD is also medial, the rectangle CF, FD is medial, and moreover the sum of the squares on CF, FD is incommensurable with the rectangle CF, FD.

+

Therefore CD is the side of the sum of two medial areas. Q. E. D.

+
+
+ PROPOSITION 71. +

If a rational and a medial area be added together, four irrational straight lines arise, namely a binomial or a first bimedial or a major or a side of a rational plus a medial area.

+

Let. AB be rational, and CD medial; I say that the side + of the area AD is a binomial or a first bimedial or a major or a side of a rational plus a medial area.

+

For AB is either greater or less than CD.

+

First, let it be greater; let a rational straight line EF be set out, let there be applied to EF the rectangle EG equal to AB, producing EH as breadth, and let HI, equal to DC, be applied to EF, producing HK as breadth.

+

+

Then, since AB is rational and is equal to EG, therefore EG is also rational.

+

And it has been applied to EF, producing EH as breadth; therefore EH is rational and commensurable in length with EF. [X. 20]

+

Again, since CD is medial and is equal to HI, therefore HI is also medial.

+

And it is applied to the rational straight line EF, producing HK as breadth; therefore HK is rational and incommensurable in length with EF [X. 22]

+

And, since CD is medial, while AB is rational, therefore AB is incommensurable with CD, so that EG is also incommensurable with HI.

+

But, as EG is to HI, so is EH to HK; [VI. 1] therefore EH is also incommensurable in length with HK. [X. 11]

+

And both are rational; therefore EH, HK are rational straight lines commensurable in square only; therefore EK is a binomial straight line, divided at H. [X. 36]

+

And, since AB is greater than CD, while AB is equal to EG and CD to HI, therefore EG is also greater than HI; therefore EH is also greater than HK.

+

The square, then, on EH is greater than the square on HK either by the square on a straight line commensurable in length with EH or by the square on a straight line incommensurable with it.

+

First, let the square on it be greater by the square on a straight line commensurable with itself.

+

Now the greater straight line HE is commensurable in length with the rational straight line EF set out; therefore EK is a first binomial. [X. Deff. II. 1]

+

But EF is rational; and, if an area be contained by a rational straight line and the first binomial, the side of the square equal to the area is binomial. [X. 54]

+

Therefore the side + of EI is binomial; so that the side + of AD is also binomial.

+

Next, let the square on EH be greater than the square on HK by the square on a straight line incommensurable with EH.

+

Now the greater straight line EH is commensurable in length with the rational straight line EF set out; therefore EK is a fourth binomial. [X. Deff. II. 4]

+

But EF is rational; and, if an area be contained by a rational straight line and the fourth binomial, the side + of the area is the irrational straight line called major. [X. 57]

+

Therefore the side + of the area EI is major; so that the side + of the area AD is also major.

+

Next, let AB be less than CD; therefore EG is also less than HI, so that EH is also less than HK.

+

Now the square on HK is greater than the square on EH either by the square on a straight line commensurable with HK or by the square on a straight line incommensurable with it.

+

First, let the square on it be greater by the square on a straight line commensurable in length with itself.

+

Now the lesser straight line EH is commensurable in length with the rational straight line EF set out; therefore EK is a second binomial. [X. Deff. II. 2]

+

But EF is rational, and, if an area be contained by a rational straight line and the second binomial, the side of the square equal to it is a first bimedial; [X. 55] therefore the side + of the area EI is a first bimedial, so that the side + of AD is also a first bimedial.

+

Next, let the square on HK be greater than the square on HE by the square on a straight line incommensurable with HK.

+

Now the lesser straight line EH is commensurable with the rational straight line EF set out; therefore EK is a fifth binomial. [X. Deff. II. 5]

+

But EF is rational; and, if an area be contained by a rational straight line and the fifth binomial, the side of the square equal to the area is a side of a rational plus a medial area. [X. 58]

+

Therefore the side + of the area EI is a side of a rational plus a medial area, so that the side + of the area AD is also a side of a rational plus a medial area.

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 72. +

If two medial areas incommensurable with one another be added together, the remaining two irrational straight lines arise, namely either a second bimedial or a side of the sum of two medial areas.

+

For let two medial areas AB, CD incommensurable with one another be added together; I say that the side + of the area AD is either a second bimedial or a side of the sum of two medial areas.

+

+

For AB is either greater or less than CD.

+

First, if it so chance, let AB be greater than CD.

+

Let the rational straight line EF be set out, and to EF let there be applied the rectangle EG equal to AB and producing EH as breadth, and the rectangle HI equal to CD and producing HK as breadth.

+

Now, since each of the areas AB, CD is medial, therefore each of the areas EG, HI is also medial.

+

And they are applied to the rational straight line FE, producing EH, HK as breadth; therefore each of the straight lines EH, HK is rational and incommensurable in length with EF. [X. 22]

+

And, since AB is incommensurable with CD, and AB is equal to EG, and CD to HI, therefore EG is also incommensurable with HI.

+

But, as EG is to HI, so is EH to HK; [VI. 1] therefore EH is incommensurable in length with HK. [X. 11]

+

Therefore EH, HK are rational straight lines commensurable in square only; therefore EK is binomial. [X. 36]

+

But the square on EH is greater than the square on HK either by the square on a straight line commensurable with EH or by the square on a straight line incommensurable with it.

+

First, let the square on it be greater by the square on a straight line commensurable in length with itself.

+

Now neither of the straight lines EH, HK is commensurable in length with the rational straight line EF set out; therefore EK is a third binomial. [X. Deff. II. 3]

+

But EF is rational; and, if an area be contained by a rational straight line and the third binomial, the side + of the area is a second bimedial; [X. 56] therefore the side + of EI, that is, of AD, is a second bimedial.

+

Next, let the square on EH be greater than the square on HK by the square on a straight line incommensurable in length with EH.

+

Now each of the straight lines EH, HK is incommensurable in length with EF; therefore EK is a sixth binomial. [X. Deff. II. 6]

+

But, if an area be contained by a rational straight line and the sixth binomial, the side + of the area is the side of the sum of two medial areas; [X. 59] so that the side + of the area AD is also the side of the sum of two medial areas.

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 73. +

If from a rational straight line there be subtracted a rational straight line commensurable with the whole in square only, the remainder is irrational; and let it be called + an apotome.

+

For from the rational straight line AB let the rational straight line BC, commensurable with the whole in square only, be subtracted;

I say that the remainder AC is the irrational straight line called apotome.

+

For, since AB is incommensurable in length with BC, and, as AB is to BC, so is the square on AB to the rectangle AB, BC, therefore the square on AB is incommensurable with the rectangle AB, BC. [X. 11]

+

But the squares on AB, BC are commensurable with the square on AB, [X. 15] and twice the rectangle AB, BC is commensurable with the rectangle AB, BC. [X. 6]

+

And, inasmuch as the squares on AB, BC are equal to twice the rectangle AB, BC together with the square on CA, [II. 7] therefore the squares on AB, BC are also incommensurable with the remainder, the square on AC. [X. 13, 16]

+

But the squares on AB, BC are rational; therefore AC is irrational. [X. Def. 4]

+

And let it be called an apotome. Q. E. D.

+
+
+ PROPOSITION 74. +

If from a medial straight line there be subtracted a medial straight line which is commensurable with the whole in square only, and which contains with the whole a rational rectangle, the remainder is irrational. And let it be called a + first apotome of a medial + straight line.

+

For from the medial straight line AB let there be subtracted the medial straight line BC which is commensurable with AB in square only and with AB makes the rectangle AB, BC rational;

I say that the remainder AC is irrational; and let it be called a first apotome of a medial straight line.

+

For, since AB, BC are medial, the squares on AB, BC are also medial.

+

But twice the rectangle AB, BC is rational; therefore the squares on AB, BC are incommensurable with twice the rectangle AB, BC; therefore twice the rectangle AB, BC is also incommensurable with the remainder, the square on AC, [Cf. II. 7] since, if the whole is incommensurable with one of the magnitudes, the original magnitudes will also be incommensurable. [X. 16]

+

But twice the rectangle AB, BC is rational; therefore the square on AC is irrational; therefore AC is irrational. [X. Def. 4]

+

And let it be called a first apotome of a medial straight line.

+
+ +
+ PROPOSITION 75 +

If from a medial straight line there be subtracted a medial straight line which is commensurable with the whole in square only, and which contains with the whole a medial rectangle, the remainder is irrational; and let it be called a + second apotome of a medial + straight line.

+

For from the medial straight line AB let there be subtracted the medial straight line CB which is commensurable with the whole AB in square only and such that the rectangle AB, BC, which it contains with the whole AB, is medial; [X. 28] I say that the remainder AC is irrational; and let it be called a second apotome of a medial straight line.

+

+

For let a rational straight line DI be set out, let DE equal to the squares on AB, BC be applied to DI, producing DG as breadth, and let DH equal to twice the rectangle AB, BC be applied to DI, producing DF as breadth; therefore the remainder FE is equal to the square on AC. [II. 7]

+

Now, since the squares on AB, BC are medial and commensurable, therefore DE is also medial. [X. 15 and 23, Por.]

+

And it is applied to the rational straight line DI, producing DG as breadth; therefore DG is rational and incommensurable in length with DI. [X. 22]

+

Again, since the rectangle AB, BC is medial, therefore twice the rectangle AB, BC is also medial. [X. 23, Por.]

+

And it is equal to DH; therefore DH is also medial.

+

And it has been applied to the rational straight line DI, producing DF as breadth; therefore DF is rational and incommensurable in length with DI. [X. 22]

+

And, since AB, BC are commensurable in square only, therefore AB is incommensurable in length with BC; therefore the square on AB is also incommensurable with the rectangle AB, BC. [X. 11]

+

But the squares on AB, BC are commensurable with the square on AB, [X. 15] and twice the rectangle AB, BC is commensurable with the rectangle AB, BC; [X. 6] therefore twice the rectangle AB, BC is incommensurable with the squares on AB, BC. [X. 13]

+

But DE is equal to the squares on AB, BC, and DH to twice the rectangle AB, BC; therefore DE is incommensurable with DH.

+

But, as DE is to DH, so is GD to DF; [VI. 1] therefore GD is incommensurable with DF. [X. 11]

+

And both are rational; therefore GD, DF are rational straight lines commensurable in square only; therefore FG is an apotome. [X. 73]

+

But DI is rational, and the rectangle contained by a rational and an irrational straight line is irrational, [deduction from X. 20] and its ’side’ is irrational.

+

And AC is the ’side’ of FE; therefore AC is irrational.

+

And let it be called a second apotome of a medial straight line. Q. E. D.

+
+
+ PROPOSITION 76 +

If from a straight line there be subtracted a straight line which is incommensurable in square with the whole and which with the whole makes the squares on them added together rational, but the rectangle contained by them medial, the remainder is irrational; and let it be called + minor.

+

For from the straight line AB let there be subtracted the straight line BC which is incommensurable in square with the whole and fulfils the given conditions. [X. 33]

+

+

I say that the remainder AC is the irrational straight line called minor.

+

For, since the sum of the squares on AB, BC is rational, while twice the rectangle AB, BC is medial, therefore the squares on AB, BC are incommensurable with twice the rectangle AB, BC; and, convertendo, the squares on AB, BC are incommensurable with the remainder, the square on AC. [II. 7, X. 16]

+

But the squares on AB, BC are rational; therefore the square on AC is irrational; therefore AC is irrational.

+

And let it be called minor.

+
+
+ PROPOSITION 77 +

If from a straight line there be subtracted a straight line which is incommensurable in square with the whole, and which with the whole makes the sum of the squares on them medial, but twice the rectangle contained by them rational, the remainder is irrational: and let it be called + that which produces with a rational area a medial whole.

+

For from the straight line AB let there be subtracted the straight line BC which is incommensurable in square with AB and fulfils the given conditions; [X. 34] I say that the remainder AC is the irrational straight line aforesaid.

+

For, since the sum of the squares on AB, BC is medial, while twice the rectangle AB, BC is rational, therefore the squares on AB, BC are incommensurable with twice the rectangle AB, BC; therefore the remainder also, the square on AC, is incommensurable with twice the rectangle AB, BC. [II. 7, X. 16]

+

+

And twice the rectangle AB, BC is rational; therefore the square on AC is irrational; therefore AC is irrational.

+

And let it be called that which produces with a rational area a medial whole. Q. E. D.

+
+
+ PROPOSITION 78 +

+ If from a straight line there be subtracted a straight line which is incommensurable in square with the whole and which with the whole makes the sum of the squares on them medial, twice the rectangle contained by them medial, and further the squares on them incommensurable with twice the rectangle contained by them, the remainder is irrational; and let it be called + that which produces with a medial area a medial whole. +

+

For from the straight line AB let there be subtracted the straight line BC incommensurable in square with AB and fulfilling the given conditions; [X. 35]

I say that the remainder AC is the irrational straight line called that which produces with a medial area a medial whole.

+

For let a rational straight line DI be set out, to DI let there be applied DE equal to the squares on AB, BC, producing DG as breadth, and let DH equal to twice the rectangle AB, BC be subtracted.

+

Therefore the remainder FE is equal to the square on AC, [II. 7] so that AC is the side + of FE.

+

Now, since the sum of the squares on AB, BC is medial and is equal to DE, therefore DE is medial.

+

And it is applied to the rational straight line DI, producing DG as breadth; therefore DG is rational and incommensurable in length with DI. [X. 22]

+

Again, since twice the rectangle AB, BC is medial and is equal to DH, therefore DH is medial.

+

And it is applied to the rational straight line DI, producing DF as breadth; therefore DF is also rational and incommensurable in length with DI. [X. 22]

+

And, since the squares on AB, BC are incommensurable with twice the rectangle AB, BC, therefore DE is also incommensurable with DH.

+

But, as DE is to DH, so also is DG to DF; [VI. 1] therefore DG is incommensurable with DF. [X. 11]

+

And both are rational; therefore GD, DF are rational straight lines commensurable in square only.

+

Therefore FG is an apotome. [X. 73]

+

And FH is rational; but the rectangle contained by a rational straight line and an apotome is irrational, [deduction from X. 20] and its side + is irrational.

+

And AC is the side + of FE; therefore AC is irrational.

+

And let it be called that which produces with a medial area a medial whole. Q. E. D.

+
+
+ PROPOSITION 79 +

To an apotome only one rational straight line can be annexed which is commensurable with the whole in square only.

+

Let AB be an apotome, and BC an annex to it; therefore AC, CB are rational straight lines commensurable in square only. [X. 73]

+

+

I say that no other rational straight line can be annexed to AB which is commensurable with the whole in square only.

+

For, if possible, let BD be so annexed; therefore AD, DB are also rational straight lines commensurable in square only. [X. 73]

+

Now, since the excess of the squares on AD, DB over twice the rectangle AD, DB is also the excess of the squares on AC, CB over twice the rectangle AC, CB, for both exceed by the same, the square on AB, [II. 7] therefore, alternately, the excess of the squares on AD, DB over the squares on AC, CB is the excess of twice the rectangle AD, DB over twice the rectangle AC, CB.

+

But the squares on AD, DB exceed the squares on AC, CB by a rational area, for both are rational; therefore twice the rectangle AD, DB also exceeds twice the rectangle AC, CB by a rational area: which is impossible, for both are medial [X. 21], and a medial area does not exceed a medial by a rational area. [X. 26]

+

Therefore no other rational straight line can be annexed to AB which is commensurable with the whole in square only.

+

Therefore only one rational straight line can be annexed to an apotome which is commensurable with the whole in square only. Q. E. D.

+
+
+ PROPOSITION 80. +

To a first apotome of a medial straight line only one medial straight line can be annexed which is commensurable with the whole in square only and which contains with the whole a rational rectangle.

+

For let AB be a first apotome of a medial straight line, and let BC be an annex to AB; therefore AC, CB are medial straight lines commensurable in square only and such that the rectangle AC, CB which they contain is rational; [X. 74] I say that no other medial straight line can be annexed to AB which is commensurable with the whole in square only and which contains with the whole a rational area.

+

For, if possible, let DB also be so annexed; therefore AD, DB are medial straight lines commensurable in square only and such that the rectangle AD, DB which they contain is rational. [X. 74]

+

+

Now, since the excess of the squares on AD, DB over twice the rectangle AD, DB is also the excess of the squares on AC, CB over twice the rectangle AC, CB, for they exceed by the same, the square on AB, [II. 7] therefore, alternately, the excess of the squares on AD, DB over the squares on AC, CB is also the excess of twice the rectangle AD, DB over twice the rectangle AC, CB.

+

But twice the rectangle AD, DB exceeds twice the rectangle AC, CB by a rational area, for both are rational.

+

Therefore the squares on AD, DB also exceed the squares on AC, CB by a rational area. which is impossible, for both are medial [X. 15 and 23, Por.], and a medial area does not exceed a medial by a rational area. [X. 26]

+

Therefore etc. Q. E. D.

+
+ +
+ PROPOSITION 81. +

To a second apotome of a medial straight line only one medial straight line can be annexed which is commensurable with the whole in square only and which contains with the whole a medial rectangle.

+

Let AB be a second apotome of a medial straight line and BC an annex to AB; therefore AC, CB are medial straight lines commensurable in square only and such that the rectangle AC, CB which they contain is medial. [X. 75]

+

+

I say that no other medial straight line can be annexed to AB which is commensurable with the whole in square only and which contains with the whole a medial rectangle.

+

For, if possible, let BD also be so annexed; therefore AD, DB are also medial straight lines commensurable in square only and such that the rectangle AD, DB which they contain is medial. [X. 75]

+

Let a rational straight line EF be set out, let EG equal to the squares on AC, CB be applied to EF, producing EM as breadth, and let HG equal to twice the rectangle AC, CB be subtracted, producing HM as breadth; therefore the remainder EL is equal to the square on AB, [II. 7] so that AB is the side + of EL.

+

Again, let EI equal to the squares on AD, DB be applied to EF, producing EN as breadth.

+

But EL is also equal to the square on AB; therefore the remainder HI is equal to twice the rectangle AD, DB. [II. 7]

+

Now, since AC, CB are medial straight lines, therefore the squares on AC, CB are also medial.

+

And they are equal to EG; therefore EG is also medial. [X. 15 and 23, Por.]

+

And it is applied to the rational straight line EF, producing EM as breadth; therefore EM is rational and incommensurable in length with EF. [X. 22]

+

Again, since the rectangle AC, CB is medial, twice the rectangle AC, CB is also medial. [X. 23, Por.]

+

And it is equal to HG; therefore HG is also medial.

+

And it is applied to the rational straight line EF, producing HM as breadth; therefore HM is also rational and incommensurable in length with EF. [X. 22]

+

And, since AC, CB are commensurable in square only, therefore AC is incommensurable in length with CB.

+

But, as AC is to CB, so is the square on AC to the rectangle AC, CB; therefore the square on AC is incommensurable with the rectangle AC, CB. [X. 11]

+

But the squares on AC, CB are commensurable with the square on AC, while twice the rectangle AC, CB is commensurable with the rectangle AC, CB; [X. 6] therefore the squares on AC, CB are incommensurable with twice the rectangle AC, CB. [X. 13]

+

And EG is equal to the squares on AC, CB, while GH is equal to twice the rectangle AC, CB; therefore EG is incommensurable with HG.

+

But, as EG is to HG, so is EM to HM; [VI. 1] therefore EM is incommensurable in length with MH. [X. 11]

+

And both are rational; therefore EM, MH are rational straight lines commensurable in square only; therefore EH is an apotome, and HM an annex to it. [X. 73]

+

Similarly we can prove that HN is also an annex to it; therefore to an apotome different straight lines are annexed which are commensurable with the wholes in square only: which is impossible. [X. 79]

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 82. +

To a minor straight line only one straight line can be annexed which is incommensurable in square with the whole and which makes, with the whole, the sum of the squares on them rational but twice the rectangle contained by them medial.

+

Let AB be the minor straight line, and let BC be an annex to AB;

therefore AC, CB are straight lines incommensurable in square which make the sum of the squares on them rational, but twice the rectangle contained by them medial. [X. 76]

+

I say that no other straight line can be annexed to AB fulfilling the same conditions.

+

For, if possible, let BD be so annexed; therefore AD, DB are also straight lines incommensurable in square which fulfil the aforesaid conditions. [X. 76]

+

Now, since the excess of the squares on AD, DB over the squares on AC, CB is also the excess of twice the rectangle AD, DB over twice the rectangle AC, CB, while the squares on AD, DB exceed the squares on AC, CB by a rational area, for both are rational, therefore twice the rectangle AD, DB also exceeds twice the rectangle AC, CB by a rational area: which is impossible, for both are medial. [X. 26]

+

Therefore to a minor straight line only one straight line can be annexed which is incommensurable in square with the whole and which makes the squares on them added together rational, but twice the rectangle contained by them medial. Q. E. D.

+
+
+ PROPOSITION 83. +

To a straight line which produces with a rational area a medial whole only one straight line can be annexed which is incommensurable in square with the whole straight line and which with the whole straight line makes the sum of the squares on them medial, but twice the rectangle contained by them rational.

+

Let AB be the straight line which produces with a rational area a medial whole, and let BC be an annex to AB; therefore AC, CB are straight lines incommensurable in square which fulfil the given conditions. [X. 77]

+

+

I say that no other straight line can be annexed to AB which fulfils the same conditions.

+

For, if possible, let BD be so annexed; therefore AD, DB are also straight lines incommensurable in square which fulfil the given conditions. [X. 77]

+

Since then, as in the preceding cases, the excess of the squares on AD, DB over the squares on AC, CB is also the excess of twice the rectangle AD, DB over twice the rectangle AC, CB, while twice the rectangle AD, DB exceeds twice the rectangle AC, CB by a rational area, for both are rational, therefore the squares on AD, DB also exceed the squares on AC, CB by a rational area: which is impossible, for both are medial. [X. 26]

+

Therefore no other straight line can be annexed to AB which is incommensurable in square with the whole and which with the whole fulfils the aforesaid conditions; therefore only one straight line can be so annexed. Q. E. D.

+
+
+ PROPOSITION 84. +

To a straight line which produces with a medial area a medial whole only one straight line can be annexed which is incommensurable in square with the whole straight line and which with the whole straight line makes the sum of the squares on them medial and twice the rectangle contained by them both medial and also incommensurable with the sum of the squares on them.

+

Let AB be the straight line which produces with a medial area a medial whole, and BC an annex to it; therefore AC, CB are straight lines incommensurable in square which fulfil the aforesaid conditions. [X. 78]

+

+

I say that no other straight line can be annexed to AB which fulfils the aforesaid conditions.

+

For, if possible, let BD be so annexed, so that AD, DB are also straight lines incommensurable in square which make the squares on AD, DB added together medial, twice the rectangle AD, DB medial, and also the squares on AD, DB incommensurable with twice the rectangle AD, DB. [X. 78]

+

Let a rational straight line EF be set out, let EG equal to the squares on AC, CB be applied to EF, producing EM as breadth, and let HG equal to twice the rectangle AC, CB be applied to EF, producing HM as breadth; therefore the remainder, the square on AB [II. 7], is equal to EL; therefore AB is the side + of EL.

+

Again, let EI equal to the squares on AD, DB be applied to EF, producing EN as breadth.

+

But the square on AB is also equal to EL; therefore the remainder, twice the rectangle AD, DB [II. 7], is equal to HI.

+

Now, since the sum of the squares on AC, CB is medial and is equal to EG, therefore EG is also medial.

+

And it is applied to the rational straight line EF, producing EM as breadth; therefore EM is rational and incommensurable in length with EF. [X. 22]

+

Again, since twice the rectangle AC, CB is medial and is equal to HG, therefore HG is also medial.

+

And it is applied to the rational straight line EF, producing HM as breadth; therefore HM is rational and incommensurable in length with EF. [X. 22]

+

And, since the squares on AC, CB are incommensurable with twice the rectangle AC, CB, EG is also incommensurable with HG; therefore EM is also incommensurable in length with MH. [VI. 1, X. 11]

+

And both are rational; therefore EM, MH are rational straight lines commensurable in square only; therefore EH is an apotome, and HM an annex to it. [X. 73]

+

Similarly we can prove that EH is again an apotome and HN an annex to it.

+

Therefore to an apotome different rational straight lines are annexed which are commensurable with the wholes in square only: which was proved impossible. [X. 79]

+

Therefore no other straight line can be so annexed to AB.

+

Therefore to AB only one straight line can be annexed which is incommensurable in square with the whole and which with the whole makes the squares on them added together medial, twice the rectangle contained by them medial, and also the squares on them incommensurable with twice the rectangle contained by them. Q. E. D.

+
+
+
+ DEFINITIONS III. +
+1 +

Given a rational straight line and an apotome, if the square on the whole be greater than the square on the annex by the square on a straight line commensurable in length with the whole, and the whole be commensurable in length with the rational straight line set out, let the apotome be called a first apotome.

+
+
+2 +

But if the annex be commensurable in length with the rational straight line set out, and the square on the whole be greater than that on the annex by the square on a straight line commensurable with the whole, let the apotome be called a second apotome.

+
+
+3 +

But if neither be commensurable in length with the rational straight line set out, and the square on the whole be greater than the square on the annex by the square on a straight line commensurable with the whole, let the apotome be called a third apotome.

+
+
+4 +

Again, if the square on the whole be greater than the square on the annex by the square on a straight line incommensurable with the whole, then, if the whole be commensurable in length with the rational straight line set out, let the apotome be called a fourth apotome;

+
+
+5 +

if the annex be so commensurable, a fifth;

+
+
+6 +

and, if neither, a sixth.

+
+
+ +
+ PROPOSITIONS 85—115. +
+ PROPOSITION 85. +

To find the first apotome.

+

Let a rational straight line A be set out, and let BG be commensurable in length with A; therefore BG is also rational.

+

+

Let two square numbers DE, EF be set out, and let their difference FD not be square; therefore neither has ED to DF the ratio which a square number has to a square number.

+

Let it be contrived that, as ED is to DF, so is the square on BG to the square on GC; [X. 6, Por.] therefore the square on BG is commensurable with the square on GC. [X. 6]

+

But the square on BG is rational; therefore the square on GC is also rational; therefore GC is also rational.

+

And, since ED has not to DF the ratio which a square number has to a square number, therefore neither has the square on BG to the square on GC the ratio which a square number has to a square number; therefore BG is incommensurable in length with GC. [X. 9]

+

And both are rational; therefore BG, GC are rational straight lines commensurable in square only; therefore BC is an apotome. [X. 73]

+

I say next that it is also a first apotome.

+

For let the square on H be that by which the square on BG is greater than the square on GC.

+

Now since. as ED is to FD, so is the square on BG to the square on GC, therefore also, convertendo, [v. 19, Por.] as DE is to EF, so is the square on GB to the square on H.

+

But DE has to EF the ratio which a square number has to a square number, for each is square; therefore the square on GB also has to the square on H the ratio which a square number has to a square number; therefore BG is commensurable in length with H. [X. 9]

+

And the square on BG is greater than the square on GC by the square on a straight line commensurable in length with BG.

+

And the whole BG is commensurable in length with the rational straight line A set out.

+

Therefore BC is a first apotome. [X. Deff. III. 1]

+

Therefore the first apotome BC has been found. (Being) that which it was required to find.

+
+ +
+ PROPOSITION 86. +

To find the second apotome.

+

Let a rational straight line A be set out, and GC commensurable in length with A; therefore GC is rational.

+

+

Let two square numbers DE, EF be set out, and let their difference DF not be square.

+

Now let it be contrived that, as FD is to DE, so is the square on CG to the square on GB. [X. 6, Por.]

+

Therefore the square on CG is commensurable with the square on GB. [X. 6]

+

But the square on CG is rational; therefore the square on GB is also rational; therefore BG is rational.

+

And, since the square on GC has not to the square on GB the ratio which a square number has to a square number, CG is incommensurable in length with GB. [X. 9]

+

And both are rational; therefore CG, GB are rational straight lines commensurable in square only; therefore BC is an apotome. [X. 73]

+

I say next that it is also a second apotome.

+

For let the square on H be that by which the square on BG is greater than the square on GC.

+

Since then, as the square on BG is to the square on GC, so is the number ED to the number DF, therefore, convertendo, as the square on BG is to the square on H, so is DE to EF. [V. 19, Por.]

+

And each of the numbers DE, EF is square; therefore the square on BG has to the square on H the ratio which a square number has to a square number; therefore BG is commensurable in length with H. [X. 9]

+

And the square on BG is greater than the square on GC by the square on H; therefore the square on BG is greater than the square on GC + by the square on a straight line commensurable in length with BG.

+

And CG, the annex, is commensurable with the rational straight line A set out.

+

Therefore BC is a second apotome. [X. Deff. III. 2]

+

Therefore the second apotome BC has been found. Q. E. D.

+
+
+ PROPOSITION 87. +

To find the third apotome.

+

Let a rational straight line A be set out, let three numbers E, BC, CD be set out which have not to one another the ratio which a square number has to a square number, but let CB have to BD the ratio which a square number has to a square number.

+

+

Let it be contrived that, as E is to BC, so is the square on A to the square on FG, and, as BC is to CD, so is the square on FG to the square on GH. [X. 6, Por.]

+

Since then, as E is to BC, so is the square on A to the square on FG, therefore the square on A is commensurable with the square on FG. [X. 6]

+

But the square on A is rational; therefore the square on FG is also rational; therefore FG is rational.

+

And, since E has not to BC the ratio which a square number has to a square number, therefore neither has the square on A to the square on FG the ratio which a square number has to a square number; therefore A is incommensurable in length with FG. [X. 9]

+

Again, since, as BC is to CD, so is the square on FG to the square on GH, therefore the square on FG is commensurable with the square on GH. [X. 6]

+

But the square on FG is rational; therefore the square on GH is also rational; therefore GH is rational.

+

And, since BC has not to CD the ratio which a square number has to a square number, therefore neither has the square on FG to the square on GH the ratio which a square number has to a square number; therefore FG is incommensurable in length with GH. [X. 9]

+

And both are rational; therefore FG, GH are rational straight lines commensurable in square only; therefore FH is an apotome. [X. 73]

+

I say next that it is also a third apotome.

+

For since, as E is to BC, so is the square on A to the square on FG, and, as BC is to CD, so is the square on FG to the square on HG, therefore, ex aequali, as E is to CD, so is the square on A to the square on HG. [V. 22]

+

But E has not to CD the ratio which a square number has to a square number; therefore neither has the square on A to the square on GH the ratio which a square number has to a square number; therefore A is incommensurable in length with GH. [X. 9]

+

Therefore neither of the straight lines FG, GH is commensurable in length with the rational straight line A set out.

+

Now let the square on K be that by which the square on FG is greater than the square on GH.

+

Since then, as BC is to CD, so is the square on FG to the square on GH, therefore, convertendo, as BC is to BD, so is the square on FG to the square on K. [V. 19, Por.]

+

But BC has to BD the ratio which a square number has to a square number; therefore the square on FG also has to the square on K the ratio which a square number has to a square number.

+

Therefore FG is commensurable in length with K, [X. 9] and the square on FG is greater than the square on GH by the square on a straight line commensurable with FG.

+

And neither of the straight lines FG, GH is commensurable in length with the rational straight line A set out; therefore FH is a third apotome. [X. Deff. III. 3]

+

Therefore the third apotome FH has been found. Q. E. D.

+
+
+ PROPOSITION 88. +

To find the fourth apotome.

+

Let a rational straight line A be set out, and BG commensurable in length with it; therefore BG is also rational.

+

+

Let two numbers DF, FE be set out such that the whole DE has not to either of the numbers DF, EF the ratio which a square number has to a square number.

+

Let it be contrived that, as DE is to EF, so is the square on BG to the square on GC; [X. 6, Por.] therefore the square on BG is commensurable with the square on GC. [X. 6]

+

But the square on BG is rational; therefore the square on GC is also rational; therefore GC is rational.

+

Now, since DE has not to EF the ratio which a square number has to a square number, therefore neither has the square on BG to the square on GC the ratio which a square number has to a square number; therefore BG is incommensurable in length with GC. [X. 9]

+

And both are rational; therefore BG, GC are rational straight lines commensurable in square only; therefore BC is an apotome. [X. 73]

+

Now let the square on H be that by which the square on BG is greater than the square on GC.

+

Since then, as DE is to EF, so is the square on BG to the square on GC, therefore also, convertendo, as ED is to DF, so is the square on GB to the square on H. [v. 19, Por.]

+

But ED has not to DF the ratio which a square number has to a square number; therefore neither has the square on GB to the square on H the ratio which a square number has to a square number; therefore BG is incommensurable in length with H. [X. 9]

+

And the square on BG is greater than the square on GC by the square on H; therefore the square on BG is greater than the square on GC by the square on a straight line incommensurable with BG.

+

And the whole BG is commensurable in length with the rational straight line A set out.

+

Therefore BC is a fourth apotome. [X. Deff. III. 4]

+

Therefore the fourth apotome has been found. Q. E. D.

+
+
+ PROPOSITION 89. +

To find the fifth apotome.

+

Let a rational straight line A be set out, and let CG be commensurable in length with A; therefore CG is rational.

+

Let two numbers DF, FE be set out such that DE again has not to either of the numbers DF, FE the ratio which a square number has to a square number; and let it be contrived that, as FE is to ED, so is the square on CG to the square on GB.

+

+

Therefore the square on GB is also rational; [X. 6] therefore BG is also rational.

+

Now since, as DE is to EF, so is the square on BG to the square on GC, while DE has not to EF the ratio which a square number has to a square number, therefore neither has the square on BG to the square on GC the ratio which a square number has to a square number; therefore BG is incommensurable in length with GC. [X. 9]

+

And both are rational; therefore BG, GC are rational straight lines commensurable in square only; therefore BC is an apotome. [X. 73]

+

I say next that it is also a fifth apotome.

+

For let the square on H be that by which the square on BG is greater than the square on GC.

+

Since then, as the square on BG is to the square on GC, so is DE to EF, therefore, convertendo, as ED is to DF, so is the square on BG to the square on H. [V. 19, Por.]

+

But ED has not to DF the ratio which a square number has to a square number; therefore neither has the square on BG to the square on H the ratio which a square number has to a square number; therefore BG is incommensurable in length with H. [X. 9]

+

And the square on BG is greater than the square on GC by the square on H; therefore the square on GB is greater than the square on GC by the square on a straight line incommensurable in length with GB.

+

And the annex CG is commensurable in length with the rational straight line A set out; therefore BC is a fifth apotome. [X. Deff. III. 5]

+

Therefore the fifth apotome BC has been found. Q. E. D.

+
+ +
+ PROPOSITION 90. +

To find the sixth apotome.

+

Let a rational straight line A be set out, and three numbers E, BC, CD not having to one another the ratio which a square number has to a square number; and further let CB also not have to BD the ratio which a square number has to a square number.

+

+

Let it be contrived that, as E is to BC, so is the square on A to the square on FG, and, as BC is to CD, so is the square on FG to the square on GH. [X. 6, Por.]

+

Now since, as E is to BC, so is the square on A to the square on FG, therefore the square on A is commensurable with the square on FG. [X. 6]

+

But the square on A is rational; therefore the square on FG is also rational; therefore FG is also rational.

+

And, since E has not to BC the ratio which a square number has to a square number, therefore neither has the square on A to the square on FG the ratio which a square number has to a square number; therefore A is incommensurable in length with FG. [X. 9]

+

Again, since, as BC is to CD, so is the square on FG to the square on GH, therefore the square on FG is commensurable with the square on GH. [X. 6]

+

But the square on FG is rational; therefore the square on GH is also rational; therefore GH is also rational.

+

And, since BC has not to CD the ratio which a square number has to a square number, therefore neither has the square on FG to the square on GH the ratio which a square number has to a square number; therefore FG is incommensurable in length with GH. [X. 9]

+

And both are rational; therefore FG, GH are rational straight lines commensurable in square only; therefore FH is an apotome. [X. 73]

+

I say next that it is also a sixth apotome.

+

For since, as E is to BC, so is the square on A to the square on FG, and, as BC is to CD, so is the square on FG to the square on GH, therefore, ex aequali, as E is to CD, so is the square on A to the square on GH. [v. 22]

+

But E has not to CD the ratio which a square number has to a square number; therefore neither has the square on A to the square on GH the ratio which a square number has to a square number; therefore A is incommensurable in length with GH; [X. 9] therefore neither of the straight lines FG, GH is commensurable in length with the rational straight line A.

+

Now let the square on K be that by which the square on FG is greater than the square on GH.

+

Since then, as BC is to CD, so is the square on FG to the square on GH, therefore, convertendo, as CB is to BD, so is the square on FG to the square on K. [v. 19, Por.]

+

But CB has not to BD the ratio which a square number has to a square number; therefore neither has the square on FG to the square on K the ratio which a square number has to a square number; therefore FG is incommensurable in length with K. [X. 9]

+

And the square on FG is greater than the square on GH by the square on K; therefore the square on FG is greater than the square on GH by the square on a straight line incommensurable in length with FG.

+

And neither of the straight lines FG, GH is commensurable with the rational straight line A set out.

+

Therefore FH is a sixth apotome. [X. Deff. III. 6]

+

Therefore the sixth apotome FH has been found. Q. E. D.

+
+
+ PROPOSITION 91. +

If an area be contained by a rational straight line and a first apotome, the side + of the area is an apotome.

+

For let the area AB be contained by the rational straight line AC and the first apotome AD;

+

I say that the side + of the area AB is an apotome.

+

For, since AD is a first apotome, let DG be its annex; therefore AG, GD are rational straight lines commensurable in square only. [X. 73]

+

And the whole AG is commensurable with the rational straight line AC set out, and the square on AG is greater than the square on GD by the square on a straight line commensurable in length with AG; [X. Deff. III. 1] if therefore there be applied to AG a parallelogram equal to the fourth part of the square on DG and deficient by a square figure, it divides it into commensurable parts. [X. 17]

+

+

Let DG be bisected at E, let there be applied to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF, FG; therefore AF is commensurable with FG.

+

And through the points E, F, G let EH, FI, GK be drawn parallel to AC.

+

Now, since AF is commensurable in length with FG, therefore AG is also commensurable in length with each of the straight lines AF, FG. [X. 15]

+

But AG is commensurable with AC; therefore each of the straight lines AF, FG is commensurable in length with AC. [X. 12]

+

And AC is rational; therefore each of the straight lines AF, FG is also rational, so that each of the rectangles AI, FK is also rational. [X. 19]

+

Now, since DE is commensurable in length with EG, therefore DG is also commensurable in length with each of the straight lines DE, EG. [X. 15]

+

But DG is rational and incommensurable in length with AC; therefore each of the straight lines DE, EG is also rational and incommensurable in length with AC; [X. 13] therefore each of the rectangles DH, EK is medial. [X. 21]

+

Now let the square LM be made equal to AI, and let there be subtracted the square NO having a common angle with it, the angle LPM, and equal to FK; therefore the squares LM, NO are about the same diameter. [VI. 26]

+

Let PR be their diameter, and let the figure be drawn.

+

Since then the rectangle contained by AF, FG is equal to the square on EG, therefore, as AF is to EG, so is EG to FG. [VI. 17]

+

But, as AF is to EG, so is AI to EK, and, as EG is to FG, so is EK to KF; [VI. 1] therefore EK is a mean proportional between AI, KF. [V. 11]

+

But MN is also a mean proportional between LM, NO, as was before proved, [Lemma after X. 53] and AI is equal to the square LM, and KF to NO; therefore MN is also equal to EK.

+

But EK is equal to DH, and MN to LO; therefore DK is equal to the gnomon UVW and NO.

+

But AK is also equal to the squares LM, NO; therefore the remainder AB is equal to ST.

+

But ST is the square on LN; therefore the square on LN is equal to AB; therefore LN is the side + of AB.

+

I say next that LN is an apotome.

+

For, since each of the rectangles AI, FK is rational, and they are equal to LM, NO, therefore each of the squares LM, NO, that is, the squares on LP, PN respectively, is also rational; therefore each of the straight lines LP, PN is also rational.

+

Again, since DH is medial and is equal to LO, therefore LO is also medial.

+

Since then LO is medial, while NO is rational, therefore LO is incommensurable with NO.

+

But, as LO is to NO, so is LP to PN; [VI. 1] therefore LP is incommensurable in length with PN. [X. 11]

+

And both are rational; therefore LP, PN are rational straight lines commensurable in square only; therefore LN is an apotome. [X. 73]

+

And it is the side + of the area AB; therefore the side + of the area AB is an apotome.

+

Therefore etc.

+
+
+ PROPOSITION 92. +

If an area be contained by a rational straight line and a second apotome, the side + of the area is a first apotome of a medial straight line.

+

For let the area AB be contained by the rational straight line AC and the second apotome AD; I say that the side + of the area AB is a first apotome of a medial straight line.

+

+

For let DG be the annex to AD; therefore AG, GD are rational straight lines commensurable in square only, [X. 73] and the annex DG is commensurable with the rational straight line AC set out, while the square on the whole AG is greater than the square on the annex GD by the square on a straight line commensurable in length with AG. [X. Deff. III. 2]

+

Since then the square on AG is greater than the square on GD by the square on a straight line commensurable with AG, therefore, if there be applied to AG a parallelogram equal to the fourth part of the square on GD and deficient by a square figure, it divides it into commensurable parts. [X. 17]

+

Let then DG be bisected at E, let there be applied to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF, FG; therefore AF is commensurable in length with FG.

+

Therefore AG is also commensurable in length with each of the straight lines AF, FG. [X. 15]

+

But AG is rational and incommensurable in length with AC; therefore each of the straight lines AF, FG is also rational and incommensurable in length with AC; [X. 13] therefore each of the rectangles AI, FK is medial. [X. 21]

+

Again, since DE is commensurable with EG, therefore DG is also commensurable with each of the straight lines DE, EG. [X. 15]

+

But DG is commensurable in length with AC.

+

Therefore each of the rectangles DH, EK is rational. [X. 19]

+

Let then the square LM be constructed equal to AI, and let there be subtracted NO equal to FK and being about the same angle with LM, namely the angle LPM; therefore the squares LM, NO are about the same diameter. [VI. 26]

+

Let PR be their diameter, and let the figure be drawn.

+

Since then AI, FK are medial and are equal to the squares on LP, PN, the squares on LP, PN are also medial; therefore LP, PN are also medial straight lines commensurable in square only.

+

And, since the rectangle AF, FG is equal to the square on EG, therefore, as AF is to EG, so is EG to FG, [VI. 17] while, as AF is to EG, so is AI to EK, and, as EG is to FG, so is EK to FK; [VI. 1] therefore EK is a mean proportional between AI, FK. [V. 11]

+

But MN is also a mean proportional between the squares LM, NO, and AI is equal to LM, and FK to NO; therefore MN is also equal to EK.

+

But DH is equal to EK, and LO equal to MN; therefore the whole DK is equal to the gnomon UVW and NO.

+

Since then the whole AK is equal to LM, NO, and, in these, DK is equal to the gnomon UVW and NO, therefore the remainder AB is equal to TS.

+

But TS is the square on LN; therefore the square on LN is equal to the area AB; therefore LN is the side + of the area AB.

+

I say that LN is a first apotome of a medial straight line.

+

For, since EK is rational and is equal to LO, therefore LO, that is, the rectangle LP, PN, is rational.

+

But NO was proved medial; therefore LO is incommensurable with NO.

+

But, as LO is to NO, so is LP to PN; [VI. 1] therefore LP, PN are incommensurable in length. [X. 11]

+

Therefore LP, PN are medial straight lines commensurable in square only which contain a rational rectangle; therefore LN is a first apotome of a medial straight line. [X. 74]

+

And it is the side + of the area AB.

+

Therefore the side + of the area AB is a first apotome of a medial straight line. Q. E. D.

+
+ +
+ PROPOSITION 93. +

If an area be contained by a rational straight line and a third apotome, the side + of the area is a second apotome of a medial straight line.

+

For let the area AB be contained by the rational straight line AC and the third apotome AD; I say that the side + of the area AB is a second apotome of a medial straight line.

+

For let DG be the annex to AD; therefore AG, GD are rational straight lines commensurable in square only, and neither of the straight lines AG, GD is commensurable in length with the rational straight line AC set out, while the square on the whole AG is greater than the square on the annex DG by the square on a straight line commensurable with AG. [X. Deff. III. 3]

+

+

Since then the square on AG is greater than the square on GD by the square on a straight line commensurable with AG, therefore, if there be applied to AG a parallelogram equal to the fourth part of the square on DG and deficient by a square figure, it will divide it into commensurable parts. [X. 17]

+

Let then DG be bisected at E, let there be applied to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF, FG.

+

Let EH, FI, GK be drawn through the points E, F, G parallel to AC.

+

Therefore AF, FG are commensurable; therefore AI is also commensurable with FK. [VI. 1, X. 11]

+

And, since AF, FG are commensurable in length, therefore AG is also commensurable in length with each of the straight lines AF, FG. [X. 15]

+

But AG is rational and incommensurable in length with AC; so that AF, FG are so also. [X. 13]

+

Therefore each of the rectangles AI, FK is medial. [X. 21]

+

Again, since DE is commensurable in length with EG, therefore DG is also commensurable in length with each of the straight lines DE, EG. [X. 15]

+

But GD is rational and incommensurable in length with AC; therefore each of the straight lines DE, EG is also rational and incommensurable in length with AC; [X. 13] therefore each of the rectangles DH, EK is medial. [X. 21]

+

And, since AG, GD are commensurable in square only, therefore AG is incommensurable in length with GD.

+

But AG is commensurable in length with AF, and DG with EG; therefore AF is incommensurable in length with EG. [X. 13]

+

But, as AF is to EG, so is AI to EK; [VI. 1] therefore AI is incommensurable with EK. [X. 11]

+

Now let the square LM be constructed equal to AI, and let there be subtracted NO equal to FK and being about the same angle with LM; therefore LM, NO are about the same diameter. [VI. 26]

+

Let PR be their diameter, and let the figure be drawn.

+

Now, since the rectangle AF, FG is equal to the square on EG, therefore, as AF is to EG, so is EG to FG. [VI. 17]

+

But, as AF is to EG, so is AI to EK, and, as EG is to FG, so is EK to FK; [VI. 1] therefore also, as AI is to EK, so is EK to FK; [V. 11] therefore EK is a mean proportional between AI, FK.

+

But MN is also a mean proportional between the squares LM, NO, and AI is equal to LM, and FK to NO; therefore EK is also equal to MN.

+

But MN is equal to LO, and EK equal to DH; therefore the whole DK is also equal to the gnomon UVW and NO.

+

But AK is also equal to LM, NO; therefore the remainder AB is equal to ST, that is, to the square on LN; therefore LN is the side + of the area AB.

+

I say that LN is a second apotome of a medial straight line.

+

For, since AI, FK were proved medial, and are equal to the squares on LP, PN, therefore each of the squares on LP, PN is also medial; therefore each of the straight lines LP, PN is medial.

+

And, since AI is commensurable with FK, [VI. 1, X. 11] therefore the square on LP is also commensurable with the square on PN.

+

Again, since AI was proved incommensurable with EK, therefore LM is also incommensurable with MN, that is, the square on LP with the rectangle LP, PN; so that LP is also incommensurable in length with PN; [VI. 1, X. 11] therefore LP, PN are medial straight lines commensurable in square only.

+

I say next that they also contain a medial rectangle.

+

For, since EK was proved medial, and is equal to the rectangle LP, PN, therefore the rectangle LP, PN is also medial, so that LP, PN are medial straight lines commensurable in square only which contain a medial rectangle.

+

Therefore LN is a second apotome of a medial straight line; [X. 75] and it is the side + of the area AB.

+

Therefore the side + of the area AB is a second apotome of a medial straight line. Q. E. D.

+
+ +
+ PROPOSITION 94. +

If an area be contained by a rational straight line and a fourth apotome, the side + of the area is minor.

+

For let the area AB be contained by the rational straight line AC and the fourth apotome AD; I say that the side + of the area AB is minor.

+

For let DG be the annex to AD; therefore AG, GD are rational straight lines commensurable in square only, AG is commensurable in length with the rational straight line AC set out, and the square on the whole AG is greater than the square on the annex DG by the square on a straight line incommensurable in length with AG, [X. Deff. III. 4]

+

+

Since then the square on AG is greater than the square on GD by the square on a straight line incommensurable in length with AG, therefore, if there be applied to AG a parallelogram equal to the fourth part of the square on DG and deficient by a square figure, it will divide it into incommensurable parts. [X. 18]

+

Let then DG be bisected at E, let there be applied to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF, FG; therefore AF is incommensurable in length with FG.

+

Let EH, FI, GK be drawn through E, F, G parallel to AC, BD.

+

Since then AG is rational and commensurable in length with AC, therefore the whole AK is rational. [X. 19]

+

Again, since DG is incommensurable in length with AC, and both are rational, therefore DK is medial. [X. 21]

+

Again, since AF is incommensurable in length with FG, therefore AI is also incommensurable with FK. [VI. 1, X. 11]

+

Now let the square LM be constructed equal to AI, and let there be subtracted NO equal to FK and about the same angle, the angle LPM.

+

Therefore the squares LM, NO are about the same diameter. [VI. 26]

+

Let PR be their diameter, and let the figure be drawn.

+

Since then the rectangle AF, FG is equal to the square on EG, therefore, proportionally, as AF is to EG, so is EG to FG. [VI. 17]

+

But, as AF is to EG, so is AI to EK, and, as EG is to FG, so is EK to FK; [VI. 1] therefore EK is a mean proportional between AI, FK. [V. 11]

+

But MN is also a mean proportional between the squares LM, NO, and AI is equal to LM, and FK to NO; therefore EK is also equal to MN.

+

But DH is equal to EK, and LO is equal to MN; therefore the whole DK is equal to the gnomon UVW and NO.

+

Since, then, the whole AK is equal to the squares LM, NO, and, in these, DK is equal to the gnomon UVW and the square NO, therefore the remainder AB is equal to ST, that is, to the square on LN; therefore LN is the side + of the area AB.

+

I say that LN is the irrational straight line called minor.

+

For, since AK is rational and is equal to the squares on LP, PN, therefore the sum of the squares on LP, PN is rational.

+

Again, since DK is medial, and DK is equal to twice the rectangle LP, PN, therefore twice the rectangle LP, PN is medial.

+

And, since AI was proved incommensurable with FK, therefore the square on LP is also incommensurable with the square on PN.

+

Therefore LP, PN are straight lines incommensurable in square which make the sum of the squares on them rational, but twice the rectangle contained by them medial.

+

Therefore LN is the irrational straight line called minor; [X. 76] and it is the side + of the area AB.

+

Therefore the side + of the area AB is minor. Q. E. D.

+
+
+ PROPOSITION 95. +

+ If an area be contained by a rational straight line and a fifth apotome, the side + of the area is a straight line which produces with a rational area a medial whole. +

+

For let the area AB be contained by the rational straight line AC and the fifth apotome AD; I say that the side + of the area AB is a straight line which produces with a rational area a medial whole.

+

For let DG be the annex to AD; therefore AG, GD are rational straight lines commensurable in square only,

the annex GD is commensurable in length with the rational straight line AC set out, and the square on the whole AG is greater than the square on the annex DG by the square on a straight line incommensurable with AG. [X. Deff. III. 5]

+

Therefore, if there be applied to AG a parallelogram equal to the fourth part of the square on DG and deficient by a square figure, it will divide it into incommensurable parts. [X. 18]

+

Let then DG be bisected at the point E, let there be applied to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF, FG; therefore AF is incommensurable in length with FG.

+

Now, since AG is incommensurable in length with CA, and both are rational, therefore AK is medial. [X. 21]

+

Again, since DG is rational and commensurable in length with AC, DK is rational. [X. 19]

+

Now let the square LM be constructed equal to AI, and let the square NO equal to FK and about the same angle, the angle LPM, be subtracted; therefore the squares LM, NO are about the same diameter. [VI. 26]

+

Let PR be their diameter, and let the figure be drawn.

+

Similarly then we can prove that LN is the side + of the area AB.

+

I say that LN is the straight line which produces with a rational area a medial whole.

+

For, since AK was proved medial and is equal to the squares on LP, PN, therefore the sum of the squares on LP, PN is medial.

+

Again, since DK is rational and is equal to twice the rectangle LP, PN, the latter is itself also rational.

+

And, since AI is incommensurable with FK, therefore the square on LP is also incommensurable with the square on PN; therefore LP, PN are straight lines incommensurable in square which make the sum of the squares on them medial but twice the rectangle contained by them rational.

+

Therefore the remainder LN is the irrational straight line called that which produces with a rational area a medial whole; [X. 77] and it is the side + of the area AB.

+

Therefore the side + of the area AB is a straight line which produces with a rational area a medial whole. Q. E. D.

+
+ +
+ PROPOSITION 96. +

If an area be contained by a rational straight line and a sixth apotome, the side + of the area is a straight line which produces with a medial area a medial whole.

+

For let the area AB be contained by the rational straight line AC and the sixth apotome AD; I say that the side + of the area AB is a straight line which produces with a medial area a medial whole.

+

+

For let DG be the annex to AD; therefore AG, GD are rational straight lines commensurable in square only, neither of them is commensurable in length with the rational straight line AC set out, and the square on the whole AG is greater than the square on the annex DG by the square on a straight line incommensurable in length with AG. [X. Deff. III. 6]

+

Since then the square on AG is greater than the square on GD by the square on a straight line incommensurable in length with AG, therefore, if there be applied to AG a parallelogram equal to the fourth part of the square on DG and deficient by a square figure, it will divide it into incommensurable parts. [X. 18]

+

Let then DG be bisected at E, let there be applied to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF, FG; therefore AF is incommensurable in length with FG.

+

But, as AF is to FG, so is AI to FK. [VI. 1] therefore AI is incommensurable with FK. [X. 11]

+

And, since AG, AC are rational straight lines commensurable in square only, AK is medial. [X. 21]

+

Again, since AC, DG are rational straight lines and incommensurable in length, DK is also medial. [X. 21]

+

Now, since AG, GD are commensurable in square only, therefore AG is incommensurable in length with GD.

+

But, as AG is to GD, so is AK to KD; [VI. 1] therefore AK is incommensurable with KD. [X. 11]

+

Now let the square LM be constructed equal to AI, and let NO equal to FK, and about the same angle, be subtracted; therefore the squares LM, NO are about the same diameter. [VI. 26]

+

Let PR be their diameter, and let the figure be drawn.

+

Then in manner similar to the above we can prove that LN is the side + of the area AB.

+

I say that LN is a straight line which produces with a medial area a medial whole.

+

For, since AK was proved medial and is equal to the squares on LP, PN, therefore the sum of the squares on LP, PN is medial.

+

Again, since DK was proved medial and is equal to twice the rectangle LP, PN, twice the rectangle LP, PN is also medial.

+

And, since AK was proved incommensurable with DK, the squares on LP, PN are also incommensurable with twice the rectangle LP, PN.

+

And, since AI is incommensurable with FK, therefore the square on LP is also incommensurable with the square on PN; therefore LP, PN are straight lines incommensurable in square which make the sum of the squares on them medial, twice the rectangle contained by them medial, and further the squares on them incommensurable with twice the rectangle contained by them.

+

Therefore LN is the irrational straight line called that which produces with a medial area a medial whole; [X. 78] and it is the side + of the area AB. +

+

Therefore the side + of the area is a straight line which produces with a medial area a medial whole. Q. E. D.

+
+ +
+ PROPOSITION 97. +

+ The square on an apotome applied to a rational straight line produces as breadth a first apotome. +

+

Let AB be an apotome, and CD rational, and to CD let there be applied CE equal to the square on AB and producing CF as breadth; I say that CF is a first apotome.

+

+

For let BG be the annex to AB; therefore AG, GB are rational straight lines commensurable in square only. [X. 73]

+

To CD let there be applied CH equal to the square on AG, and KL equal to the square on BG.

+

Therefore the whole CL is equal to the squares on AG, GB, and, in these, CE is equal to the square on AB; therefore the remainder FL is equal to twice the rectangle AG, GB. [II. 7]

+

Let FM be bisected at the point N, and let NO be drawn through N parallel to CD; therefore each of the rectangles FO, LN is equal to the rectangle AG, GB.

+

Now, since the squares on AG, GB are rational, and DM is equal to the squares on AG, GB,. therefore DM is rational.

+

And it has been applied to the rational straight line CD, producing CM as breadth; therefore CM is rational and commensurable in length with CD. [X. 20]

+

Again, since twice the rectangle AG, GB is medial, and FL is equal to twice the rectangle AG, GB, therefore FL is medial.

+

And it is applied to the rational straight line CD, producing FM as breadth; therefore FM is rational and incommensurable in length with CD. [X. 22]

+

And, since the squares on AG, GB are rational, while twice the rectangle AG, GB is medial, therefore the squares on AG, GB are incommensurable with twice the rectangle AG, GB.

+

And CL is equal to the squares on AG, GB, and FL to twice the rectangle AG, GB; therefore DM is incommensurable with FL.

+

But, as DM is to FL, so is CM to FM; [VI. 1] therefore CM is incommensurable in length with FM. [X. 11]

+

And both are rational; therefore CM, MF are rational straight lines commensurable in square only; therefore CF is an apotome. [X. 73]

+

I say next that it is also a first apotome.

+

For, since the rectangle AG, GB is a mean proportional between the squares on AG, GB, and CH is equal to the square on AG, KL equal to the square on BG, and NL equal to the rectangle AG, GB, therefore NL is also a mean proportional between CH, KL; therefore, as CH is to NL, so is NL to KL.

+

But, as CH is to NL, so is CK to NM, and, as NL is to KL, so is NM to KM; [VI. 1] therefore the rectangle CK, KM is equal to the square on NM [VI. 17], that is, to the fourth part of the square on FM.

+

And, since the square on AG is commensurable with the square on GB, CH is also commensurable with KL.

+

But, as CH is to KL, so is CK to KM; [VI. 1] therefore CK is commensurable with KM. [X. 11]

+

Since then CM, MF are two unequal straight lines, and to CM there has been applied the rectangle CK, KM equal to the fourth part of the square on FM and deficient by a square figure, while CK is commensurable with KM, therefore the square on CM is greater than the square on MF by the square on a straight line commensurable in length with CM. [X. 17]

+

And CM is commensurable in length with the rational straight line CD set out; therefore CF is a first apotome. [X. Deff. III. 1]

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 98. +

+ The square on a first apotome of a medial straight line applied to a rational straight line produces as breadth a second apotome. +

+

Let AB be a first apotome of a medial straight line and CD a rational straight line, and to CD let there be applied CE equal to the square on AB, producing CF as breadth; I say that CF is a second apotome.

+

For let BG be the annex to AB;. therefore AG, GB are medial straight lines commensurable in square only which contain a rational rectangle. [X. 74]

+

+

To CD let there be applied CH equal to the square on AG, producing CK as breadth, and KL equal to the square on GB, producing KM as breadth; therefore the whole CL is equal to the squares on AG, GB; therefore CL is also medial. [X. 15 and 23, Por.]

+

And it is applied to the rational straight line CD, producing CM as breadth; therefore CM is rational and incommensurable in length with CD. [X. 22]

+

Now, since CL is equal to the squares on AG, GB, and, in these, the square on AB is equal to CE, therefore the remainder, twice the rectangle AG, GB, is equal to FL. [II. 7]

+

But twice the rectangle AG, GB is rational; therefore FL is rational.

+

And it is applied to the rational straight line FE, producing FM as breadth; therefore FM is also rational and commensurable in length with CD. [X. 20]

+

Now, since the sum of the squares on AG, GB, that is, CL, is medial, while twice the rectangle AG, GB, that is, FL, is rational, therefore CL is incommensurable with FL.

+

But, as CL is to FL, so is CM to FM; [VI. 1] therefore CM is incommensurable in length with FM. [X. 11]

+

And both are rational; therefore CM, MF are rational straight lines commensurable in square only; therefore CF is an apotome. [X. 73]

+

I say next that it is also a second apotome.

+

For let FM be bisected at N, and let NO be drawn through N parallel to CD; therefore each of the rectangles FO, NL is equal to the rectangle AG, GB.

+

Now, since the rectangle AG, GB is a mean proportional between the squares on AG, GB, and the square on AG is equal to CH, the rectangle AG, GB to NL, and the square on BG to KL, therefore NL is also a mean proportional between CH, KL; therefore, as CH is to NL, so is NL to KL.

+

But, as CH is to NL, so is CK to NM, and, as NL is to KL, so is NM to MK; [VI. 1] therefore, as CK is to NM, so is NM, so is KM; [V. 11] therefore the rectangle CK, KM is equal to the square on NM [VI. 17], that is, to the fourth part of the square on FM.

+

Since the CM, MF are two unequal straight lines, and the rectangle CK, KM equal to the fourth part of the square on MF and deficient by a square figure has been applied to the greater, CM, and divides it into commensurable parts, therefore the square on CM is greater than the square on MF by the square on a straight line commensurable in length with CM. [X. 17]

+

And the annex FM is commensurable in length with the rational straight line CD set out; therefore CF is a second apotome. [X. Deff. III. 2]

+

Therefore etc. Q. E. D.

+
+ +
+ PROPOSITION 99. +

The square on a second apotome of a medial straight line applied to a rational straight line produces as breadth a third apotome.

+

Let AB be a second apotome of a medial straight line, and CD rational, and to CD let there be applied CE equal to the square on AB, producing CF as breadth; I say that CF is a third apotome.

+

+

For let BG be the annex to AB; therefore AG, GB are medial straight lines commensurable in square only which contain a medial rectangle. [X. 75]

+

Let CH equal to the square on AG be applied to CD, producing CK as breadth, and let KL equal to the square on BG be applied to KH, producing KM as breadth; therefore the whole CL is equal to the squares on AG, GB; therefore CL is also medial. [X. 15 and 23, Por.]

+

And it is applied to the rational straight line CD, producing CM as breadth; therefore CM is rational and incommensurable in length with CD. [X. 22]

+

Now, since the whole CL is equal to the squares on AG, GB, and, in these, CE is equal to the square on AB, therefore the remainder LF is equal to twice the rectangle AG, GB. [II. 7]

+

Let then FM be bisected at the point N, and let NO be drawn parallel to CD; therefore each of the rectangles FO, NL is equal to the rectangle AG, GB.

+

But the rectangle AG, GB is medial; therefore FL is also medial.

+

And it is applied to the rational straight line EF, producing FM as breadth; therefore FM is also rational and incommensurable in length with CD. [X. 22]

+

And, since AG, GB are commensurable in square only, therefore AG is incommensurable in length with GB; therefore the square on AG is also incommensurable with the rectangle AG, GB. [VI. 1, X. 11]

+

But the squares on AG, GB are commensurable with the square on AG, and twice the rectangle AG, GB with the rectangle AG, GB; therefore the squares on AG, GB are incommensurable with twice the rectangle AG, GB. [X. 13]

+

But CL is equal to the squares on AG, GB, and FL is equal to twice the rectangle AG, GB; therefore CL is also incommensurable with FL.

+

But, as CL is to FL, so is CM to FM; [VI. 1] therefore CM is incommensurable in length with FM. [X. 11]

+

And both are rational; therefore CM, MF are rational straight lines commensurable in square only; therefore CF is an apotome. [X. 73]

+

I say next that it is also a third apotome.

+

For, since the square on AG is commensurable with the square on GB, therefore CH is also commensurable with KL, so that CK is also commensurable with KM. [VI. 1, X. 11]

+

And, since the rectangle AG, GB is a mean proportional between the squares on AG, GB, and CH is equal to the square on AG, KL equal to the square on GB, and NL equal to the rectangle AG, GB, therefore NL is also a mean proportional between CH, KL; therefore, as CH is to NL, so is NL to KL.

+

But, as CH is to NL, so is CK to NM, and, as NL is to KL, so is NM to KM; [VI. 1] therefore, as CK is to MN, so is MN to KM; [V. 11] therefore the rectangle CK, KM is equal to [the square on MN, that is, to] the fourth part of the square on FM.

+

Since then CM, MF are two unequal straight lines, and a parallelogram equal to the fourth part of the square on FM and deficient by a square figure has been applied to CM, and divides it into commensurable parts, therefore the square on CM is greater than the square on MF by the square on a straight line commensurable with CM. [X. 17]

+

And neither of the straight lines CM, MF is commensurable in length with the rational straight line CD set out; therefore CF is a third apotome. [X. Deff. III. 3]

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 100. +

+ The square on a minor straight line applied to a rational straight line produces as breadth a fourth apotome. +

+

Let AB be a minor and CD a rational straight line, and to the rational straight line CD let CE be applied equal to the square on AB and producing CF as breadth; I say that CF is a fourth apotome.

+

+

For let BG be the annex to AB; therefore AG, GB are straight lines incommensurable in square which make the sum of the squares on AG, GB rational, but twice the rectangle AG, GB medial. [X. 76]

+

To CD let there be applied CH equal to the square on AG and producing CK as breadth, and KL equal to the square on BG, producing KM as breadth; therefore the whole CL is equal to the squares on AG, GB.

+

And the sum of the squares on AG, GB is rational; therefore CL is also rational.

+

And it is applied to the rational straight line CD, producing CM as breadth; therefore CM is also rational and commensurable in length with CD. [X. 20]

+

And, since the whole CL is equal to the squares on AG, GB, and, in these, CE is equal to the square on AB, therefore the remainder FL is equal to twice the rectangle AG, GB. [II. 7]

+

Let then FM be bisected at the point N, and let NO be drawn through N parallel to either of the straight lines CD, ML; therefore each of the rectangles FO, NL is equal to the rectangle AG, GB.

+

And, since twice the rectangle AG, GB is medial and is equal to FL, therefore FL is also medial.

+

And it is applied to the rational straight line FE, producing FM as breadth; therefore FM is rational and incommensurable in length with CD. [X. 22]

+

And, since the sum of the squares on AG, GB is rational, while twice the rectangle AG, GB is medial, the squares on AG, GB are incommensurable with twice the rectangle AG, GB.

+

But CL is equal to the squares on AG, GB, and FL equal to twice the rectangle AG, GB; therefore CL is incommensurable with FL.

+

But, as CL is to FL, so is CM to MF; [VI. 1] therefore CM is incommensurable in length with MF. [X. 11]

+

And both are rational; therefore CM, MF are rational straight lines commensurable in square only; therefore CF is an apotome. [X. 73]

+

I say that it is also a fourth apotome.

+

For, since AG, GB are incommensurable in square, therefore the square on AG is also incommensurable with the square on GB.

+

And CH is equal to the square on AG, and KL equal to the square on GB; therefore CH is incommensurable with KL.

+

But, as CH is to KL, so is CK to KM; [VI. 1] therefore CK is incommensurable in length with KM. [X. 11]

+

And, since the rectangle AG, GB is a mean proportional between the squares on AG, GB, and the square on AG is equal to CH, the square on GB to KL, and the rectangle AG, GB to NL, therefore NL is a mean proportional between CH, KL; therefore, as CH is to NL, so is NL to KL.

+

But, as CH is to NL, so is CK to NM, and, as NL is to KL, so is NM to KM; [VI. 1] therefore, as CK is to MN, so is MN to KM; [V. 11] therefore the rectangle CK, KM is equal to the square on MN [VI. 17], that is, to the fourth part of the square on FM.

+

Since then CM, MF are two unequal straight lines, and the rectangle CK, KM equal to the fourth part of the square on MF and deficient by a square figure has been applied to CM and divides it into incommensurable parts, therefore the square on CM is greater than the square on MF by the square on a straight line incommensurable with CM. [X. 18]

+

And the whole CM is commensurable in length with the rational straight line CD set out; therefore CF is a fourth apotome. [X. Deff. III. 4]

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 101. +

+ The square on the straight line which produces with a rational area a medial whole, if applied to a rational straight line, produces as breadth a fifth apotome. +

+

Let AB be the straight line which produces with a rational area a medial whole, and CD a rational straight line, and to CD let CE be applied equal to the square on AB and producing CF as breadth; I say that CF is a fifth apotome.

+

+

For let BG be the annex to AB; therefore AG, GB are straight lines incommensurable in square which make the sum of the squares on them medial but twice the rectangle contained by them rational. [X. 77]

+

To CD let there be applied CH equal to the square on AG, and KL equal to the square on GB; therefore the whole CL is equal to the squares on AG, GB.

+

But the sum of the squares on AG, GB together is medial; therefore CL is medial.

+

And it is applied to the rational straight line CD, producing CM as breadth; therefore CM is rational and incommensurable with CD. [X. 22]

+

And, since the whole CL is equal to the squares on AG, GB, and, in these, CE is equal to the square on AB, therefore the remainder FL is equal to twice the rectangle AG, GB. [II. 7]

+

Let then FM be bisected at N, and through N let NO be drawn parallel to either of the straight lines CD, ML; therefore each of the rectangles FO, NL is equal to the rectangle AG, GB:

+

And, since twice the rectangle AG, GB is rational and equal to FL, therefore FL is rational.

+

And it is applied to the rational straight line EF, producing FM as breadth; therefore FM is rational and commensurable in length with CD. [X. 20]

+

Now, since CL is medial, and FL rational, therefore CL is incommensurable with FL.

+

But, as CL is to FL, so is CM to MF; [VI. 1] therefore CM is incommensurable in length with MF. [X. 11]

+

And both are rational; therefore CM, MF are rational straight lines commensurable in square only; therefore CF is an apotome. [X. 73]

+

I say next that it is also a fifth apotome.

+

For we can prove similarly that the rectangle CK, KM is equal to the square on NM, that is, to the fourth part of the square on FM.

+

And, since the square on AG is incommensurable with the square on GB, while the square on AG is equal to CH, and the square on GB to KL, therefore CH is incommensurable with KL.

+

But, as CH is to KL, so is CK to KM; [VI. 1] therefore CK is incommensurable in length with KM. [X. 11]

+

Since then CM, MF are two unequal straight lines, and a parallelogram equal to the fourth part of the square on FM and deficient by a square figure has been applied to CM, and divides it into incommensurable parts, therefore the square on CM is greater than the square on MF by the square on a straight line incommensurable with CM. [X. 18]

+

And the annex FM is commensurable with the rational straight line CD set out; therefore CF is a fifth apotome. [X. Deff. III. 5] Q. E. D.

+
+ +
+ PROPOSITION 102. +

+ The square on the straight line which produces with a medial area a medial whole, if applied to a rational straight line, produces as breadth a sixth apotome. +

+

Let AB be the straight line which produces with a medial area a medial whole, and CD a rational straight line, and to CD let CE be applied equal to the square on AB and producing CF as breadth; I say that CF is a sixth apotome.

+

+

For let BG be the annex to AB; therefore AG, GB are straight lines incommensurable in square which make the sum of the squares on them medial, twice the rectangle AG, GB medial, and the squares on AG, GB incommensurable with twice the rectangle AG, GB. [X. 78]

+

Now to CD let there be applied CH equal to the square on AG and producing CK as breadth, and KL equal to the square on BG; therefore the whole CL is equal to the squares on AG, GB; therefore CL is also medial.

+

And it is applied to the rational straight line CD, producing CM as breadth; therefore CM is rational and incommensurable in length with CD. [X. 22]

+

Since now CL is equal to the squares on AG, GB, and, in these, CE is equal to the square on AB, therefore the remainder FL is equal to twice the rectangle AG, GB. [II. 7]

+

And twice the rectangle AG, GB is medial; therefore FL is also medial.

+

And it is applied to the rational straight line FE, producing FM as breadth; therefore FM is rational and incommensurable in length with CD. [X. 22]

+

And, since the squares on AG, GB are incommensurable with twice the rectangle AG, GB, and CL is equal to the squares on AG, GB, and FL equal to twice the rectangle AG, GB, therefore CL is incommensurable with FL.

+

But, as CL is to FL, so is CM to MF; [VI. 1] therefore CM is incommensurable in length with MF. [X. 11]

+

And both are rational.

+

Therefore CM, MF are rational straight lines commensurable in square only; therefore CF is an apotome. [X. 73]

+

I say next that it is also a sixth apotome.

+

For, since FL is equal to twice the rectangle AG, GB, let FM be bisected at N, and let NO be drawn through N parallel to CD; therefore each of the rectangles FO, NL is equal to the rectangle AG, GB.

+

And, since AG, GB are incommensurable in square, therefore the square on AG is incommensurable with the square on GB.

+

But CH is equal to the square on AG, and KL is equal to the square on GB; therefore CH is incommensurable with KL.

+

But, as CH is to KL, so is CK to KM; [VI. 1] therefore CK is incommensurable with KM. [X. 11]

+

And, since the rectangle AG, GB is a mean proportional between the squares on AG, GB, and CH is equal to the square on AG, KL equal to the square on GB, and NL equal to the rectangle AG, GB, therefore NL is also a mean proportional between CH, KL; therefore, as CH is to NL, so is NL to KL.

+

And for the same reason as before the square on CM is greater than the square on MF by the square on a straight line incommensurable with CM. [X. 18]

+

And neither of them is commensurable with the rational straight line CD set out; therefore CF is a sixth apotome. [X. Deff. III. 6] Q. E. D.

+
+
+ PROPOSITION 103. +

A straight line commensurable in length with an apotome is an apotome and the same in order.

+

Let AB be an apotome, and let CD be commensurable in length with AB;

I say that CD is also an apotome and the same in order with AB.

+

For, since AB is an apotome, let BE be the annex to it; therefore AE, EB are rational straight lines commensurable in square only. [X. 73]

+

Let it be contrived that the ratio of BE to DF is the same as the ratio of AB to CD; [VI. 12] therefore also, as one is to one, so are all to all; [V. 12] therefore also, as the whole AE is to the whole CF, so is AB to CD.

+

But AB is commensurable in length with CD.

+

Therefore AE is also commensurable with CF, and BE with DF. [X. 11]

+

And AE, EB are rational straight lines commensurable in square only; therefore CF, FD are also rational straight lines commensurable in square only. [X. 13]

+

Now since, as AE is to CF, so is BE to DF, alternately therefore, as AE is to EB, so is CF to FD. [V. 16]

+

And the square on AE is greater than the square on EB either by the square on a straight line commensurable with AE or by the square on a straight line incommensurable with it.

+

If then the square on AE is greater than the square on EB by the square on a straight line commensurable with AE, the square on CF will also be greater than the square on FD by the square on a straight line commensurable with CF. [X. 14]

+

And, if AE is commensurable in length with the rational straight line set out, CF is so also, [X. 12] if BE, then DF also, [id.] and, if neither of the straight lines AE, EB, then neither of the straight lines CF, FD. [X. 13]

+

But, if the square on AE is greater than the square on EB by the square on a straight line incommensurable with AE, the square on CF will also be greater than the square on FD by the square on a straight line incommensurable with CF. [X. 14]

+

And, if AE is commensurable in length with the rational straight line set out, CF is so also, if BE, then DF also, [X. 12] and, if neither of the straight lines AE, EB, then neither of the straight lines CF, FD. [X. 13]

+

Therefore CD is an apotome and the same in order with AB. Q. E. D.

+
+
+ PROPOSITION 104. +

A straight line commensurable with an apotome of a medial straight line is an apotome of a medial straight line and the same in order.

+

Let AB be an apotome of a medial straight line, and let CD be commensurable in length with AB; I say that CD is also an apotome of a medial straight line and the same in order with AB.

+

+

For, since AB is an apotome of a medial straight line, let EB be the annex to it.

+

Therefore AE, EB are medial straight lines commensurable in square only. [X. 74, 75]

+

Let it be contrived that, as AB is to CD, so is BE to DF; [VI. 12] therefore AE is also commensurable with CF, and BE with DF. [V. 12, X. 11]

+

But AE, EB are medial straight lines commensurable in square only; therefore CF, FD are also medial straight lines [X. 23] commensurable in square only; [X. 13] therefore CD is an apotome of a medial straight line. [X. 74, 75]

+

I say next that it is also the same in order with AB.

+

Since, as AE is to EB, so is CF to FD, therefore also, as the square on AE is to the rectangle AE, EB, so is the square on CF to the rectangle CF, FD.

+

But the square on AE is commensurable with the square on CF; therefore the rectangle AE, EB is also commensurable with the rectangle CF, FD. [V. 16, X. 11]

+

Therefore, if the rectangle AE, EB is rational, the rectangle CF, FD will also be rational, [X. Def. 4] and if the rectangle AE, EB is medial, the rectangle CF, FD is also medial. [X. 23, Por.]

+

Therefore CD is an apotome of a medial straight line and the same in order with AB. [X. 74, 75] Q. E. D.

+
+
+ PROPOSITION 105. +

A straight line commensurable with a minor straight line is minor.

+

Let AB be a minor straight line, and CD commensurable with AB; I say that CD is also minor.

+

+

Let the same construction be made as before; then, since AE, EB are incommensurable in square, [X. 76] therefore CF, FD are also incommensurable in square. [X. 13]

+

Now since, as AE is to EB, so is CF to FD, [V. 12, V. 16] therefore also, as the square on AE is to the square on EB, so is the square on CF to the square on FD. [VI. 22]

+

Therefore, componendo, as the squares on AE, EB are to the square on EB, so are the squares on CF, FD to the square on FD. [V. 18]

+

But the square on BE is commensurable with the square on DF; therefore the sum of the squares on AE, EB is also commensurable with the sum of the squares on CF, FD. [V. 16, X. 11]

+

But the sum of the squares on AE, EB is rational; [X. 76] therefore the sum of the squares on CF, FD is also rational. [X. Def. 4]

+

Again, since, as the square on AE is to the rectangle AE, EB, so is the square on CF to the rectangle CF, FD, while the square on AE is commensurable with the square on CF, therefore the rectangle AE, EB is also commensurable with the rectangle CF, FD.

+

But the rectangle AE, EB is medial; [X. 76] therefore the rectangle CF, FD is also medial; [X. 23, Por.] therefore CF, FD are straight lines incommensurable in square which make the sum of the squares on them rational, but the rectangle contained by them medial.

+

Therefore CD is minor. [X. 76] Q. E. D.

+
+
+ PROPOSITION 106. +

A straight line commensurable with that which produces with a rational area a medial whole is a straight line which produces with a rational area a medial whole.

+

Let AB be a straight line which produces with a rational area a medial whole, and CD commensurable with AB; I say that CD is also a straight line which produces with a rational area a medial whole.

+

+

For let BE be the annex to AB; therefore AE, EB are straight lines incommensurable in square which make the sum of the squares on AE, EB medial, but the rectangle contained by them rational. [X. 77]

+

Let the same construction be made.

+

Then we can prove, in manner similar to the foregoing, that CF, FD are in the same ratio as AE, EB, the sum of the squares on AE, EB is commensurable with the sum of the squares on CF, FD, and the rectangle AE, EB with the rectangle CF, FD; so that CF, FD are also straight lines incommensurable in square which make the sum of the squares on CF, FD medial, but the rectangle contained by them rational.

+

Therefore CD is a straight line which produces with a rational area a medial whole. [X. 77] Q. E. D.

+
+
+ PROPOSITION 107. +

A straight line commensurable with that which produces with a medial area a medial whole is itself also a straight line which produces with a medial area a medial whole.

+

Let AB be a straight line which produces with a medial area a medial whole, and let CD be commensurable with AB; I say that CD is also a straight line which produces with a medial area a medial whole.

+

+

For let BE be the annex to AB, and let the same construction be made; therefore AE, EB are straight lines incommensurable in square which make the sum of the squares on them medial, the rectangle contained by them medial, and further the sum of the squares on them incommensurable with the rectangle contained by them. [X. 78]

+

Now, as was proved, AE, EB are commensurable with CF, FD, the sum of the squares on AE, EB with the sum of the squares on CF, FD, and the rectangle AE, EB with the rectangle CF, FD; therefore CF, FD are also straight lines incommensurable in square which make the sum of the squares on them medial, the rectangle contained by them medial, and further the sum of the squares on them incommensurable with the rectangle contained by them.

+

Therefore CD is a straight line which produces with a medial area a medial whole. [X. 78]

+
+ +
+ PROPOSITION 108. +

If from a rational area a medial area be subtracted, the side + of the remaining area becomes one of two irrational straight lines, either an apotome or a minor straight line.

+

For from the rational area BC let the medial area BD be subtracted; I say that the side + of the remainder EC becomes one of two irrational straight lines, either an apotome or a minor straight line.

+

+

For let a rational straight line FG be set out, to FG let there be applied the rectangular parallelogram GH equal to BC, and let GK equal to DB be subtracted; therefore the remainder EC is equal to LH.

+

Since then BC is rational, and BD medial, while BC is equal to GH, and BD to GK, therefore GH is rational, and GK medial.

+

And they are applied to the rational straight line FG; therefore FH is rational and commensurable in length with FG, [X. 20] while FK is rational and incommensurable in length with FG; [X. 22] therefore FH is incommensurable in length with FK. [X. 13]

+

Therefore FH, FK are rational straight lines commensurable in square only; therefore KH is an apotome [X. 73], and KF the annex to it.

+

Now the square on HF is greater than the square on FK by the square on a straight line either commensurable with HF or not commensurable.

+

First, let the square on it be greater by the square on a straight line commensurable with it.

+

Now the whole HF is commensurable in length with the rational straight line FG set out; therefore KH is a first apotome. [X. Deff. III. 1]

+

But the side + of the rectangle contained by a rational straight line and a first apotome is an apotome. [X. 91]

+

Therefore the side + of LH, that is, of EC, is an apotome.

+

But, if the square on HF is greater than the square on FK by the square on a straight line incommensurable with HF, while the whole FH is commensurable in length with the rational straight line FG set out, KH is a fourth apotome. [X. Deff. III. 4]

+

But the side + of the rectangle contained by a rational straight line and a fourth apotome is minor. [X. 94] Q. E. D.

+
+
+ PROPOSITION 109. +

If from a medial area a rational area be subtracted, there arise two other irrational straight lines, either a first apotome of a medial straight line or a straight line which produces with a rational area a medial whole.

+

For from the medial area BC let the rational area BD be subtracted.

+

I say that the side + of the remainder EC becomes one of two irrational straight lines, either a first apotome of a medial straight line or a straight line which produces with a rational area a medial whole.

+

+

For let a rational straight line FG be set out, and let the areas be similarly applied.

+

It follows then that FH is rational and incommensurable in length with FG, while KF is rational and commensurable in length with FG; therefore FH, FK are rational straight lines commensurable in square only; [X. 13] therefore KH is an apotome, and FK the annex to it. [X. 73]

+

Now the square on HF is greater than the square on FK either by the square on a straight line commensurable with HF or by the square on a straight line incommensurable with it.

+

If then the square on HF is greater than the square on FK by the square on a straight line commensurable with HF, while the annex FK is commensurable in length with the rational straight line FG set out, KH is a second apotome. [X. Deff. III. 2]

+

But FG is rational; so that the side + of LH, that is, of EC, is a first apotome of a medial straight line. [X. 92]

+

But, if the square on HF is greater than the square on FK by the square on a straight line incommensurable with HF, while the annex FK is commensurable in length with the rational straight line FG set out, KH is a fifth apotome; [X. Deff. III. 5] so that the side + of EC is a straight line which produces with a rational area a medial whole. [X. 95]

+
+
+ PROPOSITION 110. +

If from a medial area there be subtracted a medial area incommensurable with the whole, the two remaining irrational straight lines arise, either a second apotome of a medial straight line or a straight line which produces with a medial area a medial whole.

+

For, as in the foregoing figures, let there be subtracted from the medial area BC the medial area BD incommensurable with the whole;

I say that the side + of EC is one of two irrational straight lines, either a second apotome of a medial straight line or a straight line which produces with a medial area a medial whole.

+

For, since each of the rectangles BC, BD is medial, and BC is incommensurable with BD, it follows that each of the straight lines FH, FK will be rational and incommensurable in length with FG. [X. 22]

+

And, since BC is incommensurable with BD, that is, GH with GK, HF is also incommensurable with FK; [VI. 1, X. 11] therefore FH, FK are rational straight lines commensurable in square only; therefore KH is an apotome. [X. 73]

+

If then the square on FH is greater than the square on FK by the square on a straight line commensurable with FH, while neither of the straight lines FH, FK is commensurable in length with the rational straight line FG set out, KH is a third apotome. [X. Deff. III. 3]

+

But KL is rational, and the rectangle contained by a rational straight line and a third apotome is irrational, and the side + of it is irrational, and is called a second apotome of a medial straight line; [X. 93] so that the side + of LH, that is, of EC, is a second apotome of a medial straight line.

+

But, if the square on FH is greater than the square on FK by the square on a straight line incommensurable with FH, while neither of the straight lines HF, FK is commensurable in length with FG, KH is a sixth apotome. [X. Deff. III. 6]

+

But the side + of the rectangle contained by a rational straight line and a sixth apotome is a straight line which produces with a medial area a medial whole. [X. 96]

+

Therefore the side + of LH, that is, of EC, is a straight line which produces with a medial area a medial whole. Q. E. D.

+
+
+ PROPOSITION 111. +

The apotome is not the same with the binomial straight line.

+

Let AB be an apotome; I say that AB is not the same with the binomial straight line.

+

+

For, if possible, let it be so; let a rational straight line DC be set out, and to CD let there be applied the rectangle CE equal to the square on AB and producing DE as breadth.

+

Then, since AB is an apotome, DE is a first apotome. [X. 97]

+

Let EF be the annex to it; therefore DF, FE are rational straight lines commensurable in square only, the square on DF is greater than the square on FE by the square on a straight line commensurable with DF, and DF is commensurable in length with the rational straight line DC set out. [X. Deff. III. 1]

+

Again, since AB is binomial, therefore DE is a first binomial straight line. [X. 60]

+

Let it be divided into its terms at G, and let DG be the greater term; therefore DG, GE are rational straight lines commensurable in square only, the square on DG is greater than the square on GE by the square on a straight line commensurable with DG, and the greater term DG is commensurable in length with the rational straight line DC set out. [X. Deff. II. 1]

+

Therefore DF is also commensurable in length with DG; [X. 12] therefore the remainder GF is also commensurable in length with DF. [X. 15]

+

But DF is incommensurable in length with EF; therefore FG is also incommensurable in length with EF. [X. 13]

+

Therefore GF, FE are rational straight lines commensurable in square only; therefore EG is an apotome. [X. 73]

+

But it is also rational: which is impossible.

+

Therefore the apotome is not the same with the binomial straight line. Q. E. D.

+
+
+ PROPOSITION 112. +

The square on a rational straight line applied to the binomial straight line produces as breadth an apotome the terms of which are commensurable with the terms of the binomial and moreover in the same ratio; and further the apotome so arising will have the same order as the binomial straight line.

+

Let A be a rational straight line, let BC be a binomial, and let DC be its greater term; let the rectangle BC, EF be equal to the square on A;

I say that EF is an apotome the terms of which are commensurable with CD, DB, and in the same ratio, and further EF will have the same order as BC.

+

For again let the rectangle BD, G be equal to the square on A.

+

Since then the rectangle BC, EF is equal to the rectangle BD, G, therefore, as CB is to BD, so is G to EF. [VI. 16]

+

But CB is greater than BD; therefore G is also greater than EF. [V. 16, V. 14]

+

Let EH be equal to G; therefore, as CB is to BD, so is HE to EF; therefore, separando, as CD is to BD, so is HF to FE. [V. 17]

+

Let it be contrived that, as HF is to FE, so is FK to KE; therefore also the whole HK is to the whole KF as FK is to KE; for, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents. [V. 12]

+

But, as FK is to KE, so is CD to DB; [V. 11] therefore also, as HK is to KF, so is CD to DB. [id.]

+

But the square on CD is commensurable with the square on DB; [X. 36] therefore the square on HK is also commensurable with the square on KF. [VI. 22, X. 11]

+

And, as the square on HK is to the square on KF, so is HK to KE, since the three straight lines HK, KF, KE are proportional. [V. Def. 9]

+

Therefore HK is commensurable in length with KE, so that HE is also commensurable in length with EK. [X. 15]

+

Now, since the square on A is equal to the rectangle EH, BD, while the square on A is rational, therefore the rectangle EH, BD is also rational.

+

And it is applied to the rational straight line BD; therefore EH is rational and commensurable in length with BD; [X. 20] so that EK, being commensurable with it, is also rational and commensurable in length with BD.

+

Since, then, as CD is to DB, so is FK to KE, while CD, DB are straight lines commensurable in square only, therefore FK, KE are also commensurable in square only. [X. 11]

+

But KE is rational; therefore FK is also rational.

+

Therefore FK, KE are rational straight lines commensurable in square only; therefore EF is an apotome. [X. 73]

+

Now the square on CD is greater than the square on DB either by the square on a straight line commensurable with CD or by the square on a straight line incommensurable with it.

+

If then the square on CD is greater than the square on DB by the square on a straight line commensurable with CD, the square on FK is also greater than the square on KE by the square on a straight line commensurable with FK. [X. 14]

+

And, if CD is commensurable in length with the rational straight line set out, so also is FK; [X. 11, 12] if BD is so commensurable, so also is KE; [X. 12] but, if neither of the straight lines CD, DB is so commensurable, neither of the straight lines FK, KE is so.

+

But, if the square on CD is greater than the square on DB by the square on a straight line incommensurable with CD, the square on FK is also greater than the square on KE by the square on a straight line incommensurable with FK. [X. 14]

+

And, if CD is commensurable with the rational straight line set out, so also is FK; if BD is so commensurable, so also is KE; but, if neither of the straight lines CD, DB is so commensurable, neither of the straight lines FK, KE is so; so that FE is an apotome, the terms of which FK, KE are commensurable with the terms CD, DB of the binomial straight line and in the same ratio, and it has the same order as BC. Q. E. D.

+
+
+ PROPOSITION 113. +

The square on a rational straight line, if applied to an apotome, produces as, breadth the binomial straight line the terms of which are commensurable with the terms of the apotome and in the same ratio; and further the binomial so arising has the same order as the apotome.

+

Let A be a rational straight line and BD an apotome, and let the rectangle BD, KH be equal to the square on A, so that the square on the rational straight line A when applied to the apotome BD produces KH as breadth; I say that KH is a binomial straight line the terms of which are commensurable with the terms of BD and in the same ratio; and further KH has the same order as BD.

+

+

For let DC be the annex to BD; therefore BC, CD are rational straight lines commensurable in square only. [X. 73]

+

Let the rectangle BC, G be also equal to the square on A.

+

But the square on A is rational; therefore the rectangle BC, G is also rational.

+

And it has been applied to the rational straight line BC; therefore G is rational and commensurable in length with BC. [X. 20]

+

Since now the rectangle BC, G is equal to the rectangle BD, KH, therefore, proportionally, as CB is to BD, so is KH to G. [VI. 16]

+

But BC is greater than BD; therefore KH is also greater than G. [V. 16, V. 14]

+

Let KE be made equal to G; therefore KE is commensurable in length with BC.

+

And since, as CB is to BD, so is HK to KE, therefore, convertendo, as BC is to CD, so is KH to HE. [V. 19, Por.]

+

Let it be contrived that, as KH is to HE, so is HF to FE; therefore also the remainder KF is to FH as KH is to HE, that is, as BC is to CD. [V. 19]

+

But BC, CD are commensurable in square only; therefore KF, FH are also commensurable in square only. [X. 11]

+

And since, as KH is to HE, so is KF to FH, while, as KH is to HE, so is HF to FE, therefore also, as KF is to FH, so is HF to FE, [V. 11] so that also, as the first is to the third, so is the square on the first to the square on the second; [V. Def. 9] therefore also, as KF is to FE, so is the square on KF to the square on FH.

+

But the square on KF is commensurable with the square on FH, for KF, FH are commensurable in square; therefore KF is also commensurable in length with FE, [X. 11] so that KF is also commensurable in length with KE. [X. 15]

+

But KE is rational and commensurable in length with BC; therefore KF is also rational and commensurable in length with BC. [X. 12]

+

And, since, as BC is to CD, so is KF to FH, alternately, as BC is to KF, so is DC to FH. [V. 16]

+

But BC is commensurable with KF; therefore FH is also commensurable in length with CD. [X. 11]

+

But BC, CD are rational straight lines commensurable in square only; therefore KF, FH are also rational straight lines [X. Def. 3] commensurable in square only; therefore KH is binomial. [X. 36]

+

If now the square on BC is greater than the square on CD by the square on a straight line commensurable with BC, the square on KF will also be greater than the square on FH by the square on a straight line commensurable with KF. [X 14]

+

And, if BC is commensurable in length with the rational straight line set out, so also is KF; if CD is commensurable in length with the rational straight line set out, so also is FH, but, if neither of the straight lines BC, CD, then neither of the straight lines KF, FH.

+

But, if the square on BC is greater than the square on CD by the square on a straight line incommensurable with BC, the square on KF is also greater than the square on FH by the square on a straight line incommensurable with KF. [X. 14]

+

And, if BC is commensurable with the rational straight line set out, so also is KF; if CD is so commensurable, in length with the rational straight line set out, so also is FH; but, if neither of the straight lines BC, CD, then neither of the straight lines KF, FH.

+

Therefore KH is a binomial straight line, the terms of which KF, FH are commensurable with the terms BC, CD of the apotome and in the same ratio, and further KH has the same order as BD. Q. E. D.

+
+
+ PROPOSITION 114. +

If an area be contained by an apotome and the binomial straight line the terms of which are commensurable with the terms of the apotome and in the same ratio, the side + of the area is rational.

+

For let an area, the rectangle AB, CD, be contained by the apotome AB and the binomial straight line CD, and let CE be the greater term of the latter; let the terms CE, ED of the binomial straight line be commensurable with the terms AF, FB of the apotome and in the same ratio; and let the side + of the rectangle AB, CD be G; I say that G is rational.

+

+

For let a rational straight line H be set out, and to CD let there be applied a rectangle equal to the square on H and producing KL as breadth.

+

Therefore KL is an apotome.

+

Let its terms be KM, ML commensurable with the terms CE, ED of the binomial straight line and in the same ratio. [X. 112]

+

But CE, ED are also commensurable with AF, FB and in the same ratio; therefore, as AF is to FB, so is KM to ML.

+

Therefore, alternately, as AF is to KM, so is BF to LM; therefore also the remainder AB is to the remainder KL as AF is to KM. [V. 19]

+

But AF is commensurable with KM; [X. 12] therefore AB is also commensurable with KL. [X. 11]

+

And, as AB is to KL, so is the rectangle CD, AB to the rectangle CD, KL; [VI. 1] therefore the rectangle CD, AB is also commensurable with the rectangle CD, KL. [X. 11]

+

But the rectangle CD, KL is equal to the square on H; therefore the rectangle CD, AB is commensurable with the square on H.

+

But the square on G is equal to the rectangle CD, AB; therefore the square on G is commensurable with the square on H.

+

But the square on H is rational; therefore the square on G is also rational; therefore G is rational.

+

And it is the side + of the rectangle CD, AB.

+

Therefore etc.

+ +

Porism. And it is made manifest to us by this also that it is possible for a rational area to be contained by irrational straight lines. Q. E. D.

+ +
+
+ PROPOSITION 115. +

From a medial straight line there arise irrational straight lines infinite in number, and none of them is the same as any of the preceding.

+

Let A be a medial straight line; I say that from A there arise irrational straight lines infinite in number, and none of them is the same as any of the preceding.

+

Let a rational straight line B be set out,

and let the square on C be equal to the rectangle B, A; therefore C is irrational; [X. Def. 4] for that which is contained by an irrational and a rational straight line is irrational. [deduction from X. 20]

+

And it is not the same with any of the preceding; for the square on none of the preceding, if applied to a rational straight line produces as breadth a medial straight line.

+

Again, let the square on D be equal to the rectangle B, C; therefore the square on D is irrational. [deduction from X. 20]

+

Therefore D is irrational; [X. Def. 4] and it is not the same with any of the preceding, for the square on none of the preceding, if applied to a rational straight line, produces C as breadth.

+

Similarly, if this arrangement proceeds ad infinitum, it is manifest that from the medial straight line there arise irrational straight lines infinite in number, and none is the same with any of the preceding. Q. E. D.

+
+
+
+
+ + BOOK XI. +
+ DEFINITIONS. +
+1 +

A solid is that which has length, breadth, and depth.

+
+
+2 +

An extremity of a solid is a surface.

+
+
+3 +

A straight line is at right angles to a plane, when it makes right angles with all the straight lines which meet it and are in the plane.

+
+
+4 +

A plane is at right angles to a plane when the straight lines drawn, in one of the planes, at right angles to the common section of the planes are at right angles to the remaining plane.

+
+
+5 +

The inclination of a straight line to a plane is, assuming a perpendicular drawn from the extremity of the straight line which is elevated above the plane to the plane, and a straight line joined from the point thus arising to the extremity of the straight line which is in the plane, the angle contained by the straight line so drawn and the straight line standing up.

+
+
+6 +

The inclination of a plane to a plane is the acute angle contained by the straight lines drawn at right angles to the common section at the same point, one in each of the planes.

+
+
+7 +

A plane is said to be similarly inclined to a plane as another is to another when the said angles of the inclinations are equal to one another.

+
+
+8 +

Parallel planes are those which do not meet.

+
+ +
+9 +

Similar solid figures are those contained by similar planes equal in multitude.

+
+
+10 +

Equal and similar solid figures are those contained by similar planes equal in multitude and in magnitude.

+
+
+11 +

A solid angle is the inclination constituted by more than two lines which meet one another and are not in the same surface, towards all the lines.

+

Otherwise: A solid angle is that which is contained by more than two plane angles which are not in the same plane and are constructed to one point.

+
+
+12 +

A pyramid is a solid figure, contained by planes, which is constructed from one plane to one point.

+
+
+13 +

A prism is a solid figure contained by planes two of which, namely those which are opposite, are equal, similar and parallel, while the rest are parallelograms.

+
+
+14 +

When, the diameter of a semicircle remaining fixed, the semicircle is carried round and restored again to the same position from which it began to be moved, the figure so comprehended is a sphere.

+
+
+15 +

The axis of the sphere is the straight line which remains fixed and about which the semicircle is turned.

+
+
+16 +

The centre of the sphere is the same as that of the semicircle.

+
+
+17 +

A diameter of the sphere is any straight line drawn through the centre and terminated in both directions by the surface of the sphere.

+
+
+18 +

When, one side of those about the right angle in a right-angled triangle remaining fixed, the triangle is carried round and restored again to the same position from which it began to be moved, the figure so comprehended is a cone. +

+

And, if the straight line which remains fixed be equal to the remaining side about the right angle which is carried round, the cone will be right-angled; if less, obtuse-angled; and if greater, acute-angled.

+
+
+19 +

The axis of the cone is the straight line which remains fixed and about which the triangle is turned.

+
+ +
+20 +

And the base is the circle described by the straight line which is carried round.

+
+
+21 +

When, one side of those about the right angle in a rectangular parallelogram remaining fixed, the parallelogram is carried round and restored again to the same position from which it began to be moved, the figure so comprehended is a cylinder.

+
+
+22 +

The axis of the cylinder is the straight line which remains fixed and about which the parallelogram is turned.

+
+
+23 +

And the bases are the circles described by the two sides opposite to one another which are carried round.

+
+
+24 +

Similar cones and cylinders are those in which the axes and the diameters of the bases are proportional.

+
+
+25 +

A cube is a solid figure contained by six equal squares.

+
+
+26 +

An octahedron is a solid figure contained by eight equal and equilateral triangles.

+
+
+27 +

An icosahedron is a solid figure contained by twenty equal and equilateral triangles.

+
+
+28 +

A dodecahedron is a solid figure contained by twelve equal, equilateral, and equiangular pentagons.

+
+
+ + + +
+ BOOK XI. PROPOSITIONS. +
+ PROPOSITION 1. +

+ A part of a straight line cannot be in the plane of reference and a part in a plane more elevated. +

+

For, if possible, let a part AB of the straight line ABC be in the plane of reference, and a part BC in a plane more elevated.

+

There will then be in the plane of reference some straight line continuous with AB in a straight line.

+

+

Let it be BD; therefore AB is a common segment of the two straight lines ABC, ABD: which is impossible, inasmuch as, if we describe a circle with centre B and distance AB, the diameters will cut off unequal circumferences of the circle.

+

Therefore a part of a straight line cannot be in the plane of reference, and a part in a plane more elevated. Q. E. D. +1. the plane of reference, + τὸ ὑποκείμενον ἐπίπεδον, the plane laid down or assumed. + 2. more elevated, + μετεωροτέρῳ.

+
+
+ PROPOSITION 2. +

If two straight lines cut one another, they are in one plane, and every triangle is in one plane. +

+

For let the two straight lines AB, CD cut one another at the point E; I say that AB, CD are in one plane, and every triangle is in one plane.

+

+

For let points F, G be taken at random on EC, EB, let CB, FG be joined, and let FH, GK be drawn across; I say first that the triangle ECB is in one plane.

+

For, if part of the triangle ECB, either FHC or GBK, is in the plane of reference, and the rest in another, a part also of one of the straight lines EC, EB will be in the plane of reference, and a part in another.

+

But, if the part FCBG of the triangle ECB be in the plane of reference, and the rest in another, a part also of both the straight lines EC, EB will be in the plane of reference and a part in another: which was proved absurd. [XI. 1]

+

Therefore the triangle ECB is in one plane.

+

But, in whatever plane the triangle ECB is, in that plane also is each of the straight lines EC, EB, and, in whatever plane each of the straight lines EC, EB is, in that plane are AB, CD also. [XI. 1]

+

Therefore the straight lines AB, CD are in one plane, and every triangle is in one plane. Q. E. D.

+
+
+ PROPOSITION 3. +

If two planes cut one another, their common section is a straight line. +

+

For let the two planes AB, BC cut one another, and let the line DB be their common section; I say that the line DB is a straight line.

+

+

For, if not, from D to B let the straight line DEB be joined in the plane AB, and in the plane BC the straight line DFB.

+

Then the two straight lines DEB, DFB will have the same extremities, and will clearly enclose an area: which is absurd.

+

Therefore DEB, DFB are not straight lines.

+

Similarly we can prove that neither will there be any other straight line joined from D to B except DB the common section of the planes AB, BC.

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 4. +

If a straight line be set up at right angles to two straight lines which cut one another, at their common point of section, it will also be at right angles to the plane through them. +

+

For let a straight line EF be set up at right angles to the two straight lines AB, CD, which cut one another at the point E, from E; I say that EF is also at right angles to the plane through AB, CD.

+

+

For let AE, EB, CE, ED be cut off equal to one another, and let any straight line GEH be drawn across through E, at random; let AD, CB be joined, and further let FA, FG, FD, FC, FH, FB be joined from the point F taken at random lton EFgt.

+

Now, since the two straight lines AE, ED are equal to the two straight lines CE, EB, and contain equal angles, [I. 15] therefore the base AD is equal to the base CB, and the triangle AED will be equal to the triangle CEB; [I. 4] so that the angle DAE is also equal to the angle EBC.

+

But the angle AEG is also equal to the angle BEH; [I. 15] therefore AGE, BEH are two triangles which have two angles equal to two angles respectively, and one side equal to one side, namely that adjacent to the equal angles, that is to say, AE to EB; therefore they will also have the remaining sides equal to the remaining sides. [I. 26]

+

Therefore GE is equal to EH, and AG to BH.

+

And, since AE is equal to EB, while FE is common and at right angles, therefore the base FA is equal to the base FB. [I. 4]

+

For the same reason FC is also equal to FD.

+

And, since AD is equal to CB, and FA is also equal to FB, the two sides FA, AD are equal to the two sides FB, BC respectively; and the base FD was proved equal to the base FC; therefore the angle FAD is also equal to the angle FBC. [I. 8]

+

And since, again, AG was proved equal to BH, and further FA also equal to FB, the two sides FA, AG are equal to the two sides FB, BH.

+

And the angle FAG was proved equal to the angle FBH; therefore the base FG is equal to the base FH. [I. 4]

+

Now since, again, GE was proved equal to EH, and EF is common, the two sides GE, EF are equal to the two sides HE, EF; and the base FG is equal to the base FH; therefore the angle GEF is equal to the angle HEF. [I. 8]

+

Therefore each of the angles GEF, HEF is right.

+

Therefore FE is at right angles to GH drawn at random through E.

+

Similarly we can prove that FE will also make right angles with all the straight lines which meet it and are in the plane of reference.

+

But a straight line is at right angles to a plane when it makes right angles with all the straight lines which meet it and are in that same plane; [XI. Def. 3] therefore FE is at right angles to the plane of reference.

+

But the plane of reference is the plane through the straight lines AB, CD.

+

Therefore FE is at right angles to the plane through AB, CD.

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 5. +

If a straight line be set up at right angles to three straight lines which meet one another, at their common point of section, the three straight lines are in one plane. +

+

For let a straight line AB be set up at right angles to the three straight lines BC, BD, BE, at their point of meeting at B; I say that BC, BD, BE are in one plane.

+

+

For suppose they are not, but, if possible, let BD, BE be in the plane of reference and BC in one more elevated; let the plane through AB, BC be produced; it will thus make, as common section in the plane of reference, a straight line. [XI. 3]

+

Let it make BF.

+

Therefore the three straight lines AB, BC, BF are in one plane, namely that drawn through AB, BC.

+

Now, since AB is at right angles to each of the straight lines BD, BE, therefore AB is also at right angles to the plane through BD, BE. [XI. 4]

+

But the plane through BD, BE is the plane of reference; therefore AB is at right angles to the plane of reference.

+

Thus AB will also make right angles with all the straight lines which meet it and are in the plane of reference. [XI. Def. 3]

+

But BF which is in the plane of reference meets it; therefore the angle ABF is right.

+

But, by hypothesis, the angle ABC is also right; therefore the angle ABF is equal to the angle ABC.

+

And they are in one plane: which is impossible.

+

Therefore the straight line BC is not in a more elevated plane; therefore the three straight lines BC, BD, BE are in one plane.

+

Therefore, if a straight line be set up at right angles to three straight lines, at their point of meeting, the three straight lines are in one plane. Q. E. D.

+
+
+ PROPOSITION 6. +

If two straight lines be at right angles to the same plane, the straight lines will be parallel. +

+

For let the two straight lines AB, CD be at right angles to the plane of reference; I say that AB is parallel to CD.

+

+

For let them meet the plane of reference at the points B, D, let the straight line BD be joined, let DE be drawn, in the plane of reference, at right angles to BD, let DE be made equal to AB, and let BE, AE, AD be joined.

+

Now, since AB is at right angles to the plane of reference, it will also make right angles with all the straight lines which meet it and are in the plane of reference. [XI. Def. 3]

+

But each of the straight lines BD, BE is in the plane of reference and meets AB; therefore each of the angles ABD, ABE is right.

+

For the same reason each of the angles CDB, CDE is also right.

+

And, since AB is equal to DE, and BD is common, the two sides AB, BD are equal to the two sides ED, DB; and they include right angles; therefore the base AD is equal to the base BE. [I. 4]

+

And, since AB is equal to DE, while AD is also equal to BE, the two sides AB, BE are equal to the two sides ED, DA; and AE is their common base; therefore the angle ABE is equal to the angle EDA. [I. 8]

+

But the angle ABE is right; therefore the angle EDA is also right; therefore ED is at right angles to DA.

+

But it is also at right angles to each of the straight lines BD, DC; therefore ED is set up at right angles to the three straight lines BD, DA, DC at their point of meeting; therefore the three straight lines BD, DA, DC are in one plane. [XI. 5]

+

But, in whatever plane DB, DA are, in that plane is AB also, for every triangle is in one plane; [XI. 2] therefore the straight lines AB, BD, DC are in one plane.

+

And each of the angles ABD, BDC is right; therefore AB is parallel to CD. [I. 28]

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 7. +

If two straight lines be parallel and points be taken at random on each of them, the straight line joining the points is in the same plane with the parallel straight lines. +

+

Let AB, CD be two parallel straight lines, and let points E, F be taken at random on them respectively; I say that the straight line joining the points E, F is in the same plane with the parallel straight lines.

+

+

For suppose it is not, but, if possible, let it be in a more elevated plane as EGF, and let a plane be drawn through EGF; it will then make, as section in the plane of reference, a straight line. [XI. 3]

+

Let it make it, as EF; therefore the two straight lines EGF, EF will enclose an area: which is impossible.

+

Therefore the straight line joined from E to F is not in a plane more elevated; therefore the straight line joined from E to F is in the plane through the parallel straight lines AB, CD.

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 8. +

If two straight lines be parallel, and one of them be at right angles to any plane, the remaining one will also be at right angles to the same plane. +

+

Let AB, CD be two parallel straight lines, and let one of them, AB, be at right angles to the plane of reference; I say that the remaining one, CD, will also be at right angles to the same plane.

+

+

For let AB, CD meet the plane of reference at the points B, D, and let BD be joined; therefore AB, CD, BD are in one plane. [XI. 7]

+

Let DE be drawn, in the plane of reference, at right angles to BD, let DE be made equal to AB, and let BE, AE, AD be joined.

+

Now, since AB is at right angles to the plane of reference, therefore AB is also at right angles to all the straight lines which meet it and are in the plane of reference; [XI. Def. 3] therefore each of the angles ABD, ABE is right.

+

And, since the straight line BD has fallen on the parallels AB, CD, therefore the angles ABD, CDB are equal to two right angles. [I. 29]

+

But the angle ABD is right; therefore the angle CDB is also right; therefore CD is at right angles to BD.

+

And, since AB is equal to DE, and BD is common, the two sides AB, BD are equal to the two sides ED, DB; and the angle ABD is equal to the angle EDB, for each is right; therefore the base AD is equal to the base BE.

+

And, since AB is equal to DE, and BE to AD, the two sides AB, BE are equal to the two sides ED, DA respectively, and AE is their common base; therefore the angle ABE is equal to the angle EDA.

+

But the angle ABE is right; therefore the angle EDA is also right; therefore ED is at right angles to AD.

+

But it is also at right angles to DB; therefore ED is also at right angles to the plane through BD, DA. [XI. 4]

+

Therefore ED will also make right angles with all the straight lines which meet it and are in the plane through BD, DA.

+

But DC is in the plane through BD, DA, inasmuch as AB, BD are in the plane through BD, DA, [XI. 2] and DC is also in the plane in which AB, BD are.

+

Therefore ED is at right angles to DC, so that CD is also at right angles to DE.

+

But CD is also at right angles to BD.

+

Therefore CD is set up at right angles to the two straight lines DE, DB which cut one another, from the point of section at D; so that CD is also at right angles to the plane through DE, DB. [XI. 4]

+

But the plane through DE, DB is the plane of reference; therefore CD is at right angles to the plane of reference.

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 9. +

+ Straight lines which are parallel to the same straight line and are not in the same plane with it are also parallel to one another. +

+

For let each of the straight lines AB, CD be parallel to EF, not being in the same plane with it; I say that AB is parallel to CD.

+

+

For let a point G be taken at random on EF, and from it let there be drawn GH, in the plane through EF, AB, at right angles to EF, and GK in the plane through FE, CD again at right angles to EF.

+

Now, since EF is at right angles to each of the straight lines GH, GK, therefore EF is also at right angles to the plane through GH, GK. [XI. 4]

+

And EF is parallel to AB; therefore AB is also at right angles to the plane through HG, GK. [XI. 8]

+

For the same reason CD is also at right angles to the plane through HG, GK; therefore each of the straight lines AB, CD is at right angles to the plane through HG, GK.

+

But if two straight lines be at right angles to the same plane, the straight lines are parallel; [XI. 6] therefore AB is parallel to CD.

+
+ +
+ PROPOSITION 10. +

If two straight lines meeting one another be parallel to two straight lines meeting one another not in the same plane, they will contain equal angles. +

+

For let the two straight lines AB, BC meeting one another be parallel to the two straight lines DE, EF meeting one another, not in the same plane; I say that the angle ABC is equal to the angle DEF.

+

+

For let BA, BC, ED, EF be cut off equal to one another, and let AD, CF, BE, AC, DF be joined.

+

Now, since BA is equal and parallel to ED, therefore AD is also equal and parallel to BE. [I. 33]

+

For the same reason CF is also equal and parallel to BE.

+

Therefore each of the straight lines AD, CF is equal and parallel to BE.

+

But straight lines which are parallel to the same straight line and are not in the same plane with it are parallel to one another; [XI. 9] therefore AD is parallel and equal to CF.

+

And AC, DF join them; therefore AC is also equal and parallel to DF. [I. 33]

+

Now, since the two sides AB, BC are equal to the two sides DE, EF, and the base AC is equal to the base DF, therefore the angle ABC is equal to the angle DEF. [I. 8]

+

Therefore etc.

+
+
+ PROPOSITION 11. +

+ From a given elevated point to draw a straight line perpendicular to a given plane. +

+

Let A be the given elevated point, and the plane of reference the given plane; thus it is required to draw from the point A a straight line perpendicular to the plane of reference.

+

Let any straight line BC be drawn, at random, in the plane of reference, and let AD be drawn from the point A perpendicular to BC. [I. 12]

+

+

If then AD is also perpendicular to the plane of reference, that which was enjoined will have been done.

+

But, if not, let DE be drawn from the point D at right angles to BC and in the plane of reference, [I. 11] let AF be drawn from A perpendicular to DE, [I. 12] and let GH be drawn through the point F parallel to BC. [I. 31]

+

Now, since BC is at right angles to each of the straight lines DA, DE, therefore BC is also at right angles to the plane through ED, DA. [XI. 4]

+

And GH is parallel to it; but, if two straight lines be parallel, and one of them be at right angles to any plane, the remaining one will also be at right angles to the same plane; [XI. 8] therefore GH is also at right angles to the plane through ED, DA.

+

Therefore GH is also at right angles to all the straight lines which meet it and are in the plane through ED, DA. [XI. Def. 3]

+

But AF meets it and is in the plane through ED, DA; therefore GH is at right angles to FA, so that FA is also at right angles to GH.

+

But AF is also at right angles to DE; therefore AF is at right angles to each of the straight lines GH, DE.

+

But, if a straight line be set up at right angles to two straight lines which cut one another, at the point of section, it will also be at right angles to the plane through them; [XI. 4] therefore FA is at right angles to the plane through ED, GH.

+

But the plane through ED, GH is the plane of reference; therefore AF is at right angles to the plane of reference.

+

Therefore from the given elevated point A the straight line AF has been drawn perpendicular to the plane of reference. Q. E. F.

+
+
+ PROPOSITION 12. +

+ To set up a straight line at right angles to a given plane from a given point in it. +

+

Let the plane of reference be the given plane, and A the point in it; thus it is required to set up from the point A a straight line at right angles to the plane of reference.

+

+

Let any elevated point B be conceived, from B let BC be drawn perpendicular to the plane of reference, [XI. 11] and through the point A let AD be drawn parallel to BC. [I. 31]

+

Then, since AD, CB are two parallel straight lines, while one of them, BC, is at right angles to the plane of reference, therefore the remaining one, AD, is also at right angles to the plane of reference. [XI. 8]

+

Therefore AD has been set up at right angles to the given plane from the point A in it.

+
+ +
+ PROPOSITION 13. +

+ From the same point two straight lines cannot be set up at right angles to the same plane on the same side. +

+

For, if possible, from the same point A let the two straight lines AB, AC be set up at right angles to the plane of reference and on the same side, and let a plane be drawn through BA, AC; it will then make, as section through A in the plane of reference, a straight line. [XI. 3]

+

+

Let it make DAE; therefore the straight lines AB, AC, DAE are in one plane.

+

And, since CA is at right angles to the plane of reference, it will also make right angles with all the straight lines which meet it and are in the plane of reference. [XI. Def. 3]

+

But DAE meets it and is in the plane of reference; therefore the angle CAE is right.

+

For the same reason the angle BAE is also right; therefore the angle CAE is equal to the angle BAE.

+

And they are in one plane: which is impossible.

+

Therefore etc. Q. E. D.

+
+ +
+ PROPOSITION 14. +

+ Planes to which the same straight line is at right angles will be parallel. +

+

For let any straight line AB be at right angles to each of the planes CD, EF; I say that the planes are parallel.

+

+

For, if not, they will meet when produced.

+

Let them meet; they will then make, as common section, a straight line. [XI. 3]

+

Let them make GH; let a point K be taken at random on GH, and let AK, BK be joined.

+

Now, since AB is at right angles to the plane EF, therefore AB is also at right angles to BK which is a straight line in the plane EF produced; [XI. Def. 3] therefore the angle ABK is right.

+

For the same reason the angle BAK is also right.

+

Thus, in the triangle ABK, the two angles ABK, BAK are equal to two right angles: which is impossible. [I. 17]

+

Therefore the planes CD, EF will not meet when produced; therefore the planes CD, EF are parallel. [XI. Def. 8]

+

Therefore planes to which the same straight line is at right angles are parallel. Q. E. D.

+
+ +
+ PROPOSITION 15. +

If two straight lines meeting one another be parallel to two straight lines meeting one another, not being in the same plane, the planes through them are parallel. +

+

For let the two straight lines AB, BC meeting one another be parallel to the two straight lines DE, EF meeting one another, not being in the same plane; I say that the planes produced through AB, BC and DE, EF will not meet one another.

+

+

For let BG be drawn from the point B perpendicular to the plane through DE, EF [XI. 11], and let it meet the plane at the point G; through G let GH be drawn parallel to ED, and GK parallel to EF. [I. 31]

+

Now, since BG is at right angles to the plane through DE, EF, therefore it will also make right angles with all the straight lines which meet it and are in the plane through DE, EF. [XI. Def. 3]

+

But each of the straight lines GH, GK meets it and is in the plane through DE, EF; therefore each of the angles BGH, BGK is right.

+

And, since BA is parallel to GH, [XI. 9] therefore the angles GBA, BGH are equal to two right angles. [I. 29]

+

But the angle BGH is right; therefore the angle GBA is also right; therefore GB is at right angles to BA.

+

For the same reason GB is also at right angles to BC.

+

Since then the straight line GB is set up at right angles to the two straight lines BA, BC which cut one another, therefore GB is also at right angles to the plane through BA, BC. [XI. 4]

+

But planes to which the same straight line is at right angles are parallel; [XI. 14] therefore the plane through AB, BC is parallel to the plane through DE, EF.

+

Therefore, if two straight lines meeting one another be parallel to two straight lines meeting one another, not in the same plane, the planes through them are parallel. Q. E. D.

+
+ +
+ PROPOSITION 16. +

If two parallel planes be cut by any plane, their common sections are parallel. +

+

For let the two parallel planes AB, CD be cut by the plane EFGH, and let EF, GH be their common sections; I say that EF is parallel to GH.

+

+

For, if not, EF, GH will, when produced, meet either in the direction of F, H or of E, G.

+

Let them be produced, as in the direction of F, H, and let them, first, meet at K.

+

Now, since EFK is in the plane AB, therefore all the points on EFK are also in the plane AB. [XI. 1]

+

But K is one of the points on the straight line EFK; therefore K is in the plane AB.

+

For the same reason K is also in the plane CD; therefore the planes AB, CD will meet when produced.

+

But they do not meet, because they are, by hypothesis, parallel; therefore the straight lines EF, GH will not meet when produced in the direction of F, H.

+

Similarly we can prove that neither will the straight lines EF, GH meet when produced in the direction of E, G.

+

But straight lines which do not meet in either direction are parallel. [I. Def. 23]

+

Therefore EF is parallel to GH.

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 17. +

If two straight lines be cut by parallel planes, they will be cut in the same ratios. +

+

For let the two straight lines AB, CD be cut by the parallel planes GH, KL, MN at the points A, E, B and C, F, D; I say that, as the straight line AE is to EB, so is CF to FD.

+

+

For let AC, BD, AD be joined, let AD meet the plane KL at the point O, and let EO, OF be joined.

+

Now, since the two parallel planes KL, MN are cut by the plane EBDO, their common sections EO, BD are parallel. [XI. 16]

+

For the same reason, since the two parallel planes GH, KL are cut by the plane AOFC, their common sections AC, OF are parallel. [id.]

+

And, since the straight line EO has been drawn parallel to BD, one of the sides of the triangle ABD, therefore, proportionally, as AE is to EB, so is AO to OD. [VI. 2]

+

Again, since the straight line OF has been drawn parallel to AC, one of the sides of the triangle ADC, proportionally, as AO is to OD, so is CF to FD. [id.]

+

But it was also proved that, as AO is to OD, so is AE to EB; therefore also, as AE is to EB, so is CF to FD. [V. 11]

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 18. +

If a straight line be at right angles to any plane, all the planes through it will also be at right angles to the same plane. +

+

For let any straight line AB be at right angles to the plane of reference; I say that all the planes through AB are also at right angles to the plane of reference.

+

+

For let the plane DE be drawn through AB, let CE be the common section of the plane DE and the plane of reference, let a point F be taken at random on CE, and from F let FG be drawn in the plane DE at right angles to CE. [I. 11]

+

Now, since AB is at right angles to the plane of reference, AB is also at right angles to all the straight lines which meet it and are in the plane of reference; [XI. Def. 3] so that it is also at right angles to CE; therefore the angle ABF is right.

+

But the angle GFB is also right; therefore AB is parallel to FG. [I. 28]

+

But AB is at right angles to the plane of reference; therefore FG is also at right angles to the plane of reference. [XI. 8]

+

Now a plane is at right angles to a plane, when the straight lines drawn, in one of the planes, at right angles to the common section of the planes are at right angles to the remaining plane. [XI. Def. 4]

+

And FG, drawn in one of the planes DE at right angles to CE, the common section of the planes, was proved to be at right angles to the plane of reference; therefore the plane DE is at right angles to the plane of reference.

+

Similarly also it can be proved that all the planes through AB are at right angles to the plane of reference.

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 19. +

If two planes which cut one another be at right angles to any plane, their common section will also be at right angles to the same plane. +

+

For let the two planes AB, BC be at right angles to the plane of reference, and let BD be their common section; I say that BD is at right angles to the plane of reference.

+

For suppose it is not, and from the point D let DE be drawn in the plane AB at right angles to the straight line AD, and DF in the plane BC at right angles to CD.

+

+

Now, since the plane AB is at right angles to the plane of reference, and DE has been drawn in the plane AB at right angles to AD, their common section, therefore DE is at right angles to the plane of reference. [XI. Def. 4]

+

Similarly we can prove that DF is also at right angles to the plane of reference.

+

Therefore from the same point D two straight lines have been set up at right angles to the plane of reference on the same side: which is impossible. [XI. 13]

+

Therefore no straight line except the common section DB of the planes AB, BC can be set up from the point D at right angles to the plane of reference.

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 20. +

If a solid angle be contained by three plane angles, any two, taken together in any manner, are greater than the remaining one. +

+

For let the solid angle at A be contained by the three plane angles BAC, CAD, DAB; I say that any two of the angles BAC, CAD, DAB, taken together in any manner, are greater than the remaining one.

+

+

If now the angles BAC, CAD, DAB are equal to one another, it is manifest that any two are greater than the remaining one.

+

But, if not, let BAC be greater, and on the straight line AB, and at the point A on it, let the angle BAE be constructed, in the plane through BA, AC, equal to the angle DAB; let AE be made equal to AD, and let BEC, drawn across through the point E, cut the straight lines AB, AC at the points B, C; let DB, DC be joined.

+

Now, since DA is equal to AE, and AB is common, two sides are equal to two sides; and the angle DAB is equal to the angle BAE; therefore the base DB is equal to the base BE. [I. 4]

+

And, since the two sides BD, DC are greater than BC, [I. 20] and of these DB was proved equal to BE, therefore the remainder DC is greater than the remainder EC.

+

Now, since DA is equal to AE, and AC is common, and the base DC is greater than the base EC, therefore the angle DAC is greater than the angle EAC. [I. 25]

+

But the angle DAB was also proved equal to the angle BAE; therefore the angles DAB, DAC are greater than the angle BAC.

+

Similarly we can prove that the remaining angles also, taken together two and two, are greater than the remaining one.

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 21. +

+ Any solid angle is contained by plane angles less than four right angles. +

+

Let the angle at A be a solid angle contained by the plane angles BAC, CAD, DAB; I say that the angles BAC, CAD, DAB are less than four right angles.

+

For let points B, C, D be taken at random on the straight lines AB, AC, AD respectively, and let BC, CD, DB be joined.

+

+

Now, since the solid angle at B is contained by the three plane angles CBA, ABD, CBD, any two are greater than the remaining one; [XI. 20] therefore the angles CBA, ABD are greater than the angle CBD.

+

For the same reason the angles BCA, ACD are also greater than the angle BCD, and the angles CDA, ADB are greater than the angle CDB; therefore the six angles CBA, ABD, BCA, ACD, CDA, ADB are greater than the three angles CBD, BCD, CDB.

+

But the three angles CBD, BDC, BCD are equal to two right angles; [I. 32] therefore the six angles CBA, ABD, BCA, ACD, CDA, ADB are greater than two right angles.

+

And, since the three angles of each of the triangles ABC, ACD, ADB are equal to two right angles, therefore the nine angles of the three triangles, the angles CBA, ACB, BAC, ACD, CDA, CAD, ADB, DBA, BAD are equal to six right angles; and of them the six angles ABC, BCA, ACD, CDA, ADB, DBA are greater than two right angles; therefore the remaining three angles BAC, CAD, DAB containing the solid angle are less than four right angles.

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 22. +

If there be three plane angles of which two, taken together in any manner, are greater than the remaining one, and they are contained by equal straight lines, it is possible to construct a triangle out of the straight lines joining the extremities of the equal straight lines. +

+

Let there be three plane angles ABC, DEF, GHK, of which two, taken together in any manner, are greater than the remaining one, namely the angles ABC, DEF greater than the angle GHK, the angles DEF, GHK greater than the angle ABC, and, further, the angles GHK, ABC greater than the angle DEF; let the straight lines AB, BC, DE, EF, GH, HK be equal, and let AC, DF, GK be joined; I say that it is possible to construct a triangle out of straight lines equal to AC, DF, GK, that is, that any two of the straight lines AC, DF, GK are greater than the remaining one.

+

+

Now, if the angles ABC, DEF, GHK are equal to one another, it is manifest that, AC, DF, GK being equal also, it is possible to construct a triangle out of straight lines equal to AC, DF, GK.

+

But, if not, let them be unequal, and on the straight line HK, and at the point H on it, let the angle KHL be constructed equal to the angle ABC; let HL be made equal to one of the straight lines AB, BC, DE, EF, GH, HK, and let KL, GL be joined.

+

+

Now, since the two sides AB, BC are equal to the two sides KH, HL, and the angle at B is equal to the angle KHL, therefore the base AC is equal to the base KL. [I. 4]

+

And, since the angles ABC, GHK are greater than the angle DEF, while the angle ABC is equal to the angle KHL, therefore the angle GHL is greater than the angle DEF.

+

And, since the two sides GH, HL are equal to the two sides DE, EF, and the angle GHL is greater than the angle DEF, therefore the base GL is greater than the base DF. [I. 24]

+

But GK, KL are greater than GL.

+

Therefore GK, KL are much greater than DF.

+

But KL is equal to AC; therefore AC, GK are greater than the remaining straight line DF.

+

Similarly we can prove that AC, DF are greater than GK, and further DF, GK are greater than AC.

+

Therefore it is possible to construct a triangle out of straight lines equal to AC, DF, GK. Q. E. D.

+
+
+ PROPOSITION 23. +

To construct a solid angle out of three plane angles two of which, taken together in any manner, are greater than the remaining one: thus the three angles must be less than four right angles. +

+

Let the angles ABC, DEF, GHK be the three given plane angles, and let two of these, taken together in any manner, be greater than the remaining one, while, further, the three are less than four right angles; thus it is required to construct a solid angle out of angles equal to the angles ABC, DEF, GHK.

+

+

Let AB, BC, DE, EF, GH, HK be cut off equal to one another, and let AC, DF, GK be joined; it is therefore possible to construct a triangle out of straight lines equal to AC, DF, GK. [XI. 22]

+

+

Let LMN be so constructed that AC is equal to LM, DF to MN, and further GK to NL, let the circle LMN be described about the triangle LMN, let its centre be taken, and let it be O; let LO, MO, NO be joined; I say that AB is greater than LO.

+

For, if not, AB is either equal to LO, or less.

+

First, let it be equal.

+

Then, since AB is equal to LO, while AB is equal to BC, and OL to OM, the two sides AB, BC are equal to the two sides LO, OM respectively; and, by hypothesis, the base AC is equal to the base LM; therefore the angle ABC is equal to the angle LOM. [I. 8]

+

For the same reason the angle DEF is also equal to the angle MON, and further the angle GHK to the angle NOL; therefore the three angles ABC, DEF, GHK are equal to the three angles LOM, MON, NOL.

+

But the three angles LOM, MON, NOL are equal to four right angles; therefore the angles ABC, DEF, GHK are equal to four right angles.

+

But they are also, by hypothesis, less than four right angles: which is absurd.

+

Therefore AB is not equal to LO.

+

I say next that neither is AB less than LO.

+

For, if possible, let it be so, and let OP be made equal to AB, and OQ equal to BC, and let PQ be joined.

+

Then, since AB is equal to BC, OP is also equal to OQ, so that the remainder LP is equal to QM.

+

Therefore LM is parallel to PQ, [VI. 2] and LMO is equiangular with PQO; [I. 29] therefore, as OL is to LM, so is OP to PQ; [VI. 4] and alternately, as LO is to OP, so is LM to PQ. [V. 16]

+

But LO is greater than OP; therefore LM is also greater than PQ.

+

But LM was made equal to AC; therefore AC is also greater than PQ.

+

Since, then, the two sides AB, BC are equal to the two sides PO, OQ, and the base AC is greater than the base PQ, therefore the angle ABC is greater than the angle POQ. [I. 25]

+

Similarly we can prove that the angle DEF is also greater than the angle MON, and the angle GHK greater than the angle NOL.

+

Therefore the three angles ABC, DEF, GHK are greater than the three angles LOM, MON, NOL.

+

But, by hypothesis, the angles ABC, DEF, GHK are less than four right angles; therefore the angles LOM, MON, NOL are much less than four right angles.

+

But they are also equal to four right angles: which is absurd.

+

Therefore AB is not less than LO.

+

And it was proved that neither is it equal; therefore AB is greater than LO.

+

Let then OR be set up from the point O at right angles to the plane of the circle LMN, [XI. 12] and let the square on OR be equal to that area by which the square on AB is greater than the square on LO; [Lemma] let RL, RM, RN be joined.

+

Then, since RO is at right angles to the plane of the circle LMN, therefore RO is also at right angles to each of the straight lines LO, MO, NO.

+

And, since LO is equal to OM, while OR is common and at right angles, therefore the base RL is equal to the base RM. [I. 4]

+

For the same reason RN is also equal to each of the straight lines RL, RM; therefore the three straight lines RL, RM, RN are equal to one another.

+

Next, since by hypothesis the square on OR is equal to that area by which the square on AB is greater than the square on LO, therefore the square on AB is equal to the squares on LO, OR.

+

But the square on LR is equal to the squares on LO, OR, for the angle LOR is right; [I. 47] therefore the square on AB is equal to the square on RL; therefore AB is equal to RL.

+

But each of the straight lines BC, DE, EF, GH, HK is equal to AB, while each of the straight lines RM, RN is equal to RL; therefore each of the straight lines AB, BC, DE, EF, GH, HK is equal to each of the straight lines RL, RM, RN.

+

And, since the two sides LR, RM are equal to the two sides AB, BC, and the base LM is by hypothesis equal to the base AC, therefore the angle LRM is equal to the angle ABC. [I. 8]

+

For the same reason the angle MRN is also equal to the angle DEF, and the angle LRN to the angle GHK.

+

Therefore, out of the three plane angles LRM, MRN, LRN, which are equal to the three given angles ABC, DEF, GHK, the solid angle at R has been constructed, which is contained by the angles LRM, MRN, LRN. Q. E. F.

+ +

Lemma. But how it is possible to take the square on OR equal to that area by which the square on AB is greater than the square on LO, we can show as follows.

+

+

Let the straight lines AB, LO be set out, and let AB be the greater; let the semicircle ABC be described on AB, and into the semicircle ABC let AC be fitted equal to the straight line LO, not being greater than the diameter AB; [IV. 1] let CB be joined

+

Since then the angle ACB is an angle in the semicircle ACB, therefore the angle ACB is right. [III. 31]

+

Therefore the square on AB is equal to the squares on AC, CB. [I. 47]

+

Hence the square on AB is greater than the square on AC by the square on CB.

+

But AC is equal to LO.

+

Therefore the square on AB is greater than the square on LO by the square on CB.

+

If then we cut off OR equal to BC, the square on AB will be greater than the square on LO by the square on OR. Q. E. F.

+ +
+
+ PROPOSITION 24. +

If a solid be contained by parallel planes, the opposite planes in it are equal and parallelogrammic. +

+

For let the solid CDHG be contained by the parallel planes AC, GF, AH, DF, BF, AE; I say that the opposite planes in it are equal and parallelogrammic.

+

+

For, since the two parallel planes BG, CE are cut by the plane AC, their common sections are parallel. [XI. 16]

+

Therefore AB is parallel to DC.

+

Again, since the two parallel planes BF, AE are cut by the plane AC, their common sections are parallel. [XI. 16]

+

Therefore BC is parallel to AD.

+

But AB was also proved parallel to DC; therefore AC is a parallelogram.

+

Similarly we can prove that each of the planes DF, FG, GB, BF, AE is a parallelogram.

+

Let AH, DF be joined.

+

Then, since AB is parallel to DC, and BH to CF, the two straight lines AB, BH which meet one another are parallel to the two straight lines DC, CF which meet one another, not in the same plane; therefore they will contain equal angles; [XI. 10] therefore the angle ABH is equal to the angle DCF.

+

And, since the two sides AB, BH are equal to the two sides DC, CF, [I. 34] and the angle ABH is equal to the angle DCF, therefore the base AH is equal to the base DF, and the triangle ABH is equal to the triangle DCF. [I. 4]

+

And the parallelogram BG is double of the triangle ABH, and the parallelogram CE double of the triangle DCF; [I. 34] therefore the parallelogram BG is equal to the parallelogram CE.

+

Similarly we can prove that AC is also equal to GF, and AE to BF.

+

Therefore etc. Q. E. D.

+
+ +
+ PROPOSITION 25. +

If a parallelepipedal solid be cut by a plane which is parallel to the opposite planes, then, as the base is to the base, so will the solid be to the solid. +

+

For let the parallelepipedal solid ABCD be cut by the plane FG which is parallel to the opposite planes RA, DH; I say that, as the base AEFV is to the base EHCF, so is the solid ABFU to the solid EGCD.

+

+

For let AH be produced in each direction, let any number of straight lines whatever, AK, KL, be made equal to AE, and any number whatever, HM, MN, equal to EH; and let the parallelograms LP, KV, HW, MS and the solids LQ, KR, DM, MT be completed.

+

Then, since the straight lines LK, KA, AE are equal to one another, the parallelograms LP, KV, AF are also equal to one another, KO, KB, AG are equal to one another, and further LX, KQ, AR are equal to one another, for they are opposite. [XI. 24]

+

For the same reason the parallelograms EC, HW, MS are also equal to one another, HG, HI, IN are equal to one another, and further DH, MY, NT are equal to one another.

+

Therefore in the solids LQ, KR, AU three planes are equal to three planes.

+

But the three planes are equal to the three opposite; therefore the three solids LQ, KR, AU are equal to one another.

+

For the same reason the three solids ED, DM, MT are also equal to one another.

+

Therefore, whatever multiple the base LF is of the base AF, the same multiple also is the solid LU of the solid AU.

+

For the same reason, whatever multiple the base NF is of the base FH, the same multiple also is the solid NU of the solid HU.

+

And, if the base LF is equal to the base NF, the solid LU is also equal to the solid NU; if the base LF exceeds the base NF, the solid LU also exceeds the solid NU; and, if one falls short, the other falls short.

+

Therefore, there being four magnitudes, the two bases AF, FH, and the two solids AU, UH, equimultiples have been taken of the base AF and the solid AU, namely the base LF and the solid LU, and equimultiples of the base HF and the solid HU, namely the base NF and the solid NU, and it has been proved that, if the base LF exceeds the base FN, the solid LU also exceeds the solid NU, if the bases are equal, the solids are equal, and if the base falls short, the solid falls short.

+

Therefore, as the base AF is to the base FH, so is the solid AU to the solid UH. [V. Def. 5] Q. E. D.

+
+
+ PROPOSITION 26. +

On a given straight line, and at a given point on it, to construct a solid angle equal to a given solid angle. +

+

Let AB be the given straight line, A the given point on it, and the angle at D, contained by the angles EDC, EDF, FDC, the given solid angle; thus it is required to construct on the straight line AB, and at the point A on it, a solid angle equal to the solid angle at D.

+

+

For let a point F be taken at random on DF, let FG be drawn from F perpendicular to the plane through ED, DC, and let it meet the plane at G, [XI. 11] let DG be joined, let there be constructed on the straight line AB and at the point A on it the angle BAL equal to the angle EDC, and the angle BAK equal to the angle EDG, [I. 23] let AK be made equal to DG, let KH be set up from the point K at right angles to the plane through BA, AL, [XI. 12] let KH be made equal to GF, and let HA be joined; I say that the solid angle at A, contained by the angles BAL, BAH, HAL is equal to the solid angle at D contained by the angles EDC, EDF, FDC.

+

For let AB, DE be cut off equal to one another, and let HB, KB, FE, GE be joined.

+

Then, since FG is at right angles to the plane of reference, it will also make right angles with all the straight lines which meet it and are in the plane of reference; [XI. Def. 3] therefore each of the angles FGD, FGE is right.

+

For the same reason each of the angles HKA, HKB is also right.

+

And, since the two sides KA, AB are equal to the two sides GD, DE respectively, and they contain equal angles, therefore the base KB is equal to the base GE. [I. 4]

+

But KH is also equal to GF, and they contain right angles; therefore HB is also equal to FE. [I. 4]

+

Again, since the two sides AK, KH are equal to the two sides DG, GF, and they contain right angles, therefore the base AH is equal to the base FD. [I. 4]

+

But AB is also equal to DE; therefore the two sides HA, AB are equal to the two sides DF, DE.

+

And the base HB is equal to the base FE; therefore the angle BAH is equal to the angle EDF. [I. 8]

+

For the same reason the angle HAL is also equal to the angle FDC.

+

And the angle BAL is also equal to the angle EDC.

+

Therefore on the straight line AB, and at the point A on it, a solid angle has been constructed equal to the given solid angle at D. Q. E. F.

+
+
+ PROPOSITION 27. +

+ On a given straight line to describe a parallelepipedal solid similar and similarly situated to a given parallelepipedal solid. +

+

Let AB be the given straight line and CD the given parallelepipedal solid; thus it is required to describe on the given straight line AB a parallelepipedal solid similar and similarly situated to the given parallelepipedal solid CD.

+

+

For on the straight line AB and at the point A on it let the solid angle, contained by the angles BAH, HAK, KAB, be constructed equal to the solid angle at C, so that the angle BAH is equal to the angle ECF, the angle BAK equal to the angle ECG, and the angle KAH to the angle GCF; and let it be contrived that, as EC is to CG, so is BA to AK, and, as GC is to CF, so is KA to AH. [VI. 12]

+

Therefore also, ex aequali, as EC is to CF, so is BA to AH. [V. 22]

+

Let the parallelogram HB and the solid AL be completed.

+

Now since, as EC is to CG, so is BA to AK, and the sides about the equal angles ECG, BAK are thus proportional, therefore the parallelogram GE is similar to the parallelogram KB.

+

For the same reason the parallelogram KH is also similar to the parallelogram GF, and further FE to HB; therefore three parallelograms of the solid CD are similar to three parallelograms of the solid AL.

+

But the former three are both equal and similar to the three opposite parallelograms, and the latter three are both equal and similar to the three opposite parallelograms; therefore the whole solid CD is similar to the whole solid AL. [XI. Def. 9]

+

Therefore on the given straight line AB there has been described AL similar and similarly situated to the given parallelepipedal solid CD. Q. E. F.

+
+
+ PROPOSITION 28. +

If a parallelepipedal solid be cut by a plane through the diagonals of the opposite planes, the solid will be bisected by the plane. +

+

For let the parallelepipedal solid AB be cut by the plane CDEF through the diagonals CF, DE of opposite planes; I say that the solid AB will be bisected by the plane CDEF.

+

+

For, since the triangle CGF is equal to the triangle CFB, [I. 34] and ADE to DEH, while the parallelogram CA is also equal to the parallelogram EB, for they are opposite, and GE to CH, therefore the prism contained by the two triangles CGF, ADE and the three parallelograms GE, AC, CE is also equal to the prism contained by the two triangles CFB, DEH and the three parallelograms CH, BE, CE; for they are contained by planes equal both in multitude and in magnitude. [XI. Def. 10]

+

Hence the whole solid AB is bisected by the plane CDEF. Q. E. D.

+
+
+ PROPOSITION 29. +

Parallelepipedal solids which are on the same base and of the same height, and in which the extremities of the sides which stand up are on the same straight lines, are equal to one another. +

+

Let CM, CN be parallelepipedal solids on the same base AB and of the same height, and let the extremities of their sides which stand up, namely AG, AF, LM, LN, CD, CE, BH, BK, be on the same straight lines FN, DK; I say that the solid CM is equal to the solid CN.

+

+

For, since each of the figures CH, CK is a parallelogram, CB is equal to each of the straight lines DH, EK, [I. 34] hence DH is also equal to EK.

+

Let EH be subtracted from each; therefore the remainder DE is equal to the remainder HK.

+

Hence the triangle DCE is also equal to the triangle HBK, [I. 8, 4] and the parallelogram DG to the parallelogram HN. [I. 36]

+

For the same reason the triangle AFG is also equal to the triangle MLN.

+

But the parallelogram CF is equal to the parallelogram BM, and CG to BN, for they are opposite; therefore the prism contained by the two triangles AFG, DCE and the three parallelograms AD, DG, CG is equal to the prism contained by the two triangles MLN, HBK and the three parallelograms BM, HN, BN.

+

Let there be added to each the solid of which the parallelogram AB is the base and GEHM its opposite; therefore the whole parallelepipedal solid CM is equal to the whole parallelepipedal solid CN.

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 30. +

Parallelepipedal solids which are on the same base and of the same height, and in which the extremities of the sides which stand up are not on the same straight lines, are equal to one another. +

+

Let CM, CN be parallelepipedal solids on the same base AB and of the same height, and let the extremities of their sides which stand up, namely AF, AG, LM, LN, CD, CE, BH, BK, not be on the same straight lines; I say that the solid CM is equal to the solid CN.

+

+

For let NK, DH be produced and meet one another at R, and further let FM, GE be produced to P, Q; let AO, LP, CQ, BR be joined.

+

Then the solid CM, of which the parallelogram ACBL is the base, and FDHM its opposite, is equal to the solid CP, of which the parallelogram ACBL is the base, and OQRP its opposite; for they are on the same base ACBL and of the same height, and the extremities of their sides which stand up, namely AF, AO, LM, LP, CD, CQ, BH, BR, are on the same straight lines FP, DR. [XI. 29]

+

But the solid CP, of which the parallelogram ACBL is the base, and OQRP its opposite, is equal to the solid CN, of which the parallelogram ACBL is the base and GEKN its opposite; for they are again on the same base ACBL and of the same height, and the extremities of their sides which stand up, namely AG, AO, CE, CQ, LN, LP, BK, BR, are on the same straight lines GQ, NR.

+

Hence the solid CM is also equal to the solid CN.

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 31. +

+ Parallelepipedal solids which are on equal bases and of the same height are equal to one another. +

+

Let the parallelepipedal solids AE, CF, of the same height, be on equal bases AB, CD.

+

I say that the solid AE is equal to the solid CF.

+

+

First, let the sides which stand up, HK, BE, AG, LM, PQ, DF, CO, RS, be at right angles to the bases AB, CD; let the straight line RT be produced in a straight line with CR; on the straight line RT, and at the point R on it, let the angle TRU be constructed equal to the angle ALB, [I. 23] let RT be made equal to AL, and RU equal to LB, and let the base RW and the solid XU be completed.

+

Now, since the two sides TR, RU are equal to the two sides AL, LB, and they contain equal angles, therefore the parallelogram RW is equal and similar to the parallelogram HL.

+

Since again AL is equal to RT, and LM to RS, and they contain right angles, therefore the parallelogram RX is equal and similar to the parallelogram AM.

+

For the same reason LE is also equal and similar to SU; therefore three parallelograms of the solid AE are equal and similar to three parallelograms of the solid XU.

+

But the former three are equal and similar to the three opposite, and the latter three to the three opposite; [XI. 24] therefore the whole parallelepipedal solid AE is equal to the whole parallelepipedal solid XU. [XI. Def. 10]

+

Let DR, WU be drawn through and meet one another at Y, let aTb be drawn through T parallel to DY, let PD be produced to a, and let the solids YX, RI be completed.

+

Then the solid XY, of which the parallelogram RX is the base and Yc its opposite, is equal to the solid XU of which the parallelogram RX is the base and UV its opposite, for they are on the same base RX and of the same height, and the extremities of their sides which stand up, namely RY, RU, Tb, TW, Se, Sd, Xc, XV, are on the same straight lines YW, eV. [XI. 29]

+

But the solid XU is equal to AE: therefore the solid XY is also equal to the solid AE.

+

And, since the parallelogram RUWT is equal to the parallelogram YT for they are on the same base RT and in the same parallels RT, YW, [I. 35] while RUWT is equal to CD, since it is also equal to AB, therefore the parallelogram YT is also equal to CD.

+

But DT is another parallelogram; therefore, as the base CD is to DT, so is YT to DT. [V. 7]

+

And, since the parallelepipedal solid CI has been cut by the plane RF which is parallel to opposite planes, as the base CD is to the base DT, so is the solid CF to the solid RI. [XI. 25]

+

For the same reason, since the parallelepipedal solid YI has been cut by the plane RX which is parallel to opposite planes, as the base YT is to the base TD, so is the solid YX to the solid RI. [XI. 25]

+

But, as the base CD is to DT, so is YT to DT; therefore also, as the solid CF is to the solid RI, so is the solid YX to RI. [V. 11]

+

Therefore each of the solids CF, YX has to RI the same ratio; therefore the solid CF is equal to the solid YX. [V. 9]

+

But YX was proved equal to AE; therefore AE is also equal to CF.

+

Next, let the sides standing up, AG, HK, BE, LM, CN, PQ, DF, RS, not be at right angles to the bases AB, CD; I say again that the solid AE is equal to the solid CF.

+

+

For from the points K, E, G, M, Q, F, N, S let KO, ET, GU, MV, QW, FX, NY, SI be drawn perpendicular to the plane of reference, and let them meet the plane at the points O, T, U, V, W, X, Y, I, and let OT, OU, UV, TV, WX, WY, YI, IX be joined.

+

Then the solid KV is equal to the solid QI, for they are on the equal bases KM, QS and of the same height, and their sides which stand up are at right angles to their bases. [First part of this Prop.]

+

But the solid KV is equal to the solid AE, and QI to CF; for they are on the same base and of the same height, while the extremities of their sides which stand up are not on the same straight lines. [XI. 30]

+

Therefore the solid AE is also equal to the solid CF.

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 32. +

+ Parallelepipedal solids which are of the same height are to one another as their bases. +

+

Let AB, CD be parallelepipedal solids of the same height; I say that the parallelepipedal solids AB, CD are to one another as their bases, that is, that, as the base AE is to the base CF, so is the solid AB to the solid CD.

+

+

For let FH equal to AE be applied to FG, [I. 45] and on FH as base, and with the same height as that of CD, let the parallelepipedal solid GK be completed.

+

Then the solid AB is equal to the solid GK; for they are on equal bases AE, FH and of the same height. [XI. 31]

+

And, since the parallelepipedal solid CK is cut by the plane DG which is parallel to opposite planes, therefore, as the base CF is to the base FH, so is the solid CD to the solid DH. [XI. 25]

+

But the base FH is equal to the base AE, and the solid GK to the solid AB; therefore also, as the base AE is to the base CF, so is the solid AB to the solid CD.

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 33. +

+ Similar parallelepipedal solids are to one another in the triplicate ratio of their corresponding sides. +

+

Let AB, CD be similar parallelepipedal solids, and let AE be the side corresponding to CF; I say that the solid AB has to the solid CD the ratio triplicate of that which AE has to CF.

+

+

For let EK, EL, EM be produced in a straight line with AE, GE, HE, let EK be made equal to CF, EL equal to FN, and further EM equal to FR, and let the parallelogram KL and the solid KP be completed.

+

Now, since the two sides KE, EL are equal to the two sides CF, FN, while the angle KEL is also equal to the angle CFN, inasmuch as the angle AEG is also equal to the angle CFN because of the similarity of the solids AB, CD, therefore the parallelogram KL is equal ltand similargt to the parallelogram CN.

+

For the same reason the parallelogram KM is also equal and similar to CR, and further EP to DF; therefore three parallelograms of the solid KP are equal and similar to three parallelograms of the solid CD.

+

But the former three parallelograms are equal and similar to their opposites, and the latter three to their opposites; [XI. 24] therefore the whole solid KP is equal and similar to the whole solid CD. [XI. Def. 10]

+

Let the parallelogram GK be completed, and on the parallelograms GK, KL as bases, and with the same height as that of AB, let the solids EO, LQ be completed.

+

Then since; owing to the similarity of the solids AB, CD, as AE is to CF, so is EG to FN, and EH to FR, while CF is equal to EK, FN to EL, and FR to EM, therefore, as AE is to EK, so is GE to EL, and HE to EM.

+

But, as AE is to EK, so is AG to the parallelogram GK, as GE is to EL, so is GK to KL, and, as HE is to EM, so is QE to KM; [VI. 1] therefore also, as the parallelogram AG is to GK, so is GK to KL, and QE to KM.

+

But, as AG is to GK, so is the solid AB to the solid EO, as GK is to KL, so is the solid OE to the solid QL, and, as QE is to KM, so is the solid QL to the solid KP; [XI. 32] therefore also, as the solid AB is to EO, so is EO to QL, and QL to KP.

+

But, if four magnitudes be continuously proportional, the first has to the fourth the ratio triplicate of that which it has to the second; [V. Def. 10] therefore the solid AB has to KP the ratio triplicate of that which AB has to EO.

+

But, as AB is to EO, so is the parallelogram AG to GK, and the straight line AE to EK [VI. 1]; hence the solid AB has also to KP the ratio triplicate of that which AE has to EK.

+

But the solid KP is equal to the solid CD, and the straight line EK to CF; therefore the solid AB has also to the solid CD the ratio triplicate of that which the corresponding side of it, AE, has to the corresponding side CF.

+

Therefore etc. Q. E. D.

+ +

Porism. From this it is manifest that, if four straight lines be ltcontinuouslygt proportional, as the first is to the fourth, so will a parallelepipedal solid on the first be to the similar and similarly described parallelepipedal solid on the second, inasmuch as the first has to the fourth the ratio triplicate of that which it has to the second.

+ +
+
+ PROPOSITION 34. +

+ In equal parallelepipedal solids the bases are reciprocally proportional to the heights; and those parallelepipedal solids in which the bases are reciprocally proportional to the heights are equal. +

+

Let AB, CD be equal parallelepipedal solids; I say that in the parallelepipedal solids AB, CD the bases are reciprocally proportional to the heights, that is, as the base EH is to the base NQ, so is the height of the solid CD to the height of the solid AB.

+

First, let the sides which stand up, namely AG, EF, LB, HK, CM, NO, PD, QR, be at right angles to their bases; I say that, as the base EH is to the base NQ, so is CM to AG.

+

If now the base EH is equal to the base NQ, while the solid AB is also equal to the solid CD, CM will also be equal to AG.

+

For parallelepipedal solids of the same height are to one another as the bases; [XI. 32] and, as the base EH is to NQ, so will CM be to AG, and it is manifest that in the parallelepipedal solids AB, CD the bases are reciprocally proportional to the heights.

+

Next, let the base EH not be equal to the base NQ, but let EH be greater.

+

+

Now the solid AB is equal to the solid CD; therefore CM is also greater than AG.

+

Let then CT be made equal to AG, and let the parallelepipedal solid VC be completed on NQ as base and with CT as height.

+

Now, since the solid AB is equal to the solid CD, and CV is outside them, while equals have to the same the same ratio, [V. 7] therefore, as the solid AB is to the solid CV, so is the solid CD to the solid CV.

+

But, as the solid AB is to the solid CV, so is the base EH to the base NQ, for the solids AB, CV are of equal height; [XI. 32] and, as the solid CD is to the solid CV, so is the base MQ to the base TQ [XI. 25] and CM to CT [VI. 1]; therefore also, as the base EH is to the base NQ, so is MC to CT.

+

But CT is equal to AG; therefore also, as the base EH is to the base NQ, so is MC to AG.

+

Therefore in the parallelepipedal solids AB, CD the bases are reciprocally proportional to the heights.

+

Again, in the parallelepipedal solids AB, CD let the bases be reciprocally proportional to the heights, that is, as the base EH is to the base NQ, so let the height of the solid CD be to the height of the solid AB; I say that the solid AB is equal to the solid CD.

+

Let the sides which stand up be again at right angles to the bases.

+

Now, if the base EH is equal to the base NQ, and, as the base EH is to the base NQ, so is the height of the solid CD to the height of the solid AB, therefore the height of the solid CD is also equal to the height of the solid AB.

+

But parallelepipedal solids on equal bases and of the same height are equal to one another; [XI. 31] therefore the solid AB is equal to the solid CD.

+

Next, let the base EH not be equal to the base NQ, but let EH be greater; therefore the height of the solid CD is also greater than the height of the solid AB, that is, CM is greater than AG.

+

Let CT be again made equal to AG, and let the solid CV be similarly completed.

+

Since, as the base EH is to the base NQ, so is MC to AG, while AG is equal to CT, therefore, as the base EH is to the base NQ, so is CM to CT.

+

But, as the base EH is to the base NQ, so is the solid AB to the solid CV, for the solids AB, CV are of equal height; [XI. 32] and, as CM is to CT, so is the base MQ to the base QT [VI. 1] and the solid CD to the solid CV. [XI. 25]

+

Therefore also, as the solid AB is to the solid CV, so is the solid CD to the solid CV; therefore each of the solids AB, CD has to CV the same ratio.

+

Therefore the solid AB is equal to the solid CD. [V. 9]

+

Now let the sides which stand up, FE, BL, GA, HK, ON, DP, MC, RQ, not be at right angles to their bases; let perpendiculars be drawn from the points F, G, B, K, O, M, D, R to the planes through EH, NQ, and let them meet the planes at S, T, U, V, W, X, Y, a, and let the solids FV, Oa be completed; I say that, in this case too, if the solids AB, CD are equal, the bases are reciprocally proportional to the heights, that is, as the base EH is to the base NQ, so is the height of the solid CD to the height of the solid AB.

+

+

Since the solid AB is equal to the solid CD, while AB is equal to BT, for they are on the same base FK and of the same height; [XI. 29, 30] and the solid CD is equal to DX, for they are again on the same base RO and of the same height; [id.] therefore the solid BT is also equal to the solid DX.

+

Therefore, as the base FK is to the base OR, so is the height of the solid DX to the height of the solid BT. [Part 1.]

+

But the base FK is equal to the base EH, and the base OR to the base NQ; therefore, as the base EH is to the base NQ, so is the height of the solid DX to the height of the solid BT.

+

But the solids DX, BT and the solids DC, BA have the same heights respectively; therefore, as the base EH is to the base NQ, so is the height of the solid DC to the height of the solid AB.

+

Therefore in the parallelepipedal solids AB, CD the bases are reciprocally proportional to the heights.

+

Again, in the parallelepipedal solids AB, CD let the bases be reciprocally proportional to the heights, that is, as the base EH is to the base NQ, so let the height of the solid CD be to the height of the solid AB; I say that the solid AB is equal to the solid CD.

+

For, with the same construction, since, as the base EH is to the base NQ, so is the height of the solid CD to the height of the solid AB, while the base EH is equal to the base FK, and NQ to OR, therefore, as the base FK is to the base OR, so is the height of the solid CD to the height of the solid AB.

+

But the solids AB, CD and BT, DX have the same heights respectively; therefore, as the base FK is to the base OR, so is the height of the solid DX to the height of the solid BT.

+

Therefore in the parallelepipedal solids BT, DX the bases are reciprocally proportional to the heights; therefore the solid BT is equal to the solid DX. [Part 1.]

+

But BT is equal to BA, for they are on the same base FK and of the same height; [XI. 29, 30] and the solid DX is equal to the solid DC. [id.]

+

Therefore the solid AB is also equal to the solid CD. Q. E. D.

+
+ +
+ PROPOSITION 35. +

If there be two equal plane angles, and on their vertices there be set up elevated straight lines containing equal angles with the original straight lines respectively, if on the elevated straight lines points be taken at random and perpendiculars be drawn from them to the planes in which the original angles are, and if from the points so arising in the planes straight lines be joined to the vertices of the original angles, they will contain, with the elevated straight lines, equal angles.

+

Let the angles BAC, EDF be two equal rectilineal angles, and from the points A, D let the elevated straight lines AG, DM be set up containing, with the original straight lines, equal angles respectively, namely, the angle MDE to the angle GAB and the angle MDF to the angle GAC, let points G, M be taken at random on AG, DM, let GL, MN be drawn from the points G, M perpendicular to the planes through BA, AC and ED, DF, and let them meet the planes at L, N, and let LA, ND be joined; I say that the angle GAL is equal to the angle MDN.

+

+

Let AH be made equal to DM, and let HK be drawn through the point H parallel to GL.

+

But GL is perpendicular to the plane through BA, AC; therefore HK is also perpendicular to the plane through. BA, AC. [XI. 8]

+

From the points K, N let KC, NF, KB, NE be drawn perpendicular to the straight lines AC, DF, AB, DE, and let HC, CB, MF, FE be joined.

+

Since the square on HA is equal to the squares on HK, KA, and the squares on KC, CA are equal to the square on KA, [I. 47] therefore the square on HA is also equal to the squares on HK, KC, CA.

+

But the square on HC is equal to the squares on HK, KC; [I. 47] therefore the square on HA is equal to the squares on HC, CA.

+

Therefore the angle HCA is right. [I. 48]

+

For the same reason the angle DFM is also right.

+

Therefore the angle ACH is equal to the angle DFM.

+

But the angle HAC is also equal to the angle MDF.

+

Therefore MDF, HAC are two triangles which have two angles equal to two angles respectively, and one side equal to one side, namely, that subtending one of the equal angles, that is, HA equal to MD; therefore they will also have the remaining sides equal to the remaining sides respectively. [I. 26]

+

Therefore AC is equal to DF.

+

Similarly we can prove that AB is also equal to DE.

+

Since then AC is equal to DF, and AB to DE, the two sides CA, AB are equal to the two sides FD, DE.

+

But the angle CAB is also equal to the angle FDE; therefore the base BC is equal to the base EF, the triangle to the triangle, and the remaining angles to the remaining angles; [I. 4] therefore the angle ACB is equal to the angle DFE.

+

But the right angle ACK is also equal to the right angle DFN; therefore the remaining angle BCK is also equal to the remaining angle EFN.

+

For the same reason the angle CBK is also equal to the angle FEN.

+

Therefore BCK, EFN are two triangles which have two angles equal to two angles respectively, and one side equal to one side, namely, that adjacent to the equal angles, that is, BC equal to EF; therefore they will also have the remaining sides equal to the remaining sides. [I. 26]

+

Therefore CK is equal to FN.

+

But AC is also equal to DF; therefore the two sides AC, CK are equal to the two sides DF, FN; and they contain right angles.

+

Therefore the base AK is equal to the base DN. [I. 4]

+

And, since AH is equal to DM, the square on AH is also equal to the square on DM.

+

But the squares on AK, KH are equal to the square on AH, for the angle AKH is right; [I. 47] and the squares on DN, NM are equal to the square on DM, for the angle DNM is right; [I. 47] therefore the squares on AK, KH are equal to the squares on DN, NM; and of these the square on AK is equal to the square on DN; therefore the remaining square on KH is equal to the square on NM; therefore HK is equal to MN.

+

And, since the two sides HA, AK are equal to the two sides MD, DN respectively, and the base HK was proved equal to the base MN, therefore the angle HAK is equal to the angle MDN. [I. 8]

+

Therefore etc.

+ +

Porism. From this it is manifest that, if there be two equal plane angles, and if there be set up on them elevated straight lines which are equal and contain equal angles with the original straight lines respectively, the perpendiculars drawn from their extremities to the planes in which are the original angles are equal to one another. Q. E. D.

+ + +
+
+ PROPOSITION 36. +

If three straight lines be proportional, the parallelepipedal solid formed out of the three is equal to the parallelepipedal solid on the mean which is equilateral, but equiangular with the aforesaid solid. +

+

Let A, B, C be three straight lines in proportion, so that, as A is to B, so is B to C; I say that the solid formed out of A, B, C is equal to the solid on B which is equilateral, but equiangular with the aforesaid solid.

+

Let there be set out the solid angle at E contained by the angles DEG, GEF, FED, let each of the straight lines DE, GE, EF be made equal to B, and let the parallelepipedal solid EK be completed, let LM be made equal to A, and on the straight line LM, and at the point L on it, let there be constructed a solid angle equal to the solid angle at E, namely that contained by NLO, OLM, MLN; let LO be made equal to B, and LN equal to C.

+

+

Now, since, as A is to B, so is B to C, while A is equal to LM, B to each of the straight lines LO, ED, and C to LN, therefore, as LM is to EF, so is DE to LN.

+

Thus the sides about the equal angles NLM, DEF are reciprocally proportional; therefore the parallelogram MN is equal to the parallelogram DF. [VI. 14]

+

And, since the angles DEF, NLM are two plane rectilineal angles, and on them the elevated straight lines LO, EG are set up which are equal to one another and contain equal angles with the original straight lines respectively, therefore the perpendiculars drawn from the points G, O to the planes through NL, LM and DE, EF are equal to one another; [XI. 35, Por.] hence the solids LH, EK are of the same height.

+

But parallelepipedal solids on equal bases and of the same height are equal to one another; [XI. 31] therefore the solid HL is equal to the solid EK.

+

And LH is the solid formed out of A, B, C, and EK the solid on B; therefore the parallelepipedal solid formed out of A, B, C is equal to the solid on B which is equilateral, but equiangular with the aforesaid solid. Q. E. D.

+
+
+ PROPOSITION 37. +

If four straight lines be proportional, the parallelepipedal solids on them which are similar and similarly described will also be proportional; and, if the parallelepipedal solids on them which are similar and similarly described be proportional, the straight lines will themselves also be proportional. +

+

Let AB, CD, EF, GH be four straight lines in proportion, so that, as AB is to CD, so is EF to GH; and let there be described on AB, CD, EF, GH the similar and similarly situated parallelepipedal solids KA, LC, ME, NG; I say that, as KA is to LC, so is ME to NG.

+

+

For, since the parallelepipedal solid KA is similar to LC, therefore KA has to LC the ratio triplicate of that which AB has to CD. [XI. 33]

+

For the same reason ME also has to NG the ratio triplicate of that which EF has to GH. [id.]

+

And, as AB is to CD, so is EF to GH.

+

Therefore also, as AK is to LC, so is ME to NG.

+

Next, as the solid AK is to the solid LC, so let the solid ME be to the solid NG; I say that, as the straight line AB is to CD, so is EF to GH.

+

For since, again, KA has to LC the ratio triplicate of that which AB has to CD, [XI. 33] and ME also has to NG the ratio triplicate of that which EF has to GH, [id.] and, as KA is to LC, so is ME to NG, therefore also, as AB is to CD, so is EF to GH.

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 38. +

If the sides of the opposite planes of a cube be bisected, and planes be carried through the points of section, the common section of the planes and the diameter of the cube bisect one another. +

+

For let the sides of the opposite planes CF, AH of the cube AF be bisected at the points K, L, M, N, O, Q, P, R, and through the points of section let the planes KN, OR be carried; let US be the common section of the planes, and DG the diameter of the cube AF.

+

I say that UT is equal to TS, and DT to TG.

+

For let DU, UE, BS, SG be joined.

+

Then, since DO is parallel to PE, the alternate angles DOU, UPE are equal to one another. [I. 29]

+

And, since DO is equal to PE, and OU to UP, and they contain equal angles, therefore the base DU is equal to the base UE, the triangle DOU is equal to the triangle PUE, and the remaining angles are equal to the remaining angles; [I. 4] therefore the angle OUD is equal to the angle PUE.

+

+

For this reason DUE is a straight line. [I. 14]

+

For the same reason, BSG is also a straight line, and BS is equal to SG.

+

Now, since CA is equal and parallel to DB, while CA is also equal and parallel to EG, therefore DB is also equal and parallel to EG. [XI. 9]

+

And the straight lines DE, BG join their extremities; therefore DE is parallel to BG. [I. 33]

+

Therefore the angle EDT is equal to the angle BGT, for they are alternate; [I. 29] and the angle DTU is equal to the angle GTS. [I. 15]

+

Therefore DTU, GTS are two triangles which have two angles equal to two angles, and one side equal to one side, namely that subtending one of the equal angles, that is, DU equal to GS, for they are the halves of DE, BG; therefore they will also have the remaining sides equal to the remaining sides. [I. 26]

+

Therefore DT is equal to TG, and UT to TS.

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 39. +

If there be two prisms of equal height, and one have a parallelogram as base and the other a triangle, and if the parallelogram be double of the triangle, the prisms will be equal. +

+

Let ABCDEF, GHKLMN be two prisms of equal height, let one have the parallelogram AF as base, and the other the triangle GHK, and let the parallelogram AF be double of the triangle GHK; I say that the prism ABCDEF is equal to the prism GHKLMN.

+

+

For let the solids AO, GP be completed.

+

Since the parallelogram AF is double of the triangle GHK, while the parallelogram HK is also double of the triangle GHK, [I. 34] therefore the parallelogram AF is equal to the parallelogram HK.

+

But parallelepipedal solids which are on equal bases and of the same height are equal to one another; [XI. 31] therefore the solid AO is equal to the solid GP.

+

And the prism ABCDEF is half of the solid AO, and the prism GHKLMN is half of the solid GP; [XI. 28] therefore the prism ABCDEF is equal to the prism GHKLMN.

+

Therefore etc. Q. E. D.

+
+
+ +
+ + BOOK XII. + HISTORICAL NOTE. +

The predominant feature of Book XII. is the use of the method of exhaustion, which is applied in Propositions 2, 3-5, 10, 11, 12, and (in a slightly different form) in Propositions 16-18. We conclude therefore that for the content of this Book Euclid was greatly indebted to Eudoxus, to whom the discovery of the method of exhaustion is attributed. The evidence for this attribution comes mainly from Archimedes. (1) In the preface to On the Sphere and Cylinder I., after stating the main results obtained by himself regarding the surface of a sphere or a segment thereof, and the volume and surface of a right cylinder with height equal to its diameter as compared with those of a sphere with the same diameter, Archimedes adds: Having now discovered that the properties mentioned are true of these figures, I cannot feel any hesitation in setting them side by side both with my former investigations and with those of the theorems of Eudoxus on solids which are held to be most irrefragably established, namely that any pyramid is one third part of the prism which has the same base with the pyramid and equal height [i.e. Eucl. XII. 7], and that any cone is one third part of the cylinder which has the same base with the cone and equal height [i.e. Eucl. XII. 10]. For, though these properties also were naturally inherent in the figures all along, yet they were in fact unknown to all the many able geometers who lived before Eudoxus and had not been observed by any one. + (2) In the preface to the treatise known as the Quadrature of the Parabola Archimedes states the lemma + assumed by him and known as the Axiom of Archimedes + (see note on X. 1 above) and proceeds: Earlier geometers (οἱ πρότερον γεωμέτραι) have also used this lemma; for it is by the use of this same lemma that they have shown that circles are to one another in the duplicate ratio of their diameters [Eucl. XII. 2], and that spheres are to one another in the triplicate ratio of their diameters [Eucl. XII. 18], and further that every pyramid is one third part of the prism which has the same base with the pyramid and equal height [Eucl. XII. 7]; also, that every cone is one third part of the cylinder which has the same base with the cone and equal height [Eucl. XII. 10] they proved by assuming a certain lemma similar to that aforesaid. + Thus in the first passage two theorems of Eucl. XII. are definitely attributed to Eudoxus; and, when Archimedes says, in the second passage, that earlier geometers + proved these two theorems by means of the lemma known as the Axiom of Archimedes + and of a lemma similar to it respectively, we can hardly suppose him to be alluding to any other proof than that given by Eudoxus. As a matter of fact, the lemma used by Euclid to prove both propositions (XII. 3-5 and 7, and XII. 10) is the theorem of Eucl. X. 1. As regards the connexion between the two lemmas + see note on X. 1.

+

We are not, however, to suppose that none of the results obtained by the method of exhaustion had been discovered before the time of Eudoxus (fl. about 368-5 B.C.). Two at least are of earlier date, those of Eucl. XII. 2 and XII. 7.

+

(a) Simplicius (Comment. in Aristot. Phys. p. 61, ed. Diels) quotes Eudemus as saying, in his History of Geometry, that Hippocrates of Chios (fl. say 430 B.C.) first laid it down (ἔθετο) that similar segments of circles are in the ratio of the squares on their bases and that he proved this (ἐδείκνυεν) by proving (ἐκ τοῦ δεῖξαι) that the squares on the diameters have the same ratio as the (whole) circles. We know nothing of the method by which Hippocrates proved this proposition; but, having regard to the evidence from Archimedes quoted above, it is not permissible to suppose that the method was the fully developed method of exhaustion as we know it.

+

(b) As regards the two theorems about the volume of a pyramid and of a cone respectively, which Eudoxus was the first to prove, we now have authentic evidence in the short treatise by Archimedes discovered by Heiberg in a MS. at Constantinople in 1906 and published in Hermes the following year (see now Archimedis opera omnia, ed. Heiberg, 2. ed., Vol. II., 1913, pp. 425-507; T. L. Heath, The Method of Archimedes, Cambridge, 1912). The said treatise, complete in all essentials, bears the title Ἀρχιμήδους περὶ τῶν μηχανικῶν θεωρημάτων πρὸς Ἐρατοσθένην ἔφοδος. This Method + (or Plan of attack +), addressed to Eratosthenes, is none other than the ἐφόδιον on which, according to Suidas, Theodosius wrote a commentary, and which is several times cited by Heron in his Metrica; its discovery adds a new and important chapter to the history of the integral calculus. In the preface to this work Archimedes alludes to the theorems which he first discovered by means of mechanical considerations, but proved afterwards by geometry, because the investigation by means of mechanics did not constitute a rigid proof; he observes, however, that the mechanical method is of great use for the discovery of theorems, and it is much easier to provide the rigid proof when the fact to be proved has once been discovered than it would be if nothing were known to begin with. He goes on: Hence too, in the case of those theorems the proof of which was first discovered by Eudoxus, namely those relating to the cone and the pyramid, that the cone is one third part of the cylinder, and the pyramid one third part of the prism, having the same base and equal height, no small part of the credit will naturally be assigned to Democritus, who was the first to make the statement (of the fact) regarding the said figure [i.e. property], though without proving it. + Hence the discovery of the two theorems must now be attributed to Democritus (fl. towards the end of 5th cent. B.C.). The words without proving it + (χωρὶς ἀποδείξεως) do not mean that Democritus gave no sort of proof, but only that he did not give a proof on the rigorous lines required later; for the same words are used by Archimedes of his own investigations by means of mechanics, which, however, do constitute a reasoned argument. The character of Archimedes' mechanical arguments combined with a passage of Plutarch about a particular question in infinitesimals said to have been raised by Democritus may perhaps give a clue to the line of Democritus' argument as regards the pyramid. The essential feature of Archimedes' mechanical arguments in this tract is that he regards an area as the sum of an infinite number of siraight lines parallel to one another and terminated by the boundary or boundaries of the closed figure the area of which is to be found, and a volume as the sum of an infinite number of plane sections parallel to one another: which is of course the same thing as taking (as we do in the integral calculus) the sum of an infinite number of strips of breadth dx (say), when dx becomes indefinitely small, or the sum of an infinite number of parallel laminae of depth dz (say), when dz becomes indefinitely small. To give only one instance, we may take the case of the area of a segment of a parabola cut off by a chord.

+

Let CBA be the parabolic segment, CE the tangent at C meeting the

diameter EBD through the middle point of the chord CA in E, so that . +

+

Draw AF parallel to ED meeting CE produced in F. Produce CB to H so that , where K is the point in which CH meets AF; and suppose CH to be a lever.

+

Let any diameter MNPO be drawn meeting the curve in P and CF, CK, CA in M, N, O respectively.

+

Archimedes then observes that (for this is proved in a lemma +), whence , so that, if a straight line TG equal to PO be placed with its middle point at H, the straight line MO with centre of gravity at N, and the straight line TG with centre of gravity at H, will balance about K.

+

Taking all other parts of diameters like PO intercepted between the curve and CA, and placing equal straight lines with their centres of gravity at H, these straight lines collected at H will balance (about K) all the lines like MO parallel to FA intercepted within the triangle CFA in the positions in which they severally lie in the figure.

+

Hence Archimedes infers that an area equal to that of the parabolic segment hung at H will balance (about K) the triangle CFA hung at its centre of gravity, the point X (a point on CK such that ), and therefore that , + from which it follows that . +

+

The same sort of argument is used for solids, plane sections taking the place of straight lines.

+

Archimedes is careful to state once more that this method of argument does not constitute a proof. Thus, at the end of the above proposition about the parabolic segment, he adds: This property is of course not proved by what has just been said; but it has furnished a sort of indication (ἔμφασίν τινα) that the conclusion is true. +

+

Let us now turn to the passage of Plutarch (De Comm. Not. adv. Stoicos XXXIX 3) about Democritus above referred to. Plutarch speaks of Democritus as having raised the question in natural philosophy (φυσικῶς): if a cone were cut by a plane parallel to the base [by which is clearly meant a plane indefinitely near to the base], what must we think of the surfaces of the sections, that they are equal or unequal? For, if they are unequal, they will make the cone irregular, as having many indentations, like steps, and unevennesses; but, if they are equal, the sections will be equal, and the cone will appear to have the property of the cylinder and to be made up of equal, not unequal circles, which is very absurd. + The phrase made up of equal...circles + (ἐξ ἴσων συγκείμενος...κύκλων) shows that Democritus already had the idea of a solid being the sum of an infinite number of parallel planes, or indefinitely thin laminae, indefinitely near together: a most important anticipation of the same thought which led to such fruitful results in Archimedes. If then one may hazard a conjecture as to Democritus' argument with regard to a pyramid, it seems probable that he would notice that, if two pyramids of the same height and equal triangular bases are respectively cut by planes parallel to the base and dividing the heights in the same ratio, the corresponding sections of the two pyramids are equal, whence he would infer that the pyramids are equal as being the sum of the same infinite number of equal plane sections or indefinitely thin laminae. (This would be a particular anticipation of Cavalieri's proposition that the areal or solid contents of two figures are equal if two sections of them taken at the same height, whatever the height may be, always give equal straight lines or equal surfaces respectively.) And Democritus would of course see that the three pyramids into which a prism on the same base and of equal height with the original pyramid is divided (as in Eucl. XII. 7) satisfy this test of equality, so that the pyramid would be one third part of the prism. The extension to a pyramid with a polygonal base would be easy. And Democritus may have stated the proposition for the cone (of course without an absolute proof) as a natural inference from the result of increasing indefinitely the number of sides in a regular polygon forming the base of a pyramid.

+
+ BOOK XII. PROPOSITIONS. +
+ PROPOSITION 1. +

+ Similar polygons inscribed in circles are to one another as the squares on the diameters. +

+

Let ABC, FGH be circles, let ABCDE, FGHKL be similar polygons inscribed in them, and let BM, GN be diameters of the circles; I say that, as the square on BM is to the square on GN, so is the polygon ABCDE to the polygon FGHKL.

+

+

For let BE, AM, GL, FN be joined.

+

Now, since the polygon ABCDE is similar to the polygon FGHKL, the angle BAE is equal to the angle GFL, and, as BA is to AE, so is GF to FL. [VI. Def. I]

+

Thus BAE, GFL are two triangles which have one angle equal to one angle, namely the angle BAE to the angle GFL, and the sides about the equal angles proportional; therefore the triangle ABE is equiangular with the triangle FGL. [VI. 6]

+

Therefore the angle AEB is equal to the angle FLG.

+

But the angle AEB is equal to the angle AMB, for they stand on the same circumference; [III. 27] and the angle FLG to the angle FNG; therefore the angle AMB is also equal to the angle FNG.

+

But the right angle BAM is also equal to the right angle GFN; [III. 31] therefore the remaining angle is equal to the remaining angle. [I. 32]

+

Therefore the triangle ABM is equiangular with the triangle FGN.

+

Therefore, proportionally, as BM is to GN, so is BA to GF. [VI. 4]

+

But the ratio of the square on BM to the square on GN is duplicate of the ratio of BM to GN, and the ratio of the polygon ABCDE to the polygon FGHKL is duplicate of the ratio of BA to GF; [VI. 20] therefore also, as the square on BM is to the square on GN, so is the polygon ABCDE to the polygon FGHKL.

+

Therefore etc. Q. E. D.

+
+ +
+ PROPOSITION 2. +

+ Circles are to one another as the squares on the diameters. +

+

Let ABCD, EFGH be circles, and BD, FH their diameters; I say that, as the circle ABCD is to the circle EFGH, so is the square on BD to the square on FH.

+

+

For, if the square on BD is not to the square on FH as the circle ABCD is to the circle EFGH, then, as the square on BD is to the square on FH, so will the circle ABCD be either to some less area than the circle EFGH, or to a greater.

+

First, let it be in that ratio to a less area S.

+

Let the square EFGH be inscribed in the circle EFGH; then the inscribed square is greater than the half of the circle EFGH, inasmuch as, if through the points E, F, G, H we draw tangents to the circle, the square EFGH is half the square circumscribed about the circle, and the circle is less than the circumscribed square; hence the inscribed square EFGH is greater than the half of the circle EFGH.

+

Let the circumferences EF, FG, GH, HE be bisected at the points K, L, M, N, and let EK, KF, FL, LG, GM, MH, HN, NE be joined; therefore each of the triangles EKF, FLG, GMH, HNE is also greater than the half of the segment of the circle about it, inasmuch as, if through the points K, L, M, N we draw tangents to the circle and complete the parallelograms on the straight lines EF, FG, GH, HE, each of the triangles EKF, FLG, GMH, HNE will be half of the parallelogram about it, while the segment about it is less than the parallelogram; hence each of the triangles EKF, FLG, GMH, HNE is greater than the half of the segment of the circle about it.

+

Thus, by bisecting the remaining circumferences and joining straight lines, and by doing this continually, we shall leave some segments of the circle which will be less than the excess by which the circle EFGH exceeds the area S.

+

For it was proved in the first theorem of the tenth book that, if two unequal magnitudes be set out, and if from the greater there be subtracted a magnitude greater than the half, and from that which is left a greater than the half, and if this be done continually, there will be left some magnitude which will be less than the lesser magnitude set out.

+

Let segments be left such as described, and let the segments of the circle EFGH on EK, KF, FL, LG, GM, MH, HN, NE be less than the excess by which the circle EFGH exceeds the area S.

+

Therefore the remainder, the polygon EKFLGMHN, is greater than the area S.

+

Let there be inscribed, also, in the circle ABCD the polygon AOBPCQDR similar to the polygon EKFLGMHN; therefore, as the square on BD is to the square on FH, so is the polygon AOBPCQDR to the polygon EKFLGMHN. [XII. 1]

+

But, as the square on BD is to the square on FH, so also is the circle ABCD to the area S; therefore also, as the circle ABCD is to the area S, so is the polygon AOBPCQDR to the polygon EKFLGMHN; [V. 11] therefore, alternately, as the circle ABCD is to the polygon inscribed in it, so is the area S to the polygon EKFLGMHN. [V. 16]

+

But the circle ABCD is greater than the polygon inscribed in it; therefore the area S is also greater than the polygon EKFLGMHN.

+

But it is also less: which is impossible.

+

Therefore, as the square on BD is to the square on FH, so is not the circle ABCD to any area less than the circle EFGH.

+

Similarly we can prove that neither is the circle EFGH to any area less than the circle ABCD as the square on FH is to the square on BD.

+

I say next that neither is the circle ABCD to any area greater than the circle EFGH as the square on BD is to the square on FH.

+

For, if possible, let it be in that ratio to a greater area S.

+

Therefore, inversely, as the square on FH is to the square on DB, so is the area S to the circle ABCD.

+

But, as the area S is to the circle ABCD, so is the circle EFGH to some area less than the circle ABCD; therefore also, as the square on FH is to the square on BD, so is the circle EFGH to some area less than the circle ABCD: [V. 11] which was proved impossible.

+

Therefore, as the square on BD is to the square on FH, so is not the circle ABCD to any area greater than the circle EFGH.

+

And it was proved that neither is it in that ratio to any area less than the circle EFGH; therefore, as the square on BD is to the square on FH, so is the circle ABCD to the circle EFGH.

+

Therefore etc. Q. E. D.

+ +

Lemma. I say that, the area S being greater than the circle EFGH, as the area S is to the circle ABCD, so is the circle EFGH to some area less than the circle ABCD.

+

For let it be contrived that, as the area S is to the circle ABCD, so is the circle EFGH to the area T.

+

I say that the area T is less than the circle ABCD.

+

For since, as the area S is to the circle ABCD, so is the circle EFGH to the area T, therefore, alternately, as the area S is to the circle EFGH, so is the circle ABCD to the area T. [V. 16]

+

But the area S is greater than the circle EFGH; therefore the circle ABCD is also greater than the area T.

+

Hence, as the area S is to the circle ABCD, so is the circle EFGH to some area less than the circle ABCD. Q. E. D.

+ +
+
+ PROPOSITION 3. +

Any pyramid which has a triangular base is divided into two pyramids equal and similar to one another, similar to the whole and having triangular bases, and into two equal prisms; and the two prisms are greater than the half of the whole pyramid. +

+

Let there be a pyramid of which the triangle ABC is the base and the point D the vertex; I say that the pyramid ABCD is divided into two pyramids equal to one another, having triangular bases and similar to the whole pyramid, and into two equal prisms; and the two prisms are greater than the half of the whole pyramid.

+

+

For let AB, BC, CA, AD, DB, DC be bisected at the points E, F, G, H, K, L, and let HE, EG, GH, HK, KL, LH, KF, FG be joined.

+

Since AE is equal to EB, and AH to DH, therefore EH is parallel to DB. [VI. 2]

+

For the same reason HK is also parallel to AB.

+

Therefore HEBK is a parallelogram; therefore HK is equal to EB. [I. 34]

+

But EB is equal to EA; therefore AE is also equal to HK.

+

But AH is also equal to HD; therefore the two sides EA, AH are equal to the two sides KH, HD respectively, and the angle EAH is equal to the angle KHD; therefore the base EH is equal to the base KD. [I. 4]

+

Therefore the triangle AEH is equal and similar to the triangle HKD.

+

For the same reason the triangle AHG is also equal and similar to the triangle HLD.

+

Now, since two straight lines EH, HG meeting one another are parallel to two straight lines KD, DL meeting one another, and are not in the same plane, they will contain equal angles. [XI. 10]

+

Therefore the angle EHG is equal to the angle KDL.

+

And, since the two straight lines EH, HG are equal to the two KD, DL respectively, and the angle EHG is equal to the angle KDL, therefore the base EG is equal to the base KL; [I. 4] therefore the triangle EHG is equal and similar to the triangle KDL.

+

For the same reason the triangle AEG is also equal and similar to the triangle HKL.

+

Therefore the pyramid of which the triangle AEG is the base and the point H the vertex is equal and similar to the pyramid of which the triangle HKL is the base and the point D the vertex. [XI. Def. 10]

+

And, since HK has been drawn parallel to AB, one of the sides of the triangle ADB, the triangle ADB is equiangular to the triangle DHK, [I. 29] and they have their sides proportional; therefore the triangle ADB is similar to the triangle DHK. [VI. Def. 1]

+

For the same reason the triangle DBC is also similar to the triangle DKL, and the triangle ADC to the triangle DLH.

+

Now, since the two straight lines BA, AC meeting one another are parallel to the two straight lines KH, HL meeting one another, not in the same plane, they will contain equal angles. [XI. 10]

+

Therefore the angle BAC is equal to the angle KHL.

+

And, as BA is to AC, so is KH to HL; therefore the triangle ABC is similar to the triangle HKL.

+

Therefore also the pyramid of which the triangle ABC is the base and the point D the vertex is similar to the pyramid of which the triangle HKL is the base and the point D the vertex.

+

But the pyramid of which the triangle HKL is the base and the point D the vertex was proved similar to the pyramid of which the triangle AEG is the base and the point H the vertex.

+

Therefore each of the pyramids AEGH, HKLD is similar to the whole pyramid ABCD.

+

Next, since BF is equal to FC, the parallelogram EBFG is double of the triangle GFC.

+

And since, if there be two prisms of equal height, and one have a parallelogram as base, and the other a triangle, and if the parallelogram be double of the triangle, the prisms are equal, [XI. 39] therefore the prism contained by the two triangles BKF, EHG, and the three parallelograms EBFG, EBKH, HKFG is equal to the prism contained by the two triangles GFC, HKL and the three parallelograms KFCL, LCGH, HKFG.

+

And it is manifest that each of the prisms, namely that in which the parallelogram EBFG is the base and the straight line HK is its opposite, and that in which the triangle GFC is the base and the triangle HKL its opposite, is greater than each of the pyramids of which the triangles AEG, HKL are the bases and the points H, D the vertices, inasmuch as, if we join the straight lines EF, EK, the prism in which the parallelogram EBFG is the base and the straight line HK its opposite is greater than the pyramid of which the triangle EBF is the base and the point K the vertex.

+

But the pyramid of which the triangle EBF is the base and the point K the vertex is equal to the pyramid of which the triangle AEG is the base and the point H the vertex; for they are contained by equal and similar planes.

+

Hence also the prism in which the parallelogram EBFG is the base and the straight line HK its opposite is greater than the pyramid of which the triangle AEG is the base and the point H the vertex.

+

But the prism in which the parallelogram EBFG is the base and the straight line HK its opposite is equal to the prism in which the triangle GFC is the base and the triangle HKL its opposite, and the pyramid of which the triangle AEG is the base and the point H the vertex is equal to the pyramid of which the triangle HKL is the base and the point D the vertex.

+

Therefore the said two prisms are greater than the said two pyramids of which the triangles AEG, HKL are the bases and the points H, D the vertices.

+

Therefore the whole pyramid, of which the triangle ABC is the base and the point D the vertex, has been divided into two pyramids equal to one another and into two equal prisms, and the two prisms are greater than the half of the whole pyramid. Q. E. D.

+
+
+ PROPOSITION 4. +

If there be two pyramids of the same height which have triangular bases, and each of them be divided into two pyramids equal to one another and similar to the whole, and into two equal prisms, then, as the base of the one pyramid is to the base of the other pyramid, so will all the prisms in the one pyramid be to all the prisms, being equal in multitude, in the other pyramid. +

+

Let there be two pyramids of the same height which have the triangular bases ABC, DEF, and vertices the points G, H, and let each of them be divided into two pyramids equal to one another and similar to the whole and into two equal prisms; [XII. 3] I say that, as the base ABC is to the base DEF, so are all the prisms in the pyramid ABCG to all the prisms, being equal in multitude, in the pyramid DEFH,

+

For, since BO is equal to OC, and AL to LC, therefore LO is parallel to AB, and the triangle ABC is similar to the triangle LOC.

+

+

For the same reason the triangle DEF is also similar to the triangle RVF.

+

And, since BC is double of CO, and EF of FV, therefore, as BC is to CO, so is EF to FV.

+

And on BC, CO are described the similar and similarly situated rectilineal figures ABC, LOC, and on EF, FV the similar and similarly situated figures DEF, RVF; therefore, as the triangle ABC is to the triangle LOC, so is the triangle DEF to the triangle RVF; [VI. 22] therefore, alternately, as the triangle ABC is to the triangle DEF, so is the triangle LOC to the triangle RVF. [V. 16]

+

But, as the triangle LOC is to the triangle RVF, so is the prism in which the triangle LOC is the base and PMN its opposite to the prism in which the triangle RVF is the base and STU its opposite; [Lemma following] therefore also, as the triangle ABC is to the triangle DEF, so is the prism in which the triangle LOC is the base and PMN its opposite to the prism in which the triangle RVF is the base and STU its opposite.

+

But, as the said prisms are to one another, so is the prism in which the parallelogram KBOL is the base and the straight line PM its opposite to the prism in which the parallelogram QEVR is the base and the straight line ST its opposite. [XI. 39; cf. XII. 3]

+

Therefore also the two prisms, that in which the parallelogram KBOL is the base and PM its opposite, and that in which the triangle LOC is the base and PMN its opposite, are to the prisms in which QEVR is the base and the straight line ST its opposite and in which the triangle RVF is the base and STU its opposite in the same ratio [V. 12]

+

Therefore also, as the base ABC is to the base DEF, so are the said two prisms to the said two prisms.

+

And similarly, if the pyramids PMNG, STUH be divided into two prisms and two pyramids, as the base PMN is to the base STU, so will the two prisms in the pyramid PMNG be to the two prisms in the pyramid STUH.

+

But, as the base PMN is to the base STU, so is the base ABC to the base DEF; for the triangles PMN, STU are equal to the triangles LOC, RVF respectively.

+

Therefore also, as the base ABC is to the base DEF, so are the four prisms to the four prisms.

+

And similarly also, if we divide the remaining pyramids into two pyramids and into two prisms, then, as the base ABC is to the base DEF, so will all the prisms in the pyramid ABCG be to all the prisms, being equal in multitude, in the pyramid DEFH. Q. E. D.

+ +

Lemma. But that, as the triangle LOC is to the triangle RVF, so is the prism in which the triangle LOC is the base and PMN its opposite to the prism in which the triangle RVF is the base and STU its opposite, we must prove as follows.

+

For in the same figure let perpendiculars be conceived drawn from G, H to the planes ABC, DEF; these are of course equal because, by hypothesis, the pyramids are of equal height.

+

Now, since the two straight lines GC and the perpendicular from G are cut by the parallel planes ABC, PMN, they will be cut in the same ratios. [XI. 17]

+

And GC is bisected by the plane PMN at N; therefore the perpendicular from G to the plane ABC will also be bisected by the plane PMN.

+

For the same reason the perpendicular from H to the plane DEF will also be bisected by the plane STU.

+

And the perpendiculars from G, H to the planes ABC, DEF are equal; therefore the perpendiculars from the triangles PMN, STU to the planes ABC, DEF are also equal.

+

Therefore the prisms in which the triangles LOC, RVF are bases, and PMN, STU their opposites, are of equal height.

+

Hence also the parallelepipedal solids described from the said prisms are of equal height and are to one another as their bases; [XI. 32] therefore their halves, namely the said prisms, are to one another as the base LOC is to the base RVF. Q. E. D.

+ +
+
+ PROPOSITION 5. +

+ Pyramids which are of the same height and have triangular bases are to one another as the bases. +

+

Let there be pyramids of the same height, of which the triangles ABC, DEF are the bases and the points G, H the vertices; I say that, as the base ABC is to the base DEF, so is the pyramid ABCG to the pyramid DEFH.

+

+

For, if the pyramid ABCG is not to the pyramid DEFH as the base ABC is to the base DEF, then, as the base ABC is to the base DEF, so will the pyramid ABCG be either to some solid less than the pyramid DEFH or to a greater.

+

Let it, first, be in that ratio to a less solid W, and let the pyramid DEFH be divided into two pyramids equal to one another and similar to the whole and into two equal prisms; then the two prisms are greater than the half of the whole pyramid. [XII. 3]

+

Again, let the pyramids arising from the division be similarly divided, and let this be done continually until there are left over from the pyramid DEFH some pyramids which are less than the excess by which the pyramid DEFH exceeds the solid W. [X. 1]

+

Let such be left, and let them be, for the sake of argument, DQRS, STUH; therefore the remainders, the prisms in the pyramid DEFH, are greater than the solid W.

+

Let the pyramid ABCG also be divided similarly, and a similar number of times, with the pyramid DEFH; therefore, as the base ABC is to the base DEF, so are the prisms in the pyramid ABCG to the prisms in the pyramid DEFH. [XII. 4]

+

But, as the base ABC is to the base DEF, so also is the pyramid ABCG to the solid W; therefore also, as the pyramid ABCG is to the solid W, so are the prisms in the pyramid ABCG to the prisms in the pyramid DEFH; [V. 11] therefore, alternately, as the pyramid ABCG is to the prisms in it, so is the solid W to the prisms in the pyramid DEFH. [V. 16]

+

But the pyramid ABCG is greater than the prisms in it; therefore the solid W is also greater than the prisms in the pyramid DEFH.

+

But it is also less: which is impossible.

+

Therefore the prism ABCG is not to any solid less than the pyramid DEFH as the base ABC is to the base DEF.

+

Similarly it can be proved that neither is the pyramid DEFH to any solid less than the pyramid ABCG as the base DEF is to the base ABC.

+

I say next that neither is the pyramid ABCG to any solid greater than the pyramid DEFH as the base ABC is to the base DEF.

+

For, if possible, let it be in that ratio to a greater solid W; therefore, inversely, as the base DEF is to the base ABC, so is the solid W to the pyramid ABCG.

+

But, as the solid W is to the solid ABCG, so is the pyramid DEFH to some solid less than the pyramid ABCG, as was before proved; [XII. 2, Lemma] therefore also, as the base DEF is to the base ABC, so is the pyramid DEFH to some solid less than the pyramid ABCG: [V. 11] which was proved absurd.

+

Therefore the pyramid ABCG is not to any solid greater than the pyramid DEFH as the base ABC is to the base DEF.

+

But it was proved that neither is it in that ratio to a less solid.

+

Therefore, as the base ABC is to the base DEF, so is the pyramid ABCG to the pyramid DEFH. Q. E. D.

+
+ +
+ PROPOSITION 6. +

+ Pyramids which are of the same height and have polygonal bases are to one another as the bases. +

+

Let there be pyramids of the same height of which the polygons ABCDE, FGHKL are the bases and the points M, N the vertices; I say that, as the base ABCDE is to the base FGHKL, so is the pyramid ABCDEM to the pyramid FGHKLN.

+

+

For let AC, AD, FH, FK be joined.

+

Since then ABCM, ACDM are two pyramids which have triangular bases and equal height, they are to one another as the bases; [XII. 5] therefore, as the base ABC is to the base ACD, so is the pyramid ABCM to the pyramid ACDM.

+

And, componendo, as the base ABCD is to the base ACD, so is the pyramid ABCDM to the pyramid ACDM. [V. 18]

+

But also, as the base ACD is to the base ADE, so is the pyramid ACDM to the pyramid ADEM. [XII. 5]

+

Therefore, ex aequali, as the base ABCD is to the base ADE, so is the pyramid ABCDM to the pyramid ADEM. [V. 22]

+

And again componendo, as the base ABCDE is to the base ADE, so is the pyramid ABCDEM to the pyramid ADEM. [V. 18]

+

Similarly also it can be proved that, as the base FGHKL is to the base FGH, so is the pyramid FGHKLN to the pyramid FGHN.

+

And, since ADEM, FGHN are two pyramids which have triangular bases and equal height, therefore, as the base ADE is to the base FGH, so is the pyramid ADEM to the pyramid FGHN. [XII. 5]

+

But, as the base ADE is to the base ABCDE, so was the pyramid ADEM to the pyramid ABCDEM.

+

Therefore also, ex aequali, as the base ABCDE is to the base FGH, so is the pyramid ABCDEM to the pyramid FGHN. [V. 22]

+

But further, as the base FGH is to the base FGHKL, so also was the pyramid FGHN to the pyramid FGHKLN.

+

Therefore also, ex aequali, as the base ABCDE is to the base FGHKL, so is the pyramid ABCDEM to the pyramid FGHKLN. [V. 22] Q. E. D.

+
+ +
+ PROPOSITION 7. +

+ Any prism which has a triangular base is divided into three pyramids equal to one another which have triangular bases. +

+

Let there be a prism in which the triangle ABC is the base and DEF its opposite; I say that the prism ABCDEF is divided into three pyramids equal to one another, which have triangular bases.

+

+

For let BD, EC, CD be joined.

+

Since ABED is a parallelogram, and BD is its diameter, therefore the triangle ABD is equal to the triangle EBD; [I. 34] therefore also the pyramid of which the triangle ABD is the base and the point C the vertex is equal to the pyramid of which the triangle DEB is the base and the point C the vertex. [XII. 5]

+

But the pyramid of which the triangle DEB is the base and the point C the vertex is the same with the pyramid of which the triangle EBC is the base and the point D the vertex; for they are contained by the same planes.

+

Therefore the pyramid of which the triangle ABD is the base and the point C the vertex is also equal to the pyramid of which the triangle EBC is the base and the point D the vertex.

+

Again, since FCBE is a parallelogram, and CE is its diameter, the triangle CEF is equal to the triangle CBE. [I. 34]

+

Therefore also the pyramid of which the triangle BCE is the base and the point D the vertex is equal to the pyramid of which the triangle ECF is the base and the point D the vertex. [XII. 5]

+

But the pyramid of which the triangle BCE is the base and the point D the vertex was proved equal to the pyramid of which the triangle ABD is the base and the point C the vertex; therefore also the pyramid of which the triangle CEF is the base and the point D the vertex is equal to the pyramid of which the triangle ABD is the base and the point C the vertex; therefore the prism ABCDEF has been divided into three pyramids equal to one another which have triangular bases.

+

And, since the pyramid of which the triangle ABD is the base and the point C the vertex is the same with the pyramid of which the triangle CAB is the base and the point D the vertex, for they are contained by the same planes, while the pyramid of which the triangle ABD is the base and the point C the vertex was proved to be a third of the prism in which the triangle ABC is the base and DEF its opposite, therefore also the pyramid of which the triangle ABC is the base and the point D the vertex is a third of the prism which has the same base, the triangle ABC, and DEF as its opposite.

+ +

Porism. From this it is manifest that any pyramid is a third part of the prism which has the same base with it and equal height. Q. E. D.

+ +
+
+ PROPOSITION 8. +

+ Similar pyramids which have triangular bases are in the triplicate ratio of their corresponding sides. +

+

Let there be similar and similarly situated pyramids of which the triangles ABC, DEF, are the bases and the points G, H the vertices; I say that the pyramid ABCG has to the pyramid DEFH the ratio triplicate of that which BC has to EF.

+

+

For let the parallelepipedal solids BGML, EHQP be completed.

+

Now, since the pyramid ABCG is similar to the pyramid DEFH, therefore the angle ABC is equal to the angle DEF, the angle GBC to the angle HEF, and the angle ABG to the angle DEH; and, as AB is to DE, so is BC to EF, and BG to EH.

+

And since, as AB is to DE, so is BC to EF, and the sides are proportional about equal angles, therefore the parallelogram BM is similar to the parallelogram EQ.

+

For the same reason BN is also similar to ER, and BK to EO; therefore the three parallelograms MB, BK, BN are similar to the three EQ, EO, ER.

+

But the three parallelograms MB, BK, BN are equal and similar to their three opposites, and the three EQ, EO, ER are equal and similar to their three opposites. [XI. 24]

+

Therefore the solids BGML, EHQP are contained by similar planes equal in multitude.

+

Therefore the solid BGML is similar to the solid EHQP.

+

But similar parallelepipedal solids are in the triplicate ratio of their corresponding sides. [XI. 33]

+

Therefore the solid BGML has to the solid EHQP the ratio triplicate of that which the corresponding side BC has to the corresponding side EF.

+

But, as the solid BGML is to the solid EHQP, so is the pyramid ABCG to the pyramid DEFH, inasmuch as the pyramid is a sixth part of the solid, because the prism which is half of the parallelepipedal solid [XI. 28] is also triple of the pyramid. [XII. 7]

+

Therefore the pyramid ABCG also has to the pyramid DEFH the ratio triplicate of that which BC has to EF. Q. E. D.

+ +

Porism. From this it is manifest that similar pyramids which have polygonal bases are also to one another in the triplicate ratio of their corresponding sides.

+

For, if they are divided into the pyramids contained in them which have triangular bases, by virtue of the fact that the similar polygons forming their bases are also divided into similar triangles equal in multitude and corresponding to the wholes [VI. 20], then, as the one pyramid which has a triangular base in the one complete pyramid is to the one pyramid which has a triangular base in the other complete pyramid, so also will all the pyramids which have triangular bases contained in the one pyramid be to all the pyramids which have triangular bases contained in the other pyramid [V. 12], that is, the pyramid itself which has a polygonal base to the pyramid which has a polygonal base.

+

But the pyramid which has a triangular base is to the pyramid which has a triangular base in the triplicate ratio of the corresponding sides; therefore also the pyramid which has a polygonal base has to the pyramid which has a similar base the ratio triplicate of that which the side has to the side.

+ +
+ +
+ PROPOSITION 9. +

+ In equal pyramids which have triangular bases the bases are reciprocally proportional to the heights; and those pyramids in which the bases are reciprocally proportional to the heights are equal. +

+

For let there be equal pyramids which have the triangular bases ABC, DEF and vertices the points G, H; I say that in the pyramids ABCG, DEFH the bases are reciprocally proportional to the heights, that is, as the base ABC is to the base DEF, so is the height of the pyramid DEFH to the height of the pyramid ABCG.

+

+

For let the parallelepipedal solids BGML, EHQP be completed.

+

Now, since the pyramid ABCG is equal to the pyramid DEFH, and the solid BGML is six times the pyramid ABCG, and the solid EHQP six times the pyramid DEFH, therefore the solid BGML is equal to the solid EHQP.

+

But in equal parallelepipedal solids the bases are reciprocally proportional to the heights; [XI. 34] therefore, as the base BM is to the base EQ, so is the height of the solid EHQP to the height of the solid BGML.

+

But, as the base BM is to EQ, so is the triangle ABC to the triangle DEF. [I. 34]

+

Therefore also, as the triangle ABC is to the triangle DEF, so is the height of the solid EHQP to the height of the solid BGML. [V. 11]

+

But the height of the solid EHQP is the same with the height of the pyramid DEFH, and the height of the solid BGML is the same with the height of the pyramid ABCG, therefore, as the base ABC is to the base DEF, so is the height of the pyramid DEFH to the height of the pyramid ABCG.

+

Therefore in the pyramids ABCG, DEFH the bases are reciprocally proportional to the heights.

+

Next, in the pyramids ABCG, DEFH let the bases be reciprocally proportional to the heights; that is, as the base ABC is to the base DEF, so let the height of the pyramid DEFH be to the height of the pyramid ABCG; I say that the pyramid ABCG is equal to the pyramid DEFH.

+

For, with the same construction, since, as the base ABC is to the base DEF, so is the height of the pyramid DEFH to the height of the pyramid ABCG, while, as the base ABC is to the base DEF, so is the parallelogram BM to the parallelogram EQ, therefore also, as the parallelogram BM is to the parallelogram EQ, so is the height of the pyramid DEFH to the height of the pyramid ABCG. [V. 11]

+

But the height of the pyramid DEFH is the same with the height of the parallelepiped EHQP, and the height of the pyramid ABCG is the same with the height of the parallelepiped BGML; therefore, as the base BM is to the base EQ, so is the height of the parallelepiped EHQP to the height of the parallelepiped BGML.

+

But those parallelepipedal solids in which the bases are reciprocally proportional to the heights are equal; [XI. 34] therefore the parallelepipedal solid BGML is equal to the parallelepipedal solid EHQP.

+

And the pyramid ABCG is a sixth part of BGML, and the pyramid DEFH a sixth part of the parallelepiped EHQP; therefore the pyramid ABCG is equal to the pyramid DEFH.

+

Therefore etc Q. E. D.

+
+
+ PROPOSITION 10. +

+ Any cone is a third part of the cylinder which has the same base with it and equal height. +

+

For let a cone have the same base, namely the circle ABCD, with a cylinder and equal height; I say that the cone is a third part of the cylinder, that is, that the cylinder is triple of the cone.

+

For if the cylinder is not triple of the cone, the cylinder will be either greater than triple or less than triple of the cone.

+

+

First let it be greater than triple, and let the square ABCD be inscribed in the circle ABCD; [IV. 6] then the square ABCD is greater than the half of the circle ABCD.

+

From the square ABCD let there be set up a prism of equal height with the cylinder.

+

Then the prism so set up is greater than the half of the cylinder, inasmuch as, if we also circumscribe a square about the circle ABCD [IV. 7], the square inscribed in the circle ABCD is half of that circumscribed about it, and the solids set up from them are parallelepipedal prisms of equal height, while parallelepipedal solids which are of the same height are to one another as their bases; [XI. 32] therefore also the prism set up on the square ABCD is half of the prism set up from the square circumscribed about the circle ABCD; [cf. XI. 28, or XII. 6 and 7, Por.] and the cylinder is less than the prism set up from the square circumscribed about the circle ABCD; therefore the prism set up from the square ABCD and of equal height with the cylinder is greater than the half of the cylinder.

+

Let the circumferences AB, BC, CD, DA be bisected at the points E, F, G, H, and let AE, EB, BF, FC, CG, GD, DH, HA be joined; then each of the triangles AEB, BFC, CGD, DHA is greater than the half of that segment of the circle ABCD which is about it, as we proved before. [XII. 2]

+

On each of the triangles AEB, BFC, CGD, DHA let prisms be set up of equal height with the cylinder; then each of the prisms so set up is greater than the half part of that segment of the cylinder which is about it, inasmuch as, if we draw through the points E, F, G, H parallels to AB, BC, CD, DA, complete the parallelograms on AB, BC, CD, DA, and set up from them parallelepipedal solids of equal height with the cylinder, the prisms on the triangles AEB, BFC, CGD, DHA are halves of the several solids set up; and the segments of the cylinder are less than the parallelepipedal solids set up; hence also the prisms on the triangles AEB, BFC, CGD, DHA are greater than the half of the segments of the cylinder about them.

+

Thus, bisecting the circumferences that are left, joining straight lines, setting up on each of the triangles prisms of equal height with the cylinder, and doing this continually, we shall leave some segments of the cylinder which will be less than the excess by which the cylinder exceeds the triple of the cone. [X. 1]

+

Let such segments be left, and let them be AE, EB, BF, FC, CG, GD, DH, HA; therefore the remainder, the prism of which the polygon AEBFCGDH is the base and the height is the same as that of the cylinder, is greater than triple of the cone.

+

But the prism of which the polygon AEBFCGDH is the base and the height the same as that of the cylinder is triple of the pyramid of which the polygon AEBFCGDH is the base and the vertex is the same as that of the cone; [XII. 7, Por.] therefore also the pyramid of which the polygon AEBFCGDH is the base and the vertex is the same as that of the cone is greater than the cone which has the circle ABCD as base.

+

But it is also less, for it is enclosed by it: which is impossible.

+

Therefore the cylinder is not greater than triple of the cone.

+

I say next that neither is the cylinder less than triple of the cone,

+

For, if possible, let the cylinder be less than triple of the cone, therefore, inversely, the cone is greater than a third part of the cylinder.

+

Let the square ABCD be inscribed in the circle ABCD; therefore the square ABCD is greater than the half of the circle ABCD.

+

Now let there be set up from the square ABCD a pyramid having the same vertex with the cone; therefore the pyramid so set up is greater than the half part of the cone, seeing that, as we proved before, if we circumscribe a square about the circle, the square ABCD will be half of the square circumscribed about the circle, and if we set up from the squares parallelepipedal solids of equal height with the cone, which are also called prisms, the solid set up from the square ABCD will be half of that set up from the square circumscribed about the circle; for they are to one another as their bases. [XI. 32]

+

Hence also the thirds of them are in that ratio; therefore also the pyramid of which the square ABCD is the base is half of the pyramid set up from the square circumscribed about the circle.

+

And the pyramid set up from the square about the circle is greater than the cone, for it encloses it.

+

Therefore the pyramid of which the square ABCD is the base and the vertex is the same with that of the cone is greater than the half of the cone.

+

Let the circumferences AB, BC, CD, DA be bisected at the points E, F, G, H, and let AE, EB, BF, FC, CG, GD, DH, HA be joined; therefore also each of the triangles AEB, BFC, CGD, DHA is greater than the half part of that segment of the circle ABCD which is about it.

+

Now, on each of the triangles AEB, BFC, CGD, DHA let pyramids be set up which have the same vertex as the cone; therefore also each of the pyramids so set up is, in the same manner, greater than the half part of that segment of the cone which is about it.

+

Thus, by bisecting the circumferences that are left, joining straight lines, setting up on each of the triangles a pyramid which has the same vertex as the cone, and doing this continually, we shall leave some segments of the cone which will be less than the excess by which the cone exceeds the third part of the cylinder. [X. 1]

+

Let such be left, and let them be the segments on AE, EB, BF, FC, CG, GD, DH, HA; therefore the remainder, the pyramid of which the polygon AEBFCGDH is the base and the vertex the same with that of the cone, is greater than a third part of the cylinder.

+

But the pyramid of which the polygon AEBFCGDH is the base and the vertex the same with that of the cone is a third part of the prism of which the polygon AEBFCGDH is the base and the height is the same with that of the cylinder; therefore the prism of which the polygon AEBFCGDH is the base and the height is the same with that of the cylinder is greater than the cylinder of which the circle ABCD is the base.

+

But it is also less, for it is enclosed by it: which is impossible.

+

Therefore the cylinder is not less than triple of the cone.

+

But it was proved that neither is it greater than triple; therefore the cylinder is triple of the cone; hence the cone is a third part of the cylinder.

+

Therefore etc. Q. E. D.

+
+ +
+ PROPOSITION 11. +

+ Cones and cylinders which are of the same height are to one another as their bases. +

+

Let there be cones and cylinders of the same height, let the circles ABCD, EFGH be their bases, KL, MN their axes and AC, EG the diameters of their bases; I say that, as the circle ABCD is to the circle EFGH, so is the cone AL to the cone EN.

+

+

For, if not, then, as the circle ABCD is to the circle EFGH, so will the cone AL be either to some solid less than the cone EN or to a greater.

+

First, let it be in that ratio to a less solid O, and let the solid X be equal to that by which the solid O is less than the cone EN; therefore the cone EN is equal to the solids O, X.

+

Let the square EFGH be inscribed in the circle EFGH; therefore the square is greater than the half of the circle.

+

Let there be set up from the square EFGH a pyramid of equal height with the cone; therefore the pyramid so set up is greater than the half of the cone, inasmuch as, if we circumscribe a square about the circle, and set up from it a pyramid of equal height with the cone, the inscribed pyramid is half of the circumscribed pyramid, for they are to one another as their bases, [XII. 6] while the cone is less than the circumscribed pyramid.

+

Let the circumferences EF, FG, GH, HE be bisected at the points P, Q, R, S, and let HP, PE, EQ, QF, FR, RG, GS, SH be joined.

+

Therefore each of the triangles HPE, EQF, FRG, GSH is greater than the half of that segment of the circle which is about it.

+

On each of the triangles HPE, EQF, FRG, GSH let there be set up a pyramid of equal height with the cone; therefore, also, each of the pyramids so set up is greater than the half of that segment of the cone which is about it.

+

Thus, bisecting the circumferences which are left, joining straight lines, setting up on each of the triangles pyramids of equal height with the cone, and doing this continually, we shall leave some segments of the cone which will be less than the solid X. [X. 1]

+

Let such be left, and let them be the segments on HP, PE, EQ, QF, FR, RG, GS, SH; therefore the remainder, the pyramid of which the polygon HPEQFRGS is the base and the height the same with that of the cone, is greater than the solid O.

+

Let there also be inscribed in the circle ABCD the polygon DTAUBVCW similar and similarly situated to the polygon HPEQFRGS, and on it let a pyramid be set up of equal height with the cone AL.

+

Since then, as the square on AC is to the square on EG, so is the polygon DTAUBVCW to the polygon HPEQFRGS, [XII. 1] while, as the square on AC is to the square on EG, so is the circle ABCD to the circle EFGH, [XII. 2] therefore also, as the circle ABCD is to the circle EFGH, so is the polygon DTAUBVCW to the polygon HPEQFRGS.

+

But, as the circle ABCD is to the circle EFGH, so is the cone AL to the solid O, and, as the polygon DTAUBVCW is to the polygon HPEQFRGS, so is the pyramid of which the polygon DTAUBVCW is the base and the point L the vertex to the pyramid of which the polygon HPEQFRGS is the base and the point N the vertex. [XII. 6]

+

Therefore also, as the cone AL is to the solid O, so is the pyramid of which the polygon DTAUBVCW is the base and the point L the vertex to the pyramid of which the polygon HPEQFRGS is the base and the point N the vertex; [V. 11] therefore, alternately, as the cone AL is to the pyramid in it, so is the solid O to the pyramid in the cone EN. [V. 16]

+

But the cone AL is greater than the pyramid in it; therefore the solid O is also greater than the pyramid in the cone EN.

+

But it is also less: which is absurd.

+

Therefore the cone AL is not to any solid less than the cone EN as the circle ABCD is to the circle EFGH.

+

Similarly we can prove that neither is the cone EN to any solid less than the cone AL as the circle EFGH is to the circle ABCD.

+

I say next that neither is the cone AL to any solid greater than the cone EN as the circle ABCD is to the circle EFGH.

+

For, if possible, let it be in that ratio to a greater solid O; therefore, inversely, as the circle EFGH is to the circle ABCD, so is the solid O to the cone AL.

+

But, as the solid O is to the cone AL, so is the cone EN to some solid less than the cone AL; therefore also, as the circle EFGH is to the circle ABCD, so is the cone EN to some solid less than the cone AL: which was proved impossible.

+

Therefore the cone AL is not to any solid greater than the cone EN as the circle ABCD is to the circle EFGH.

+

But it was proved that neither is it in this ratio to a less solid; therefore, as the circle ABCD is to the circle EFGH, so is the cone AL to the cone EN.

+

But, as the cone is to the cone, so is the cylinder to the cylinder, for each is triple of each; [XII. 10]

+

Therefore also, as the circle ABCD is to the circle EFGH, so are the cylinders on them which are of equal height.

+

Therefore etc. Q. E. D.

+
+ +
+ PROPOSITION 12. +

+ Similar cones and cylinders are to one another in the triplicate ratio of the diameters in their bases. +

+

Let there be similar cones and cylinders, let the circles ABCD, EFGH be their bases, BD, FH the diameters of the bases, and KL, MN the axes of the cones and cylinders; I say that the cone of which the circle ABCD is the base and the point L the vertex has to the cone of which the circle EFGH is the base and the point N the vertex the ratio triplicate of that which BD has to FH.

+

+

For, if the cone ABCDL has not to the cone EFGHN the ratio triplicate of that which BD has to FH, the cone ABCDL will have that triplicate ratio either to some solid less than the cone EFGHN or to a greater.

+

First, let it have that triplicate ratio to a less solid O.

+

Let the square EFGH be inscribed in the circle EFGH; [IV. 6] therefore the square EFGH is greater than the half of the circle EFGH.

+

Now let there be set up on the square EFGH a pyramid having the same vertex with the cone; therefore the pyramid so set up is greater than the half part of the cone.

+

Let the circumferences EF, FG, GH, HE be bisected at the points P, Q, R, S, and let EP, PF, FQ, QG, GR, RH, HS, SE be joined.

+

Therefore each of the triangles EPF, FQG, GRH, HSE is also greater than the half part of that segment of the circle EFGH which is about it.

+

Now on each of the triangles EPF, FQG, GRH, HSE let a pyramid be set up having the same vertex with the cone; therefore each of the pyramids so set up is also greater than the half part of that segment of the cone which is about it.

+

Thus, bisecting the circumferences so left, joining straight lines, setting up on each of the triangles pyramids having the same vertex with the cone, and doing this continually, we shall leave some segments of the cone which will be less than the excess by which the cone EFGHN exceeds the solid O. [X. 1]

+

Let such be left, and let them be the segments on EP, PF, FQ, QG, GR, RH, HS, SE; therefore the remainder, the pyramid of which the polygon EPFQGRHS is the base and the point N the vertex, is greater than the solid O.

+

Let there be also inscribed in the circle ABCD the polygon ATBUCVDW similar and similarly situated to the polygon EPFQGRHS, and let there be set up on the polygon ATBUCVDW a pyramid having the same vertex with the cone; of the triangles containing the pyramid of which the polygon ATBUCVDW is the base and the point L the vertex let LBT be one, and of the triangles containing the pyramid of which the polygon EPFQGRHS is the base and the point N the vertex let NFP be one; and let KT, MP be joined.

+

Now, since the cone ABCDL is similar to the cone EFGHN, therefore, as BD is to FH, so is the axis KL to the axis MN. [XI. Def. 24]

+

But, as BD is to FH, so is BK to FM; therefore also, as BK is to FM, so is KL to MN.

+

And, alternately, as BK is to KL, so is FM to MN. [V. 16]

+

And the sides are proportional about equal angles, namely the angles BKL, FMN; therefore the triangle BKL is similar to the triangle FMN. [VI. 6]

+

Again, since, as BK is to KT, so is FM to MP, and they are about equal angles, namely the angles BKT, FMP, inasmuch as, whatever part the angle BKT is of the four right angles at the centre K, the same part also is the angle FMP of the four right angles at the centre M; since then the sides are proportional about equal angles, therefore the triangle BKT is similar to the triangle FMP. [VI. 6]

+

Again, since it was proved that, as BK is to KL, so is FM to MN, while BK is equal to KT, and FM to PM, therefore, as TK is to KL, so is PM to MN; and the sides are proportional about equal angles, namely the angles TKL, PMN, for they are right; therefore the triangle LKT is similar to the triangle NMP. [VI. 6]

+

And since, owing to the similarity of the triangles LKB, NMF, as LB is to BK, so is NF to FM, and, owing to the similarity of the triangles BKT, FMP, as KB is to BT, so is MF to FP, therefore, ex aequali, as LB is to BT, so is NF to FP. [V. 22]

+

Again since, owing to the similarity of the triangles LTK, NPM, as LT is to TK, so is NP to PM, and, owing to the similarity of the triangles TKB, PMF, as KT is to TB, so is MP to PF; therefore, ex aequali, as LT is to TB, so is NP to PF. [V. 22]

+

But it was also proved that, as TB is to BL, so is PF to FN.

+

Therefore, ex aequali, as TL is to LB, so is PN to NF. [V. 22]

+

Therefore in the triangles LTB, NPF the sides are proportional; therefore the triangles LTB, NPF are equiangular; [VI. 5] hence they are also similar. [VI. Def. I]

+

Therefore the pyramid of which the triangle BKT is the base and the point L the vertex is also similar to the pyramid of which the triangle FMP is the base and the point N the vertex, for they are contained by similar planes equal in multitude. [XI. Def. 9]

+

But similar pyramids which have triangular bases are to one another in the triplicate ratio of their corresponding sides. [XII. 8]

+

Therefore the pyramid BKTL has to the pyramid FMPN the ratio triplicate of that which BK has to FM.

+

Similarly, by joining straight lines from A, W, D, V, C, U to K, and from E, S, H, R, G, Q to M, and setting up on each of the triangles pyramids which have the same vertex with the cones, we can prove that each of the similarly arranged pyramids will also have to each similarly arranged pyramid the ratio triplicate of that which the corresponding side BK has to the corresponding side FM, that is, which BD has to FH.

+

And, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents; [V. 12] therefore also, as the pyramid BKTL is to the pyramid FMPN, so is the whole pyramid of which the polygon ATBUCVDW is the base and the point L the vertex to the whole pyramid of which the polygon EPFQGRHS is the base and the point N the vertex; hence also the pyramid of which ATBUCVDW is the base and the point L the vertex has to the pyramid of which the polygon EPFQGRHS is the base and the point N the vertex the ratio triplicate of that which BD has to FH.

+

But, by hypothesis, the cone of which the circle ABCD + is the base and the point L the vertex has also to the solid O the ratio triplicate of that which BD has to FH; therefore, as the cone of which the circle ABCD is the base and the point L the vertex is to the solid O, so is the pyramid of which the polygon ATBUCVDW is the base and L the vertex to the pyramid of which the polygon EPFQGRHS is the base and the point N the vertex; therefore, alternately, as the cone of which the circle ABCD is the base and L the vertex is to the pyramid contained in it of which the polygon ATBUCVDW is the base and L the vertex, so is the solid O to the pyramid of which the polygon EPFQGRHS is the base and N the vertex. [V. 16]

+

But the said cone is greater than the pyramid in it; for it encloses it.

+

Therefore the solid O is also greater than the pyramid of which the polygon EPFQGRHS is the base and N the vertex.

+

But it is also less: which is impossible.

+

Therefore the cone of which the circle ABCD is the base and L the vertex has not to any solid less than the cone of which the circle EFGH is the base and the point N the vertex the ratio triplicate of that which BD has to FH:

+

Similarly we can prove that neither has the cone EFGHN to any solid less than the cone ABCDL the ratio triplicate of that which FH has to BD.

+

I say next that neither has the cone ABCDL to any solid greater than the cone EFGHN the ratio triplicate of that which BD has to FH.

+

For, if possible, let it have that ratio to a greater solid O.

+

Therefore, inversely, the solid O has to the cone ABCDL the ratio triplicate of that which FH has to BD.

+

But, as the solid O is to the cone ABCDL, so is the cone EFGHN to some solid less than the cone ABCDL.

+

Therefore the cone EFGHN also has to some solid less than the cone ABCDL the ratio triplicate of that which FH has to BD: which was proved impossible.

+

Therefore the cone ABCDL has not to any solid greater than the cone EFGHN the ratio triplicate of that which BD has to FH.

+

But it was proved that neither has it this ratio to a less solid than the cone EFGHN.

+

Therefore the cone ABCDL has to the cone EFGHN the ratio triplicate of that which BD has to FH.

+

But, as the cone is to the cone, so is the cylinder to the cylinder, for the cylinder which is on the same base as the cone and of equal height with it is triple of the cone; [XII. 10] therefore the cylinder also has to the cylinder the ratio triplicate of that which BD has to FH.

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 13. +

+ If a cylinder be cut by a plane which is parallel to its opposite planes, then, as the cylinder is to the cylinder, so will the axis be to the axis. +

+

For let the cylinder AD be cut by the plane GH which is parallel to the opposite planes AB, CD, and let the plane GH meet the axis at the point K; I say that, as the cylinder BG is to the cylinder GD, so is the axis EK to the axis KF.

+

+

For let the axis EF be produced in both directions to the points L, M, and let there be set out any number whatever of axes EN, NL equal to the axis EK, and any number whatever FO, OM equal to FK; and let the cylinder PW on the axis LM be conceived of which the circles PQ, VW are the bases.

+

Let planes be carried through the points N, O parallel to AB, CD and to the bases of the cylinder PW, and let them produce the circles RS, TU about the centres N, O.

+

Then, since the axes LN, NE, EK are equal to one another, therefore the cylinders QR, RB, BG are to one another as their bases. [XII. 11]

+

But the bases are equal; therefore the cylinders QR, RB, BG are also equal to one another.

+

Since then the axes LN, NE, EK are equal to one another, and the cylinders QR, RB, BG are also equal to one another, and the multitude of the former is equal to the multitude of the latter, therefore, whatever multiple the axis KL is of the axis EK, the same multiple also will the cylinder QG be of the cylinder GB.

+

For the same reason, whatever multiple the axis MK is of the axis KF, the same multiple also is the cylinder WG of the cylinder GD.

+

And, if the axis KL is equal to the axis KM, the cylinder QG will also be equal to the cylinder GW, if the axis is greater than the axis, the cylinder will also be greater than the cylinder, and if less, less.

+

Thus, there being four magnitudes, the axes EK, KF and the cylinders BG, GD, there have been taken equimultiples of the axis EK and of the cylinder BG, namely the axis LK and the cylinder QG, and equimultiples of the axis KF and of the cylinder GD, namely the axis KM and the cylinder GW; and it has been proved that, if the axis KL is in excess of the axis KM, the cylinder QG is also in excess of the cylinder GW, if equal, equal, and if less, less.

+

Therefore, as the axis EK is to the axis KF, so is the cylinder BG to the cylinder GD. [V. Def. 5] Q. E. D.

+
+
+ PROPOSITION 14. +

+ Cones and cylinders which are on equal bases are to one another as their heights. +

+

For let EB, FD be cylinders on equal bases, the circles AB, CD; I say that, as the cylinder EB is to the cylinder FD, so is the axis GH to the axis KL.

+

+

For let the axis KL be produced to the point N, let LN be made equal to the axis GH, and let the cylinder CM be conceived about LN as axis.

+

Since then the cylinders EB, CM are of the same height, they are to one another as their bases. [XII. 11]

+

But the bases are equal to one another; therefore the cylinders EB, CM are also equal.

+

And, since the cylinder FM has been cut by the plane CD which is parallel to its opposite planes, therefore, as the cylinder CM is to the cylinder FD, so is the axis LN to the axis KL. [XII. 13]

+

But the cylinder CM is equal to the cylinder EB, and the axis LN to the axis GH; therefore, as the cylinder EB is to the cylinder FD, so is the axis GH to the axis KL.

+

But, as the cylinder EB is to the cylinder FD, so is the cone ABG to the cone CDK. [XII. 10]

+

Therefore also, as the axis GH is to the axis KL, so is the cone ABG to the cone CDK and the cylinder EB to the cylinder FD. Q. E. D.

+
+ +
+ PROPOSITION 15. +

+ In equal cones and cylinders the bases are reciprocally proportional to the heights; and those cones and cylinders in which the bases are reciprocally proportional to the heights are equal. +

+

Let there be equal cones and cylinders of which the circles ABCD, EFGH are the bases; let AC, EG be the diameters of the bases, and KL, MN the axes, which are also the heights of the cones or cylinders; let the cylinders AO, EP be completed.

+

I say that in the cylinders AO, EP the bases are reciprocally proportional to the heights, that is, as the base ABCD is to the base EFGH, so is the height MN to the height KL.

+

+

For the height LK is either equal to the height MN or not equal.

+

First, let it be equal.

+

Now the cylinder AO is also equal to the cylinder EP.

+

But cones and cylinders which are of the same height are to one another as their bases; [XII. 11] therefore the base ABCD is also equal to the base EFGH.

+

Hence also, reciprocally, as the base ABCD is to the base EFGH, so is the height MN to the height KL.

+

Next, let the height LK not be equal to MN, but let MN be greater; from the height MN let QN be cut off equal to KL, through the point Q let the cylinder EP be cut by the plane TUS parallel to the planes of the circles EFGH, RP, and let the cylinder ES be conceived erected from the circle EFGH as base and with height NQ.

+

Now, since the cylinder AO is equal to the cylinder EP, therefore, as the cylinder AO is to the cylinder ES, so is the cylinder EP to the cylinder ES. [V. 7]

+

But, as the cylinder AO is to the cylinder ES, so is the base ABCD to the base EFGH, for the cylinders AO, ES are of the same height; [XII. 11] and, as the cylinder EP is to the cylinder ES, so is the height MN to the height QN, for the cylinder EP has been cut by a plane which is parallel to its opposite planes. [XII. 13]

+

Therefore also, as the base ABCD is to the base EFGH, so is the height MN to the height QN. [V. 11]

+

But the height QN is equal to the height KL; therefore, as the base ABCD is to the base EFGH, so is the height MN to the height KL.

+

Therefore in the cylinders AO, EP the bases are reciprocally proportional to the heights.

+

Next, in the cylinders AO, EP let the bases be reciprocally proportional to the heights, that is, as the base ABCD is to the base EFGH, so let the height MN be to the height KL; I say that the cylinder AO is equal to the cylinder EP.

+

For, with the same construction, since, as the base ABCD is to the base EFGH, so is the height MN to the height KL, while the height KL is equal to the height QN, therefore, as the base ABCD is to the base EFGH, so is the height MN to the height QN +

+

But, as the base ABCD is to the base EFGH, so is the cylinder AO to the cylinder ES, for they are of the same height; [XII. 11] and, as the height MN is to QN, so is the cylinder EP to the cylinder ES; [XII. 13] therefore, as the cylinder AO is to the cylinder ES, so is the cylinder EP to the cylinder ES. [V. 11]

+

Therefore the cylinder AO is equal to the cylinder EP. [V. 9]

+

And the same is true for the cones also. Q. E. D.

+
+
+ PROPOSITION 16. +

Given two circles about the same centre, to inscribe in the greater circle an equilateral polygon with an even number of sides which does not touch the lesser circle. +

+

Let ABCD, EFGH be the two given circles about the same centre K; thus it is required to inscribe in the greater circle ABCD an equilateral polygon with an even number of sides which does not touch the circle EFGH.

+

+

For let the straight line BKD be drawn through the centre K, and from the point G let GA be drawn at right angles to the straight line BD and carried through to C; therefore AC touches the circle EFGH. [III. 16, Por.]

+

Then, bisecting the circumference BAD, bisecting the half of it, and doing this continually, we shall leave a circumference less than AD. [X. 1]

+

Let such be left, and let it be LD; from L let LM be drawn perpendicular to BD and carried through to N, and let LD, DN be joined; therefore LD is equal to DN. [III. 3, I. 4]

+

Now, since LN is parallel to AC, and AC touches the circle EFGH, therefore LN does not touch the circle EFGH; therefore LD, DN are far from touching the circle EFGH.

+

If then we fit into the circle ABCD straight lines equal to the straight line LD and placed continuously, there will be inscribed in the circle ABCD an equilateral polygon with an even number of sides which does not touch the lesser circle EFGH. Q. E. F.

+
+ +
+ PROPOSITION 17. +

Given two spheres about the same centre, to inscribe in the greater sphere a polyhedral solid which does not touch the lesser sphere at its surface. +

+

Let two spheres be conceived about the same centre A; thus it is required to inscribe in the greater sphere a polyhedral solid which does not touch the lesser sphere at its surface.

+

+

Let the spheres be cut by any plane through the centre; then the sections will be circles, inasmuch as the sphere was produced by the diameter remaining fixed and the semicircle being carried round it; [XI. Def. 14] hence, in whatever position we conceive the semicircle to be, the plane carried through it will produce a circle on the circumference of the sphere.

+

And it is manifest that this circle is the greatest possible, inasmuch as the diameter of the sphere, which is of course the diameter both of the semicircle and of the circle, is greater than all the straight lines drawn across in the circle or the sphere.

+

Let then BCDE be the circle in the greater sphere, and FGH the circle in the lesser sphere; let two diameters in them, BD, CE, be drawn at right angles to one another; then, given the two circles BCDE, FGH about the same centre, let there be inscribed in the greater circle BCDE an equilateral polygon with an even number of sides which does not touch the lesser circle FGH, let BK, KL, LM + ME be its sides in the quadrant BE. let KA be joined and carried through to N, let AO be set up from the point A at right angles to the plane of the circle BCDE, and let it meet the surface of the sphere at O, and through AO and each of the straight lines BD, KN let planes be carried; they will then make greatest circles on the surface of the sphere, for the reason stated.

+

Let them make such, and in them let BOD, KON be the semicircles on BD, KN.

+

Now, since OA is at right angles to the plane of the circle BCDE, therefore all the planes through OA are also at right angles to the plane of the circle BCDE; [XI. 18] hence the semicircles BOD, KON are also at right angles to the plane of the circle BCDE.

+

And, since the semicircles BED, BOD, KON are equal, for they are on the equal diameters BD, KN, therefore the quadrants BE, BO, KO are also equal to one another.

+

Therefore there are as many straight lines in the quadrants BO, KO equal to the straight lines BK, KL, LM, ME as there are sides of the polygon in the quadrant BE.

+

Let them be inscribed, and let them be BP, PQ, QR, RO and KS, ST, TU, UO, let SP, TQ, UR be joined, and from P, S let perpendiculars be drawn to the plane of the circle BCDE; [XI. 11] these will fall on BD, KN, the common sections of the planes, inasmuch as the planes of BOD, KON are also at right angles to the plane of the circle BCDE. [cf. XI. Def. 4]

+

Let them so fall, and let them be PV, SW, and let WV be joined.

+

Now since, in the equal semicircles BOD, KON, equal straight lines BP, KS have been cut off, and the perpendiculars PV, SW have been drawn, therefore PV is equal to SW, and BV to KW. [III. 27, I. 26]

+

But the whole BA is also equal to the whole KA; therefore the remainder VA is also equal to the remainder WA; therefore, as BV is to VA, so is KW to WA; therefore WV is parallel to KB. [VI. 2]

+

And, since each of the straight lines PV, SW is at right angles to the plane of the circle BCDE, therefore PV is parallel to SW. [XI. 6]

+

But it was also proved equal to it; therefore WV, SP are also equal and parallel. [I. 33]

+

And, since WV is parallel to SP, while WV is parallel to KB, therefore SP is also parallel to KB. [XI. 9]

+

And BP, KS join their extremities; therefore the quadrilateral KBPS is in one plane, inasmuch as, if two straight lines be parallel, and points be taken at random on each of them, the straight line joining the points is in the same plane with the parallels. [XI. 7]

+

For the same reason each of the quadrilaterals SPQT, TQRU is also in one plane.

+

But the triangle URO is also in one plane. [XI. 2]

+

If then we conceive straight lines joined from the points P, S, Q, T, R, U to A, there will be constructed a certain polyhedral solid figure between the circumferences BO, KO, consisting of pyramids of which the quadrilaterals KBPS, SPQT, TQRU and the triangle URO are the bases and the point A the vertex.

+

And, if we make the same construction in the case of each of the sides KL, LM, ME as in the case of BK, and further in the case of the remaining three quadrants, there will be constructed a certain polyhedral figure inscribed in the sphere and contained by pyramids, of which the said quadrilaterals and the triangle URO, and the others corresponding to them, are the bases and the point A the vertex.

+

I say that the said polyhedron will not touch the lesser sphere at the surface on which the circle FGH is.

+

Let AX be drawn from the point A perpendicular to the plane of the quadrilateral KBPS, and let it meet the plane at the point X; [XI. 11] let XB, XK be joined.

+

Then, since AX is at right angles to the plane of the quadrilateral KBPS, therefore it is also at right angles to all the straight lines which meet it and are in the plane of the quadrilateral. [XI. Def. 3]

+

Therefore AX is at right angles to each of the straight lines BX, XK.

+

And, since AB is equal to AK, the square on AB is also equal to the square on AK.

+

And the squares on AX, XB are equal to the square on AB, for the angle at X is right; [I. 47] and the squares on AX, XK are equal to the square on AK. [id.]

+

Therefore the squares on AX, XB are equal to the squares on AX, XK.

+

Let the square on AX be subtracted from each; therefore the remainder, the square on BX, is equal to the remainder, the square on XK; therefore BX is equal to XK.

+

Similarly we can prove that the straight lines joined from X to P, S are equal to each of the straight lines BX, XK.

+

Therefore the circle described with centre X and distance one of the straight lines XB, XK will pass through P, S also, and KBPS will be a quadrilateral in a circle.

+

Now, since KB is greater than WV, while WV is equal to SP, therefore KB is greater than SP.

+

But KB is equal to each of the straight lines KS, BP; therefore each of the straight lines KS, BP is greater than SP.

+

And, since KBPS is a quadrilateral in a circle, and KB, BP, KS are equal, and PS less, and BX is the radius of the circle, therefore the square on KB is greater than double of the square on BX.

+

Let KZ be drawn from K perpendicular to BV.

+

Then, since BD is less than double of DZ, and, as BD is to DZ, so is the rectangle DB, BZ to the rectangle DZ, ZB, if a square be described upon BZ and the parallelogram on ZD be completed, then the rectangle DB, BZ is also less than double of the rectangle DZ, ZB.

+

And, if KD be joined, the rectangle DB, BZ is equal to the square on BK, and the rectangle DZ, ZB equal to the square on KZ; [III. 31, VI. 8 and Por.] therefore the square on KB is less than double of the square on KZ.

+

But the square on KB is greater than double of the square on BX; therefore the square on KZ is greater than the square on BX.

+

And, since BA is equal to KA, the square on BA is equal to the square on AK.

+

And the squares on BX, XA are equal to the square on BA, and the squares on KZ, ZA equal to the square on KA; [I. 47] therefore the squares on BX, XA are equal to the squares on KZ, ZA, and of these the square on KZ is greater than the square on BX; therefore the remainder, the square on ZA, is less than the square on XA.

+

Therefore AX is greater than AZ; therefore AX is much greater than AG.

+

And AX is the perpendicular on one base of the polyhedron, and AG on the surface of the lesser sphere; hence the polyhedron will not touch the lesser sphere on its surface.

+

Therefore, given two spheres about the same centre, a polyhedral solid has been inscribed in the greater sphere which does not touch the lesser sphere at its surface. Q. E. F.

+ +

Porism. But if in another sphere also a polyhedral solid be inscribed similar to the solid in the sphere BCDE, the polyhedral solid in the sphere BCDE has to the polyhedral solid in the other sphere the ratio triplicate of that which the diameter of the sphere BCDE has to the diameter of the other sphere.

+

For, the solids being divided into their pyramids similar in multitude and arrangement, the pyramids will be similar.

+

But similar pyramids are to one another in the triplicate ratio of their corresponding sides; [XII. 8, Por.] therefore the pyramid of which the quadrilateral KBPS is the base, and the point A the vertex, has to the similarly arranged pyramid in the other sphere the ratio triplicate of that which the corresponding side has to the corresponding side, that is, of that which the radius AB of the sphere about A as centre has to the radius of the other sphere.

+

Similarly also each pyramid of those in the sphere about A as centre has to each similarly arranged pyramid of those in the other sphere the ratio triplicate of that which AB has to the radius of the other sphere.

+

And, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents; [V. 12] hence the whole polyhedral solid in the sphere about A as centre has to the whole polyhedral solid in the other sphere the ratio triplicate of that which AB has to the radius of the other sphere, that is, of that which the diameter BD has to the diameter of the other sphere. Q. E. D.

+ +
+
+ PROPOSITION 18. +

+ Spheres are to one another in the triplicate ratio of their respective diameters. +

+

Let the spheres ABC, DEF be conceived, and let BC, EF be their diameters; I say that the sphere ABC has to the sphere DEF the ratio triplicate of that which BC has to EF.

+

For, if the sphere ABC has not to the sphere DEF the ratio triplicate of that which BC has to EF, then the sphere ABC will have either to some less sphere than the sphere DEF, or to a greater, the ratio triplicate of that which BC has to EF.

+

First, let it have that ratio to a less sphere GHK, let DEF be conceived about the same centre with GHK, let there be inscribed in the greater sphere DEF a polyhedral solid which does not touch the lesser sphere GHK at its surface, [XII. 17] and let there also be inscribed in the sphere ABC a polyhedral solid similar to the polyhedral solid in the sphere DEF; therefore the polyhedral solid in ABC has to the polyhedral solid in DEF the ratio triplicate of that which BC has to EF. [XII. 17, Por.]

+

+

But the sphere ABC also has to the sphere GHK the ratio triplicate of that which BC has to EF; therefore, as the sphere ABC is to the sphere GHK, so is the polyhedral solid in the sphere ABC to the polyhedral solid in the sphere DEF; and, alternately, as the sphere ABC is to the polyhedron in it, so is the sphere GHK to the polyhedral solid in the sphere DEF. [V. 16]

+

But the sphere ABC is greater than the polyhedron in it; therefore the sphere GHK is also greater than the polyhedron in the sphere DEF.

+

But it is also less, for it is enclosed by it.

+

Therefore the sphere ABC has not to a less sphere than the sphere DEF the ratio triplicate of that which the diameter BC has to EF.

+

Similarly we can prove that neither has the sphere DEF to a less sphere than the sphere ABC the ratio triplicate of that which EF has to BC.

+

I say next that neither has the sphere ABC to any greater sphere than the sphere DEF the ratio triplicate of that which BC has to EF.

+

For, if possible, let it have that ratio to a greater, LMN; therefore, inversely, the sphere LMN has to the sphere ABC the ratio triplicate of that which the diameter EF has to the diameter BC.

+

But, inasmuch as LMN is greater than DEF, therefore, as the sphere LMN is to the sphere ABC, so is the sphere DEF to some less sphere than the sphere ABC, as was before proved. [XII. 2, Lemma]

+

Therefore the sphere DEF also has to some less sphere than the sphere ABC the ratio triplicate of that which EF has to BC: which was proved impossible.

+

Therefore the sphere ABC has not to any sphere greater than the sphere DEF the ratio triplicate of that which BC has to EF.

+

But it was proved that neither has it that ratio to a less sphere.

+

Therefore the sphere ABC has to the sphere DEF the ratio triplicate of that which BC has to EF. Q. E. D.

+
+
+
+
+ + BOOK XIII. + HISTORICAL NOTE. +

I have already given, in the note to IV. 10, the evidence upon which the construction of the five regular solids is attributed to the Pythagoreans. Some of them, the cube, the tetrahedron (which is nothing but a pyramid), and the octahedron (which is only a double pyramid with a square base), cannot but have been known to the Egyptians. And it appears that dodecahedra have been found, of bronze or other material, which may belong to periods earlier than Pythagoras' time by some centuries (for references see Cantor's Geschichte der Mathematik I_{3}, pp. 175-6).

+

It is true that the author of the scholium No. I to Eucl. XIII. says that the Book is about the five so-called Platonic figures, which however do not belong to Plato, three of the aforesaid five figures being due to the Pythagoreans, namely the cube, the pyramid and the dodecahedron, while the octahedron and the icosahedron are due to Theaetetus. + This statement (taken probably from Geminus) may perhaps rest on the fact that Theaetetus was the first to write at any length about the two last-mentioned solids. We are told indeed by Suidas (S. V. Θεαίτητος) that Theaetetus first wrote on the ’five solids’ as they are called. + This no doubt means that Theaetetus was the first to write a complete and systematic treatise on all the regular solids; it does not exclude the possibility that Hippasus or others had already written on the dodecahedron. The fact that Theaetetus wrote upon the regular solids agrees very well with the evidence which we possess of his contributions to the theory of irrationals, the connexion between which and the investigation of the regular solids is seen in Euclid's Book XIII.

+

Theaetetus flourished about 380 B.C., and his work on the regular solids was soon followed by another, that of Aristaeus, an elder contemporary of Euclid, who also wrote an important book on Solid Loci, i.e. on conics treated as loci. This Aristaeus (known as the elder +) wrote in the period about 320 B.C. We hear of his Comparison of the five regular solids from Hypsicles (2nd cent. B.C.), the writer of the short book commonly included in the editions of the Elements as Book XIV. Hypsicles gives in this Book some six propositions supplementing Eucl. XIII.; and he introduces the second of the propositions (Heiberg's Euclid, Vol. v. p. 6) as follows:

+

The same circle circumscribes both the pentagon of the dodecahedron and the triangle of the icosahedron when both are inscribed in the same sphere. This is proved by Aristaeus in the book entitled Comparison of the five figures. +

+

Hypsicles proceeds (pp. 7 sqq.) to give a proof of this theorem. Allman pointed out (Greek Geometry from Thales to Euclid, 1889, pp. 201-2) that this proof depends on eight theorems, six of which appear in Euclid's Book XIII. (in Propositions 8, 10, 12, 15, 16 with Por., 17); two other propositions not mentioned by Allman are also used, namely XIII. 4 and 9. This seems, as Allman says, to confirm the inference of Bretschneider (p. 171) that, as Aristaeus' work was the newest and latest in which, before Euclid's time, this subject was treated, we have in Eucl. XIII. at least a partial recapitulation of the contents of the treatise of Aristaeus.

+

After Euclid, Apollonius wrote on the comparison of the dodecahedron and the icosahedron inscribed in one and the same sphere. This we also learn from Hypsicles, who says in the next words following those about Aristaeus above quoted: But it is proved by Apollonius in the second edition of his Comparison of the dodecahedron with the icosahedron that, as the surface of the dodecahedron is to the surface of the icosahedron [inscribed in the same sphere], so is the dodecahedron itself [i.e. its volume] to the icosahedron, because the perpendicular is the same from the centre of the sphere to the pentagon of the dodecahedron and to the triangle of the icosahedron. +

+
+ BOOK XIII. PROPOSITIONS. +
+ PROPOSITION 1. +

If a straight line be cut in extreme and mean ratio, the square on the greater segment added to the half of the whole is five times the square on the half. +

+

For let the straight line AB be cut in extreme and mean ratio at the point C, and let AC be the greater segment; let the straight line AD be produced in a straight line with CA, and let AD be made half of AB; I say that the square on CD is five times the square on AD.

+

+

For let the squares AE, DF be described on AB, DC, and let the figure in DF be drawn; let FC be carried through to G.

+

Now, since AB has been cut in extreme and mean ratio at C, therefore the rectangle AB, BC is equal to the square on AC. [VI. Def. 3, VI. 17]

+

And CE is the rectangle AB, BC, and FH the square on AC; therefore CE is equal to FH.

+

And, since BA is double of AD, while BA is equal to KA, and AD to AH, therefore KA is also double of AH.

+

But, as KA is to AH, so is CK to CH; [VI. 1] therefore CK is double of CH.

+

But LH, HC are also double of CH.

+

Therefore KC is equal to LH, HC.

+

But CE was also proved equal to HF; therefore the whole square AE is equal to the gnomon MNO.

+

And, since BA is double of AD, the square on BA is quadruple of the square on AD, that is, AE is quadruple of DH.

+

But AE is equal to the gnomon MNO; therefore the gnomon MNO is also quadruple of AP; therefore the whole DF is five times AP.

+

And DF is the square on DC, and AP the square on DA; therefore the square on CD is five times the square on DA.

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 2. +

If the square on a straight line be five times the square on a segment of it, then, when the double of the said segment is cut in extreme and mean ratio, the greater segment is the remaining part of the original straight line. +

+

For let the square on the straight line AB be five times the square on the segment AC of it, and let CD be double of AC; I say that, when CD is cut in extreme and mean ratio, the greater segment is CB.

+

+

Let the squares AF, CG be described on AB, CD respectively, let the figure in AF be drawn, and let BE be drawn through.

+

Now, since the square on BA is five times the square on AC, AF is five times AH.

+

Therefore the gnomon MNO is quadruple of AH.

+

And, since DC is double of CA, therefore the square on DC is quadruple of the square on CA, that is, CG is quadruple of AH.

+

But the gnomon MNO was also proved quadruple of AH; therefore the gnomon MNO is equal to CG.

+

And, since DC is double of CA, while DC is equal to CK, and AC to CH, therefore KB is also double of BH. [VI. 1]

+

But LH, HB are also double of HB; therefore KB is equal to LH, HB.

+

But the whole gnomon MNO was also proved equal to the whole CG; therefore the remainder HF is equal to BG.

+

And BG is the rectangle CD, DB, for CD is equal to DG; and HF is the square on CB; therefore the rectangle CD, DB is equal to the square on CB.

+

Therefore, as DC is to CB, so is CB to BD.

+

But DC is greater than CB; therefore CB is also greater than BD.

+

Therefore, when the straight line CD is cut in extreme and mean ratio, CB is the greater segment.

+

Therefore etc. Q. E. D.

+ +

Lemma. That the double of AC is greater than BC is to be proved thus.

+

If not, let BC be, if possible, double of CA.

+

Therefore the square on BC is quadruple of the square on CA; therefore the squares on BC, CA are five times the square on CA.

+

But, by hypothesis, the square on BA is also five times the square on CA; therefore the square on BA is equal to the squares on BC, CA: which is impossible. [II. 4]

+

Therefore CB is not double of AC.

+

Similarly we can prove that neither is a straight line less than CB double of CA; for the absurdity is much greater.

+

Therefore the double of AC is greater than CB. Q. E. D.

+ +
+
+ PROPOSITION 3. +

If a straight line be cut in extreme and mean ratio, the square on the lesser segment added to the half of the greater segment is five times the square on the half of the greater segment. +

+

For let any straight line AB be cut in extreme and mean ratio at the point C, let AC be the greater segment, and let AC be bisected at D; I say that the square on BD is five times the square on DC.

+

+

For let the square AE be described on AB, and let the figure be drawn double.

+

Since AC is double of DC, therefore the square on AC is quadruple of the square on DC, that is, RS is quadruple of FG.

+

And, since the rectangle AB, BC is equal to the square on AC, and CE is the rectangle AB, BC, therefore CE is equal to RS.

+

But RS is quadruple of FG; therefore CE is also quadruple of FG.

+

Again, since AD is equal to DC, HK is also equal to KF.

+

Hence the square GF is also equal to the square HL.

+

Therefore GK is equal to KL, that is, MN to NE; hence MF is also equal to FE.

+

But MF is equal to CG; therefore CG is also equal to FE.

+

Let CN be added to each; therefore the gnomon OPQ is equal to CE.

+

But CE was proved quadruple of GF; therefore the gnomon OPQ is also quadruple of the square FG.

+

Therefore the gnomon OPQ and the square FG are five times FG.

+

But the gnomon OPQ and the square FG are the square DN.

+

And DN is the square on DB, and GF the square on DC.

+

Therefore the square on DB is five times the square on DC. Q. E. D.

+
+
+ PROPOSITION 4. +

If a straight line be cut in extreme and mean ratio, the square on the whole and the square on the lesser segment together are triple of the square on the greater segment. +

+

Let AB be a straight line, let it be cut in extreme and mean ratio at C, and let AC be the greater segment; I say that the squares on AB, BC are triple of the square on CA.

+

+

For let the square ADEB be described on AB, and let the figure be drawn.

+

Since then AB has been cut in extreme and mean ratio at C, and AC is the greater segment, therefore the rectangle AB, BC is equal to the square on AC. [VI. Def. 3, VI. 17]

+

And AK is the rectangle AB, BC, and HG the square on AC; therefore AK is equal to HG.

+

And, since AF is equal to FE, let CK be added to each; therefore the whole AK is equal to the whole CE; therefore AK, CE are double of AK.

+

But AK, CE are the gnomon LMN and the square CK; therefore the gnomon LMN and the square CK are double of AK.

+

But, further, AK was also proved equal to HG; therefore the gnomon LMN and the squares CK, HG are triple of the square HG.

+

And the gnomon LMN and the squares CK, HG are the whole square AE and CK, which are the squares on AB, BC, while HG is the square on AC.

+

Therefore the squares on AB, BC are triple of the square on AC. Q. E. D.

+
+
+ PROPOSITION 5. +

If a straight line be cut in extreme and mean ratio, and there be added to it a straight line equal to the greater segment, the whole straight line has been cut in extreme and mean ratio, and the original straight line is the greater segment. +

+

For let the straight line AB be cut in extreme and mean ratio at the point C, let AC be the greater segment, and let AD be equal to AC.

+

+

I say that the straight line DB has been cut in extreme and mean ratio at A, and the original straight line AB is the greater segment.

+

For let the square AE be described on AB, and let the figure be drawn.

+

Since AB has been cut in extreme and mean ratio at C, therefore the rectangle AB, BC is equal to the square on AC. [VI. Def. 3, VI. 17]

+

And CE is the rectangle AB, BC, and CH the square on AC; therefore CE is equal to HC.

+

But HE is equal to CE, and DH is equal to HC; therefore DH is also equal to HE.

+

Therefore the whole DK is equal to the whole AE.

+

And DK is the rectangle BD, DA, for AD is equal to DL; and AE is the square on AB; therefore the rectangle BD, DA is equal to the square on AB.

+

Therefore, as DB is to BA, so is BA to AD. [VI. 17]

+

And DB is greater than BA; therefore BA is also greater than AD. [V. 14]

+

Therefore DB has been cut in extreme and mean ratio at A, and AB is the greater segment. Q. E. D.

+
+
+ PROPOSITION 6. +

If a rational straight line be cut in extreme and mean ratio, each of the segments is the irrational straight line called apotome. +

+

Let AB be a rational straight line, let it be cut in extreme and mean ratio at C, and let AC be the greater segment; I say that each of the straight lines AC, CB is the irrational straight line called apotome.

+

+

For let BA be produced, and let AD be made half of BA.

+

Since then the straight line AB has been cut in extreme and mean ratio, and to the greater segment AC is added AD which is half of AB, therefore the square on CD is five times the square on DA. [XIII. 1]

+

Therefore the square on CD has to the square on DA the ratio which a number has to a number; therefore the square on CD is commensurable with the square on DA. [X. 6]

+

But the square on DA is rational, for DA is rational, being half of AB which is rational; therefore the square on CD is also rational; [X. Def. 4] therefore CD is also rational.

+

And, since the square on CD has not to the square on DA the ratio which a square number has to a square number, therefore CD is incommensurable in length with DA; [X. 9] therefore CD, DA are rational straight lines commensurable in square only; therefore AC is an apotome. [X. 73]

+

Again, since AB has been cut in extreme and mean ratio, and AC is the greater segment, therefore the rectangle AB, BC is equal to the square on AC. [VI. Def. 3, VI. 17]

+

Therefore the square on the apotome AC, if applied to the rational straight line AB, produces BC as breadth.

+

But the square on an apotome, if applied to a rational straight line, produces as breadth a first apotome; [X. 97] therefore CB is a first apotome.

+

And CA was also proved to be an apotome.

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 7. +

If three angles of an equilateral pentagon, taken either in order or not in order, be equal, the pentagon will be equiangular. +

+

For in the equilateral pentagon ABCDE let, first, three angles taken in order, those at A, B, C, be equal to one another; I say that the pentagon ABCDE is equiangular.

+

+

For let AC, BE, FD be joined.

+

Now, since the two sides CB, BA are equal to the two sides BA, AE respectively, and the angle CBA is equal to the angle BAE, therefore the base AC is equal to the base BE, the triangle ABC is equal to the triangle ABE, and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend, [I. 4] that is, the angle BCA to the angle BEA, and the angle ABE to the angle CAB; hence the side AF is also equal to the side BF. [I. 6]

+

But the whole AC was also proved equal to the whole BE; therefore the remainder FC is also equal to the remainder FE.

+

But CD is also equal to DE.

+

Therefore the two sides FC, CD are equal to the two sides FE, ED; and the base FD is common to them; therefore the angle FCD is equal to the angle FED. [I. 8]

+

But the angle BCA was also proved equal to the angle AEB; therefore the whole angle BCD is also equal to the whole angle AED.

+

But, by hypothesis, the angle BCD is equal to the angles at A, B; therefore the angle AED is also equal to the angles at A, B.

+

Similarly we can prove that the angle CDE is also equal to the angles at A, B, C; therefore the pentagon ABCDE is equiangular.

+

Next, let the given equal angles not be angles taken in order, but let the angles at the points A, C, D be equal; I say that in this case too the pentagon ABCDE is equiangular.

+

For let BD be joined.

+

Then, since the two sides BA, AE are equal to the two sides BC, CD, and they contain equal angles, therefore the base BE is equal to the base BD, the triangle ABE is equal to the triangle BCD, and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend; [I. 4] therefore the angle AEB is equal to the angle CDB.

+

But the angle BED is also equal to the angle BDE, since the side BE is also equal to the side BD. [I. 5]

+

Therefore the whole angle AED is equal to the whole angle CDE.

+

But the angle CDE is, by hypothesis, equal to the angles at A, C; therefore the angle AED is also equal to the angles at A, C.

+

For the same reason the angle ABC is also equal to the angles at A, C, D.

+

Therefore the pentagon ABCDE is equiangular. Q. E. D.

+
+
+ PROPOSITION 8. +

If in an equilateral and equiangular pentagon straight lines subtend two angles taken in order, they cut one another in extreme and mean ratio, and their greater segments are equal to the side of the pentagon. +

+

For in the equilateral and equiangular pentagon ABCDE let the straight lines AC, BE, cutting one another at the point H, subtend two angles taken in order, the angles at A, B; I say that each of them has been cut in extreme and mean ratio at the point H, and their greater segments are equal to the side of the pentagon.

+

+

For let the circle ABCDE be circumscribed about the pentagon ABCDE. [IV. 14]

+

Then, since the two straight lines EA, AB are equal to the two AB, BC, and they contain equal angles, therefore the base BE is equal to the base AC, the triangle ABE is equal to the triangle ABC, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend. [I. 4]

+

Therefore the angle BAC is equal to the angle ABE; therefore the angle AHE is double of the angle BAH. [I. 32]

+

But the angle EAC is also double of the angle BAC, inasmuch as the circumference EDC is also double of the circumference CB; [III. 28, VI. 33] therefore the angle HAE is equal to the angle AHE; hence the straight line HE is also equal to EA, that is, to AB. [I. 6]

+

And, since the straight line BA is equal to AE, the angle ABE is also equal to the angle AEB. [I. 5]

+

But the angle ABE was proved equal to the angle BAH; therefore the angle BEA is also equal to the angle BAH.

+

And the angle ABE is common to the two triangles ABE and ABH; therefore the remaining angle BAE is equal to the remaining angle AHB; [I. 32] therefore the triangle ABE is equiangular with the triangle ABH; therefore, proportionally, as EB is to BA, so is AB to BH. [VI. 4]

+

But BA is equal to EH; therefore, as BE is to EH, so is EH to HB.

+

And BE is greater than EH; therefore EH is also greater than HB. [V. 14]

+

Therefore BE has been cut in extreme and mean ratio at H, and the greater segment HE is equal to the side of the pentagon.

+

Similarly we can prove that AC has also been cut in extreme and mean ratio at H, and its greater segment CH is equal to the side of the pentagon. Q. E. D.

+
+
+ PROPOSITION 9. +

If the side of the hexagon and that of the decagon inscribed in the same circle be added together, the whole straight line has been cut in extreme and mean ratio, and its greater segment is the side of the hexagon. +

+

Let ABC be a circle; of the figures inscribed in the circle ABC let BC be the side of a decagon, CD that of a hexagon, and let them be in a straight line; I say that the whole straight line BD has been cut in extreme and mean ratio, and CD is its greater segment.

+

+

For let the centre of the circle, the point E, be taken, let EB, EC, ED be joined, and let BE be carried through to A.

+

Since BC is the side of an equilateral decagon, therefore the circumference ACB is five times the circumference BC; therefore the circumference AC is quadruple of CB.

+

But, as the circumference AC is to CB, so is the angle AEC to the angle CEB; [VI. 33] therefore the angle AEC is quadruple of the angle CEB.

+

And, since the angle EBC is equal to the angle ECB, [I. 5] therefore the angle AEC is double of the angle ECB. [I. 32]

+

And, since the straight line EC is equal to CD, for each of them is equal to the side of the hexagon inscribed in the circle ABC, [IV. 15, Por.] the angle CED is also equal to the angle CDE; [I. 5] therefore the angle ECB is double of the angle EDC. [I. 32]

+

But the angle AEC was proved double of the angle ECB; therefore the angle AEC is quadruple of the angle EDC.

+

But the angle AEC was also proved quadruple of the angle BEC; therefore the angle EDC is equal to the angle BEC.

+

But the angle EBD is common to the two triangles BEC and BED; therefore the remaining angle BED is also equal to the remaining angle ECB; [I. 32] therefore the triangle EBD is equiangular with the triangle EBC.

+

Therefore, proportionally, as DB is to BE, so is EB to BC. [VI. 4]

+

But EB is equal to CD.

+

Therefore, as BD is to DC, so is DC to CB.

+

And BD is greater than DC; therefore DC is also greater than CB.

+

Therefore the straight line BD has been cut in extreme and mean ratio, and DC is its greater segment. Q. E. D.

+
+
+ PROPOSITION 10. +

If an equilateral pentagon be inscribed in a circle, the square on the side of the pentagon is equal to the squares on the side of the hexagon and on that of the decagon inscribed in the same circle. +

+

Let ABCDE be a circle, and let the equilateral pentagon ABCDE be inscribed in the circle ABCDE.

+

I say that the square on the side of the pentagon ABCDE is equal to the squares on the side of the hexagon and on that of the decagon inscribed in the circle ABCDE.

+

For let the centre of the circle, the point F, be taken, let AF be joined and carried through to the point G, let FB be joined, let FH be drawn from F perpendicular to AB and be carried through to K, let AK, KB be joined, let FL be again drawn from F perpendicular to AK, and be carried through to M, and let KN be joined.

+

Since the circumference ABCG is equal to the circumference AEDG, and in them ABC is equal to AED, therefore the remainder, the circumference CG, is equal to the remainder GD.

+

+

But CD belongs to a pentagon; therefore CG belongs to a decagon.

+

And, since FA is equal to FB, and FH is perpendicular, therefore the angle AFK is also equal to the angle KFB. [I. 5, I. 26]

+

Hence the circumference AK is also equal to KB; [III. 26] therefore the circumference AB is double of the circumference BK; therefore the straight line AK is a side of a decagon.

+

For the same reason AK is also double of KM.

+

Now, since the circumference AB is double of the circumference BK, while the circumference CD is equal to the circumference AB, therefore the circumference CD is also double of the circumference BK.

+

But the circumference CD is also double of CG; therefore the circumference CG is equal to the circumference BK.

+

But BK is double of KM, since KA is so also; therefore CG is also double of KM.

+

But, further, the circumference CB is also double of the circumference BK, for the circumference CB is equal to BA.

+

Therefore the whole circumference GB is also double of BM; hence the angle GFB is also double of the angle BFM. [VI. 33]

+

But the angle GFB is also double of the angle FAB, for the angle FAB is equal to the angle ABF.

+

Therefore the angle BFN is also equal to the angle FAB.

+

But the angle ABF is common to the two triangles ABF and BFN; therefore the remaining angle AFB is equal to the remaining angle BNF; [I. 32] therefore the triangle ABF is equiangular with the triangle BFN.

+

Therefore, proportionally, as the straight line AB is to BF, so is FB to BN; [VI. 4] therefore the rectangle AB, BN is equal to the square on BF. [VI. 17]

+

Again, since AL is equal to LK, while LN is common and at right angles, therefore the base KN is equal to the base AN; [I. 4] therefore the angle LKN is also equal to the angle LAN.

+

But the angle LAN is equal to the angle KBN; therefore the angle LKN is also equal to the angle KBN.

+

And the angle at A is common to the two triangles AKB and AKN.

+

Therefore the remaining angle AKB is equal to the remaining angle KNA; [I. 32] therefore the triangle KBA is equiangular with the triangle KNA.

+

Therefore, proportionally, as the straight line BA is to AK, so is KA to AN; [VI. 4] therefore the rectangle BA, AN is equal to the square on AK. [VI. 17]

+

But the rectangle AB, BN was also proved equal to the square on BF; therefore the rectangle AB, BN together with the rectangle BA, AN, that is, the square on BA [II. 2], is equal to the square on BF together with the square on AK.

+

And BA is a side of the pentagon, BF of the hexagon [IV. 15, Por.], and AK of the decagon.

+

Therefore etc. Q. E. D.

+
+
+ PROPOSITION 11. +

If in a circle which has its diameter rational an equilateral pentagon be inscribed, the side of the pentagon is the irrational straight line called minor. +

+

For in the circle ABCDE which has its diameter rational let the equilateral pentagon ABCDE be inscribed; I say that the side of the pentagon is the irrational straight line called minor.

+

For let the centre of the circle, the point F, be taken, let AF, FB be joined and carried through to the points, G, H, let AC be joined, and let FK be made a fourth part of AF.

+

Now AF is rational; therefore FK is also rational.

+

But BF is also rational; therefore the whole BK is rational.

+

And, since the circumference ACG is equal to the circumference ADG, and in them ABC is equal to AED, therefore the remainder CG is equal to the remainder GD.

+

And, if we join AD, we conclude that the angles at L are right, and CD is double of CL.

+

For the same reason the angles at M are also right, and AC is double of CM.

+

+

Since then the angle ALC is equal to the angle AMF, and the angle LAC is common to the two triangles ACL and AMF, therefore the remaining angle ACL is equal to the remaining angle MFA; [I. 32] therefore the triangle ACL is equiangular with the triangle AMF; therefore, proportionally, as LC is to CA, so is MF to FA.

+

And the doubles of the antecedents may be taken; therefore, as the double of LC is to CA, so is the double of MF to FA.

+

But, as the double of MF is to FA, so is MF to the half of FA; therefore also, as the double of LC is to CA, so is MF to the half of FA.

+

And the halves of the consequents may be taken; therefore, as the double of LC is to the half of CA, so is MF to the fourth of FA.

+

And DC is double of LC, CM is half of CA, and FK a fourth part of FA; therefore, as DC is to CM, so is MF to FK.

+

Componendo also, as the sum of DC, CM is to CM, so is MK to KF; [V. 18] therefore also, as the square on the sum of DC, CM is to the square on CM, so is the square on MK to the square on KF.

+

And since, when the straight line subtending two sides of the pentagon, as AC, is cut in extreme and mean ratio, the greater segment is equal to the side of the pentagon, that is, to DC, [XIII. 8] while the square on the greater segment added to the half of the whole is five times the square on the half of the whole, [XIII. 1] and CM is half of the whole AC, therefore the square on DC, CM taken as one straight line is five times the square on CM.

+

But it was proved that, as the square on DC, CM taken as one straight line is to the square on CM, so is the square on MK to the square on KF; therefore the square on MK is five times the square on KF.

+

But the square on KF is rational, for the diameter is rational; therefore the square on MK is also rational; therefore MK is rational

+

And, since BF is quadruple of FK, therefore BK is five times KF; therefore the square on BK is twenty-five times the square on KF.

+

But the square on MK is five times the square on KF; therefore the square on BK is five times the square on KM; therefore the square on BK has not to the square on KM the ratio which a square number has to a square number; therefore BK is incommensurable in length with KM. [X. 9]

+

And each of them is rational.

+

Therefore BK, KM are rational straight lines commensurable in square only.

+

But, if from a rational straight line there be subtracted a rational straight line which is commensurable with the whole in square only, the remainder is irrational, namely an apotome; therefore MB is an apotome and MK the annex to it. [X. 73]

+

I say next that MB is also a fourth apotome.

+

Let the square on N be equal to that by which the square on BK is greater than the square on KM; therefore the square on BK is greater than the square on KM by the square on N.

+

And, since KF is commensurable with FB, componendo also, KB is commensurable with FB. [X. 15]

+

But BF is commensurable with BH; therefore BK is also commensurable with BH. [X. 12]

+

And, since the square on BK is five times the square on KM, therefore the square on BK has to the square on KM the ratio which 5 has to 1.

+

Therefore, convertendo, the square on BK has to the square on N the ratio which 5 has to 4 [V. 19, Por.], and this is not the ratio which a square number has to a square number; therefore BK is incommensurable with N; [X. 9] therefore the square on BK is greater than the square on KM by the square on a straight line incommensurable with BK.

+

Since then the square on the whole BK is greater than the square on the annex KM by the square on a straight line incommensurable with BK, and the whole BK is commensurable with the rational straight line, BH, set out, therefore MB is a fourth apotome. [X. Deff. III. 4]

+

But the rectangle contained by a rational straight line and a fourth apotome is irrational, and its square root is irrational, and is called minor. [X. 94]

+

But the square on AB is equal to the rectangle HB, BM, because, when AH is joined, the triangle ABH is equiangular with the triangle ABM, and, as HB is to BA, so is AB to BM.

+

Therefore the side AB of the pentagon is the irrational straight line called minor. Q. E. D.

+
+
+ PROPOSITION 12. +

If an equilateral triangle be inscribed in a circle, the square on the side of the triangle is triple of the square on the radius of the circle. +

+

Let ABC be a circle, and let the equilateral triangle ABC be inscribed in it; I say that the square on one side of the triangle ABC is triple of the square on the radius of the circle.

+

+

For let the centre D of the circle ABC be taken, let AD be joined and carried through to E, and let BE be joined.

+

Then, since the triangle ABC is equilateral, therefore the circumference BEC is a third part of the circumference of the circle ABC.

+

Therefore the circumference BE is a sixth part of the circumference of the circle; therefore the straight line BE belongs to a hexagon; therefore it is equal to the radius DE. [IV. 15, Por.]

+

And, since AE is double of DE, the square on AE is quadruple of the square on ED, that is, of the square on BE.

+

But the square on AE is equal to the squares on AB, BE; [III. 31, I. 47] therefore the squares on AB, BE are quadruple of the square on BE.

+

Therefore, separando, the square on AB is triple of the square on BE.

+

But BE is equal to DE; therefore the square on AB is triple of the square on DE.

+

Therefore the square on the side of the triangle is triple of the square on the radius. Q. E. D.

+
+
+ PROPOSITION 13. +

To construct a pyramid, to comprehend it in a given sphere, and to prove that the square on the diameter of the sphere is one and a half times the square on the side of the pyramid. +

+

Let the diameter AB of the given sphere be set out, and let it be cut at the point C so that AC is double of CB; let the semicircle ADB be described on AB, let CD be drawn from the point C at right angles to AB, and let DA be joined; let the circle EFG which has its radius equal to DC be set out, let the equilateral triangle EFG be inscribed in the circle EFG, [IV. 2] let the centre of the circle, the point H, be taken, [III. 1] let EH, HF, HG be joined; from the point H let HK be set up at right angles to the plane of the circle EFG, [XI. 12] let HK equal to the straight line AC be cut off from HK, and let KE, KF, KG be joined.

+

+

Now, since KH is at right angles to the plane of the circle EFG, therefore it will also make right angles with all the straight lines which meet it and are in the plane of the circle EFG. [XI. Def. 3]

+

But each of the straight lines HE, HF, HG meets it: therefore HK is at right angles to each of the straight lines HE, HF, HG.

+

And, since AC is equal to HK, and CD to HE, and they contain right angles, therefore the base DA is equal to the base KE. [I. 4]

+

For the same reason each of the straight lines KF, KG is also equal to DA; therefore the three straight lines KE, KF, KG are equal to one another.

+

And, since AC is double of CB, therefore AB is triple of BC.

+

But, as AB is to BC, so is the square on AD to the square on DC, as will be proved afterwards.

+

Therefore the square on AD is triple of the square on DC.

+

But the square on FE is also triple of the square on EH, [XIII. 12] and DC is equal to EH; therefore DA is also equal to EF.

+

But DA was proved equal to each of the straight lines KE, KF, KG; therefore each of the straight lines EF, FG, GE is also equal to each of the straight lines KE, KF, KG; therefore the four triangles EFG, KEF, KFG, KEG are equilateral.

+

Therefore a pyramid has been constructed out of four equilateral triangles, the triangle EFG being its base and the point K its vertex.

+

It is next required to comprehend it in the given sphere and to prove that the square on the diameter of the sphere is one and a half times the square on the side of the pyramid.

+

For let the straight line HL be produced in a straight line with KH, and let HL be made equal to CB.

+

Now, since, as AC is to CD, so is CD to CB, [VI. 8, Por.] while AC is equal to KH, CD to HE, and CB to HL, therefore, as KH is to HE, so is EH to HL; therefore the rectangle KH, HL is equal to the square on EH. [VI. 17]

+

And each of the angles KHE. EHL is right; therefore the semicircle described on KL will pass through E also. [cf. VI. 8, III. 31.]

+

If then, KL remaining fixed, the semicircle be carried round and restored to the same position from which it began to be moved, it will also pass through the points F, G, since, if FL, LG be joined, the angles at F, G similarly become right angles; and the pyramid will be comprehended in the given sphere.

+

For KL, the diameter of the sphere, is equal to the diameter AB of the given sphere, inasmuch as KH was made equal to AC, and HL to CB.

+

I say next that the square on the diameter of the sphere is one and a half times the square on the side of the pyramid

+

For, since AC is double of CB, therefore AB is triple of BC; and, convertendo, BA is one and a half times AC.

+

But, as BA is to AC, so is the square on BA to the square on AD.

+

Therefore the square on BA is also one and a half times the square on AD.

+

And BA is the diameter of the given sphere, and AD is equal to the side of the pyramid.

+

Therefore the square on the diameter of the sphere is one and a half times the square on the side of the pyramid. Q. E. D.

+ +

Lemma. It is to be proved that, as AB is to BC, so is the square on AD to the square on DC.

+

For let the figure of the semicircle be set out, let DB be joined, let the square EC be described on AC, and let the parallelogram FB be completed.

+

+

Since then, because the triangle DAB is equiangular with the triangle DAC, as BA is to AD, so is DA to AC, [VI. 8, VI. 4] therefore the rectangle BA, AC is equal to the square on AD. [VI. 17]

+

And since, as AB is to BC, so is EB to BF, [VI. 1] and EB is the rectangle BA, AC, for EA is equal to AC, and BF is the rectangle AC, CB, therefore, as AB is to BC, so is the rectangle BA, AC to the rectangle AC, CB.

+

And the rectangle BA, AC is equal to the square on AD, and the rectangle AC, CB to the square on DC, for the perpendicular DC is a mean proportional between the segments AC, CB of the base, because the angle ADB is right. [VI. 8, Por.]

+

Therefore, as AB is to BC, so is the square on AD to the square on DC. Q. E. D.

+ +
+ +
+ PROPOSITION 14. +

To construct an octahedron and comprehend it in a sphere, as in the preceding case; and to prove that the square on the diameter of the sphere is double of the square on the side of the octahedron. +

+

Let the diameter AB of the given sphere be set out, and let it be bisected at C; let the semicircle ADB be described on AB, let CD be drawn from C at right angles to AB, let DB be joined; let the square EFGH, having each of its sides equal to DB, be set out, let HF, EG be joined, from the point K let the straight line KL be set up at right angles to the plane of the square EFGH [XI. 12], and let it be carried through to the other side of the plane, as KM; from the straight lines KL, KM let KL, KM be respectively cut off equal to one of the straight lines EK, FK, GK, HK, and let LE, LF, LG, LH, ME, MF, MG, MH be joined.

+

+

Then, since KE is equal to KH, and the angle EKH is right, therefore the square on HE is double of the square on EK. [I. 47]

+

Again, since LK is equal to KE, and the angle LKE is right, therefore the square on EL is double of the square on EK. [id.]

+

But the square on HE was also proved double of the square on EK; therefore the square on LE is equal to the square on EH; therefore LE is equal to EH.

+

For the same reason LH is also equal to HE; therefore the triangle LEH is equilateral.

+

Similarly we can prove that each of the remaining triangles of which the sides of the square EFGH are the bases, and the points L, M the vertices, is equilateral; therefore an octahedron has been constructed which is contained by eight equilateral triangles.

+

It is next required to comprehend it in the given sphere, and to prove that the square on the diameter of the sphere is double of the square on the side of the octahedron.

+

For, since the three straight lines LK, KM, KE are equal to one another, therefore the semicircle described on LM will also pass through E.

+

And for the same reason, if, LM remaining fixed, the semicircle be carried round and restored to the same position from which it began to be moved, it will also pass through the points F, G, H, and the octahedron will have been comprehended in a sphere.

+

I say next that it is also comprehended in the given sphere.

+

For, since LK is equal to KM, while KE is common, and they contain right angles, therefore the base LE is equal to the base EM. [I. 4]

+

And, since the angle LEM is right, for it is in a semicircle, [III. 31] therefore the square on LM is double of the square on LE. [I. 47]

+

Again, since AC is equal to CB, AB is double of BC.

+

But, as AB is to BC, so is the square on AB to the square on BD; therefore the square on AB is double of the square on BD.

+

But the square on LM was also proved double of the square on LE.

+

And the square on DB is equal to the square on LE, for EH was made equal to DB.

+

Therefore the square on AB is also equal to the square on LM; therefore AB is equal to LM.

+

And AB is the diameter of the given sphere; therefore LM is equal to the diameter of the given sphere.

+

Therefore the octahedron has been comprehended in the given sphere, and it has been demonstrated at the same time that the square on the diameter of the sphere is double of the square on the side of the octahedron. Q. E. D.

+
+ +
+ PROPOSITION 15. +

To construct a cube and comprehend it in a sphere, like the pyramid; and to prove that the square on the diameter of the sphere is triple of the square on the side of the cube. +

+

Let the diameter AB of the given sphere be set out, and let it be cut at C so that AC is double of CB; let the semicircle ADB be described on AB, let CD be drawn from C at right angles to AB, and let DB be joined; let the square EFGH having its side equal to DB be set out, from E, F, G, H let EK, FL, GM, HN be drawn at right angles to the plane of the square EFGH, from EK, FL, GM, HN let EK, FL, GM, HN respectively be cut off equal to one of the straight lines EF, FG, GH, HE, and let KL, LM, MN, NK be joined; therefore the cube FN has been constructed which is contained by six equal squares.

+

+

It is then required to comprehend it in the given sphere, and to prove that the square on the diameter of the sphere is triple of the square on the side of the cube.

+

For let KG, EG be joined.

+

Then, since the angle KEG is right, because KE is also at right angles to the plane EG and of course to the straight line EG also, [XI. Def. 3] therefore the semicircle described on KG will also pass through the point E.

+

Again, since GF is at right angles to each of the straight lines FL, FE, GF is also at right angles to the plane FK; hence also, if we join FK, GF will be at right angles to FK; and for this reason again the semicircle described on GK will also pass through F.

+

Similarly it will also pass through the remaining angular points of the cube.

+

If then, KG remaining fixed, the semicircle be carried round and restored to the same position from which it began to be moved, the cube will be comprehended in a sphere.

+

I say next that it is also comprehended in the given sphere.

+

For, since GF is equal to FE, and the angle at F is right, therefore the square on EG is double of the square on EF.

+

But EF is equal to EK; therefore the square on EG is double of the square on EK; hence the squares on GE, EK, that is the square on GK [I. 47], is triple of the square on EK.

+

And, since AB is triple of BC, while, as AB is to BC, so is the square on AB to the square on BD, therefore the square on AB is triple of the square on BD.

+

But the square on GK was also proved triple of the square on KE.

+

And KE was made equal to DB; therefore KG is also equal to AB.

+

And AB is the diameter of the given sphere; therefore KG is also equal to the diameter of the given sphere.

+

Therefore the cube has been comprehended in the given sphere; and it has been demonstrated at the same time that the square on the diameter of the sphere is triple of the square on the side of the cube. Q. E. D.

+
+ +
+ PROPOSITION 16. +

To construct an icosahedron and comprehend it in a sphere, like the aforesaid figures; and to prove that the side of the icosahedron is the irrational straight line called minor. +

+

Let the diameter AB of the given sphere be set out, and let it be cut at C so that AC is quadruple of CB, let the semicircle ADB be described on AB, let the straight line CD be drawn from C at right angles to AB, and let DB be joined;

let the circle EFGHK be set out and let its radius be equal to DB, let the equilateral and equiangular pentagon EFGHK be inscribed in the circle EFGHK, let the circumferences EF, FG, GH, HK, KE be bisected at the points L, M, N, O, P, and let LM, MN, NO, OP, PL, EP be joined.

+

Therefore the pentagon LMNOP is also equilateral, and the straight line EP belongs to a decagon.

+

Now from the points E, F, G, H, K let the straight lines EQ, FR, GS, HT, KU be set up at right angles to the plane of the circle, and let them be equal to the radius of the circle EFGHK, let QR, RS, ST, TU, UQ, QL, LR, RM, MS, SN, NT, TO, OU, UP, PQ be joined.

+

Now, since each of the straight lines EQ, KU is at right angles to the same plane, therefore EQ is parallel to KU. [XI. 6]

+

But it is also equal to it; and the straight lines joining those extremities of equal and parallel straight lines which are in the same direction are equal and parallel. [I. 33]

+

Therefore QU is equal and parallel to EK.

+

But EK belongs to an equilateral pentagon; therefore QU also belongs to the equilateral pentagon inscribed in the circle EFGHK.

+

For the same reason each of the straight lines QR, RS, ST, TU also belongs to the equilateral pentagon inscribed in the circle EFGHK; therefore the pentagon QRSTU is equilateral.

+

And, since QE belongs to a hexagon, and EP to a decagon, and the angle QEP is right, therefore QP belongs to a pentagon; for the square on the side of the pentagon is equal to the square on the side of the hexagon and the square on the side of the decagon inscribed in the same circle. [XIII. 10]

+

For the same reason PU is also a side of a pentagon.

+

But QU also belongs to a pentagon; therefore the triangle QPU is equilateral.

+

For the same reason each of the triangles QLR, RMS, SNT, TOU is also equilateral.

+

And, since each of the straight lines QL, QP was proved to belong to a pentagon, and LP also belongs to a pentagon, therefore the triangle QLP is equilateral.

+

For the same reason each of the triangles LRM, MSN, NTO, OUP is also equilateral.

+

Let the centre of the circle EFGHK the point V, be taken; from V let VZ be set up at right angles to the plane of the circle, let it be produced in the other direction, as VX, let there be cut off VW, the side of a hexagon, and each of the straight lines VX, WZ, being sides of a decagon, and let QZ, QW, UZ, EV, LV, LX, XM be joined.

+

Now, since each of the straight lines VW, QE is at right angles to the plane of the circle, therefore VW is parallel to QE. [XI. 6]

+

But they are also equal; therefore EV, QW are also equal and parallel. [I. 33]

+

But EV belongs to a hexagon; therefore QW also belongs to a hexagon.

+

And, since QW belongs to a hexagon, and WZ to a decagon, and the angle QWZ is right, therefore QZ belongs to a pentagon. [XIII. 10]

+

For the same reason UZ also belongs to a pentagon, inasmuch as, if we join VK, WU, they will be equal and opposite, and VK, being a radius, belongs to a hexagon; [IV. 15, Por.] therefore WU also belongs to a hexagon.

+

But WZ belongs to a decagon, and the angle UWZ is right; therefore UZ belongs to a pentagon. [XIII. 10]

+

But QU also belongs to a pentagon; therefore the triangle QUZ is equilateral.

+

For the same reason each of the remaining triangles of which the straight lines QR, RS, ST, TU are the bases, and the point Z the vertex, is also equilateral.

+

Again, since VL belongs to a hexagon, and VX to a decagon, and the angle LVX is right, therefore LX belongs to a pentagon. [XIII. 10]

+

For the same reason, if we join MV, which belongs to a hexagon, MX is also inferred to belong to a pentagon.

+

But LM also belongs to a pentagon; therefore the triangle LMX is equilateral.

+

Similarly it can be proved that each of the remaining triangles of which MN, NO, OP, PL are the bases, and the point X the vertex, is also equilateral.

+

Therefore an icosahedron has been constructed which is contained by twenty equilateral triangles.

+

It is next required to comprehend it in the given sphere, and to prove that the side of the icosahedron is the irrational straight line called minor.

+

For, since VW belongs to a hexagon, and WZ to a decagon, therefore VZ has been cut in extreme and mean ratio at W, and VW is its greater segment; [XIII. 9] therefore, as ZV is to VW, so is VW to WZ.

+

But VW is equal to VE, and WZ to VX; therefore, as ZV is to VE, so is EV to VX.

+

And the angles ZVE, EVX are right; therefore, if we join the straight line EZ, the angle XEZ will be right because of the similarity of the triangles XEZ, VEZ.

+

For the same reason, since, as ZV is to VW, so is VW to WZ, and ZV is equal to XW, and VW to WQ, therefore, as XW is to WQ, so is QW to WZ.

+

And for this reason again, if we join QX, the angle at Q will be right; [VI. 8] therefore the semicircle described on XZ will also pass through Q. [III. 31]

+

And if, XZ remaining fixed, the semicircle be carried round and restored to the same position from which it began to be moved, it will also pass through Q and the remaining angular points of the icosahedron, and the icosahedron will have been comprehended in a sphere.

+

I say next that it is also comprehended in the given sphere.

+

For let VW be bisected at A'.

+

Then, since the straight line VZ has been cut in extreme and mean ratio at W, and ZW is its lesser segment, therefore the square on ZW added to the half of the greater segment, that is WA', is five times the square on the half of the greater segment; [XIII. 3] therefore the square on ZA' is five times the square on .

+

And ZX is double of ZA', and VW double of ; therefore the square on ZX is five times the square on WV.

+

And, since AC is quadruple of CB, therefore AB is five times BC.

+

But, as AB is to BC, so is the square on AB to the square on BD; [VI. 8, V. Def. 9] therefore the square on AB is five times the square on BD.

+

But the square on ZX was also proved to be five times the square on VW.

+

And DB is equal to VW, for each of them is equal to the radius of the circle EFGHK; therefore AB is also equal to XZ.

+

And AB is the diameter of the given sphere; therefore XZ is also equal to the diameter of the given sphere.

+

Therefore the icosahedron has been comprehended in the given sphere

+

I say next that the side of the icosahedron is the irrational straight line called minor.

+

For, since the diameter of the sphere is rational, and the square on it is five times the square on the radius of the circle EFGHK, therefore the radius of the circle EFGHK is also rational; hence its diameter is also rational.

+

But, if an equilateral pentagon be inscribed in a circle which has its diameter rational, the side of the pentagon is the irrational straight line called minor. [XIII. 11]

+

And the side of the pentagon EFGHK is the side of the icosahedron.

+

Therefore the side of the icosahedron is the irrational straight line called minor.

+ +

Porism. From this it is manifest that the square on the diameter of the sphere is five times the square on the radius of the circle from which the icosahedron has been described, and that the diameter of the sphere is composed of the side of the hexagon and two of the sides of the decagon inscribed in the same circle. Q. E. D.

+ +
+ +
+ PROPOSITION 17. +

To construct a dodecahedron and comprehend it in a sphere, like the aforesaid figures, and to prove that the side of the dodecahedron is the irrational straight line called apotome. +

+

Let ABCD, CBEF, two planes of the aforesaid cube at right angles to one another, be set out, let the sides AB, BC, CD, DA, EF, EB, FC be bisected at G, H, K, L, M, N, O respectively, let GK, HL, MH, NO be joined, let the straight lines NP, PO, HQ be cut in extreme and mean ratio at the points R, S, T respectively, and let RP, PS, TQ be their greater segments; from the points R, S, T let RU, SV, TW be set up at right angles to the planes of the cube towards the outside of the cube, let them be made equal to RP, PS, TQ, and let UB, BW, WC, CV, VU be joined.

+

+

I say that the pentagon UBWCV is equilateral, and in one plane, and is further equiangular.

+

For let RB, SB, VB be joined.

+

Then, since the straight line NP has been cut in extreme and mean ratio at R, and RP is the greater segment, therefore the squares on PN, NR are triple of the square on RP. [XIII. 4]

+

But PN is equal to NB, and PR to RU; therefore the squares on BN, NR are triple of the square on RU.

+

But the square on BR is equal to the squares on BN, NR; [I. 47] therefore the square on BR is triple of the square on RU; hence the squares on BR, RU are quadruple of the square on RU.

+

But the square on BU is equal to the squares on BR, RU; therefore the square on BU is quadruple of the square on RU; therefore BU is double of RU.

+

But VU is also double of UR, inasmuch as SR is also double of PR, that is, of RU; therefore BU is equal to UV.

+

Similarly it can be proved that each of the straight lines BW, WC, CV is also equal to each of the straight lines BU, UV.

+

Therefore the pentagon BUVCW is equilateral.

+

I say next that it is also in one plane.

+

For let PX be drawn from P parallel to each of the straight lines RU, SV and towards the outside of the cube, and let XH, HW be joined; I say that XHW is a straight line.

+

For, since HQ has been cut in extreme and mean ratio at T, and QT is its greater segment, therefore, as HQ is to QT, so is QT to TH.

+

But HQ is equal to HP, and QT to each of the straight lines TW, PX; therefore, as HP is to PX, so is WT to TH.

+

And HP is parallel to TW, for each of them is at right angles to the plane BD; [XI. 6] and TH is parallel to PX, for each of them is at right angles to the plane BF. [id.]

+

But if two triangles, as XPH, HTW, which have two sides proportional to two sides be placed together at one angle so that their corresponding sides are also parallel, the remaining straight lines will be in a straight line; [VI. 32] therefore XH is in a straight line with HW.

+

But every straight line is in one plane; [XI. 1] therefore the pentagon UBWCV is in one plane.

+

I say next that it is also equiangular.

+

For, since the straight line NP has been cut in extreme and mean ratio at R, and PR is the greater segment, while PR is equal to PS, therefore NS has also been cut in extreme and mean ratio at P, and NP is the greater segment; [XIII. 5] therefore the squares on NS, SP are triple of the square on NP. [XIII. 4]

+

But NP is equal to NB, and PS to SV; therefore the squares on NS, SV are triple of the square on NB; hence the squares on VS, SN, NB are quadruple of the square on NB.

+

But the square on SB is equal to the squares on SN, NB; therefore the squares on BS, SV, that is, the square on BV —for the angle VSB is right—is quadruple of the square on NB; therefore VB is double of BN.

+

But BC is also double of BN; therefore BV is equal to BC.

+

And, since the two sides BU, UV are equal to the two sides BW, WC, and the base BV is equal to the base BC, therefore the angle BUV is equal to the angle BWC. [I. 8]

+

Similarly we can prove that the angle UVC is also equal to the angle BWC; therefore the three angles BWC, BUV, UVC are equal to one another.

+

But if in an equilateral pentagon three angles are equal to one another, the pentagon will be equiangular, [XIII. 7] therefore the pentagon BUVCW is equiangular.

+

And it was also proved equilateral; therefore the pentagon BUVCW is equilateral and equiangular, and it is on one side BC of the cube.

+

Therefore, if we make the same construction in the case of each of the twelve sides of the cube, a solid figure will have been constructed which is contained by twelve equilateral and equiangular pentagons, and which is called a dodecahedron.

+

It is then required to comprehend it in the given sphere, and to prove that the side of the dodecahedron is the irrational straight line called apotome.

+

For let XP be produced, and let the produced straight line be XZ; therefore PZ meets the diameter of the cube, and they bisect one another, for this has been proved in the last theorem but one of the eleventh book. [XI. 38]

+

Let them cut at Z; therefore Z is the centre of the sphere which comprehends the cube, and ZP is half of the side of the cube.

+

Let UZ be joined.

+

Now, since the straight line NS has been cut in extreme and mean ratio at P, and NP is its greater segment, therefore the squares on NS, SP are triple of the square on NP. [XIII. 4]

+

But NS is equal to XZ, inasmuch as NP is also equal to PZ, and XP to PS.

+

But further PS is also equal to XU, since it is also equal to RP; therefore the squares on ZX, XU are triple of the square on NP.

+

But the square on UZ is equal to the squares on ZX, XU; therefore the square on UZ is triple of the square on NP.

+

But the square on the radius of the sphere which comprehends the cube is also triple of the square on the half of the side of the cube, for it has previously been shown how to construct a cube and comprehend it in a sphere, and to prove that the square on the diameter of the sphere is triple of the square on the side of the cube. [XIII. 15]

+

But, if whole is so related to whole, so is half to half also; and NP is half of the side of the cube; therefore UZ is equal to the radius of the sphere which comprehends the cube.

+

And Z is the centre of the sphere which comprehends the cube; therefore the point U is on the surface of the sphere.

+

Similarly we can prove that each of the remaining angles of the dodecahedron is also on the surface of the sphere; therefore the dodecahedron has been comprehended in the given sphere.

+

I say next that the side of the dodecahedron is the irrational straight line called apotome.

+

For since, when NP has been cut in extreme and mean ratio, RP is the greater segment, and, when PO has been cut in extreme and mean ratio, PS is the greater segment, therefore, when the whole NO is cut in extreme and mean ratio, RS is the greater segment.

+

[Thus, since, as NP is to PR, so is PR to RN, the same is true of the doubles also, for parts have the same ratio as their equimultiples; [V. 15] therefore as NO is to RS, so is RS to the sum of NR, SO.

+

But NO is greater than RS; therefore RS is also greater than the sum of NR, SO; therefore NO has been cut in extreme and mean ratio, and RS is its greater segment.]

+

But RS is equal to UV; therefore, when NO is cut in extreme and mean ratio, UV is the greater segment.

+

And, since the diameter of the sphere is rational, and the square on it is triple of the square on the side of the cube, therefore NO, being a side of the cube, is rational.

+

[But if a rational line be cut in extreme and mean ratio, each of the segments is an irrational apotome.]

+

Therefore UV, being a side of the dodecahedron, is an irrational apotome. [XIII. 6]

+ +

Porism. From this it is manifest that, when the side of the cube is cut in extreme and mean ratio, the greater segment is the side of the dodecahedron. Q. E. D.

+ +
+
+ PROPOSITION 18. +

+ To set out the sides of the five figures and to compare them with one another. +

+

Let AB, the diameter of the given sphere, be set out, and let it be cut at C so that AC is equal to CB, and at D so that AD is double of DB; let the semicircle AEB be described on AB, from C, D let CE, DF be drawn at right angles to AB, and let AF, FB, EB be joined.

+

Then, since AD is double of DB, therefore AB is triple of BD.

+

+

Convertendo, therefore, BA is one and a half times AD.

+

But, as BA is to AD, so is the square on BA to the square on AF, [V. Def. 9, VI. 8] for the triangle AFB is equiangular with the triangle AFD; therefore the square on BA is one and a half times the square on AF.

+

But the square on the diameter of the sphere is also one and a half times the square on the side of the pyramid. [XIII. 13]

+

And AB is the diameter of the sphere; therefore AF is equal to the side of the pyramid.

+

Again, since AD is double of DB, therefore AB is triple of BD.

+

But, as AB is to BD, so is the square on AB to the square on BF; [VI. 8, V. Def. 9] therefore the square on AB is triple of the square on BF.

+

But the square on the diameter of the sphere is also triple of the square on the side of the cube. [XIII. 15]

+

And AB is the diameter of the sphere; therefore BF is the side of the cube.

+

And, since AC is equal to CB, therefore AB is double of BC.

+

But, as AB is to BC, so is the square on AB to the square on BE; therefore the square on AB is double of the square on BE.

+

But the square on the diameter of the sphere is also double of the square on the side of the octahedron. [XIII. 14]

+

And AB is the diameter of the given sphere; therefore BE is the side of the octahedron.

+

Next, let AG be drawn from the point A at right angles to the straight line AB, let AG be made equal to AB, let GC be joined, and from H let HK be drawn perpendicular to AB.

+

Then, since GA is double of AC, for GA is equal to AB, and, as GA is to AC, so is HK to KC, therefore HK is also double of KC.

+

Therefore the square on HK is quadruple of the square on KC; therefore the squares on HK, KC, that is, the square on HC, is five times the square on KC.

+

But HC is equal to CB; therefore the square on BC is five times the square on CK.

+

And, since AB is double of CB, and, in them, AD is double of DB, therefore the remainder BD is double of the remainder DC.

+

Therefore BC is triple of CD; therefore the square on BC is nine times the square on CD.

+

But the square on BC is five times the square on CK; therefore the square on CK is greater than the square on CD; therefore CK is greater than CD.

+

Let CL be made equal to CK, from L let LM be drawn at right angles to AB, and let MB be joined.

+

Now, since the square on BC is five times the square on CK, and AB is double of BC, and KL double of CK, therefore the square on AB is five times the square on KL.

+

But the square on the diameter of the sphere is also five times the square on the radius of the circle from which the icosahedron has been described. [XIII. 16, Por.]

+

And AB is the diameter of the sphere; therefore KL is the radius of the circle from which the icosahedron has been described; therefore KL is a side of the hexagon in the said circle. [IV. 15, Por.]

+

And, since the diameter of the sphere is made up of the side of the hexagon and two of the sides of the decagon inscribed in the same circle, [XIII. 16, Por.] and AB is the diameter of the sphere, while KL is a side of the hexagon, and AK is equal to LB, therefore each of the straight lines AK, LB is a side of the decagon inscribed in the circle from which the icosahedron has been described.

+

And, since LB belongs to a decagon, and ML to a hexagon, for ML is equal to KL, since it is also equal to HK, being the same distance from the centre, and each of the straight lines HK, KL is double of KC, therefore MB belongs to a pentagon. [XIII. 10]

+

But the side of the pentagon is the side of the icosahedron; [XIII. 16] therefore MB belongs to the icosahedron.

+

Now, since FB is a side of the cube, let it be cut in extreme and mean ratio at N, and let NB be the greater segment; therefore NB is a side of the dodecahedron. [XIII. 17, Por.]

+

And, since the square on the diameter of the sphere was proved to be one and a half times the square on the side AF of the pyramid, double of the square on the side BE of the octahedron and triple of the side FB of the cube, therefore, of parts of which the square on the diameter of the sphere contains six, the square on the side of the pyramid contains four, the square on the side of the octahedron three, and the square on the side of the cube two.

+

Therefore the square on the side of the pyramid is fourthirds of the square on the side of the octahedron, and double of the square on the side of the cube; and the square on the side of the octahedron is one and a half times the square on the side of the cube.

+

The said sides, therefore, of the three figures, I mean the pyramid, the octahedron and the cube, are to one another in rational ratios.

+

But the remaining two, I mean the side of the icosahedron and the side of the dodecahedron, are not in rational ratios either to one another or to the aforesaid sides; for they are irrational, the one being minor [XIII. 16] and the other an apotome [XIII. 17].

+

That the side MB of the icosahedron is greater than the side NB of the dodecahedron we can prove thus.

+

For, since the triangle FDB is equiangular with the triangle FAB, [VI. 8] proportionally, as DB is to BF, so is BF to BA. [VI. 4]

+

And, since the three straight lines are proportional, as the first is to the third, so is the square on the first to the square on the second; [V. Def. 9, VI. 20, Por.] therefore, as DB is to BA, so is the square on DB to the square on BF; therefore, inversely, as AB is to BD, so is the square on FB to the square on BD.

+

But AB is triple of BD; therefore the square on FB is triple of the square on BD.

+

But the square on AD is also quadruple of the square on DB, for AD is double of DB; therefore the square on AD is greater than the square on FB; therefore AD is greater than FB; therefore AL is by far greater than FB.

+

And, when AL is cut in extreme and mean ratio, KL is the greater segment, inasmuch as LK belongs to a hexagon, and KA to a decagon; [XIII. 9] and, when FB is cut in extreme and mean ratio, NB is the greater segment; therefore KL is greater than NB.

+

But KL is equal to LM; therefore LM is greater than NB.

+

Therefore MB, which is a side of the icosahedron, is by far greater than NB which is a side of the dodecahedron. Q. E. D.

+

I say next that no other figure, besides the said five figures, can be constructed which is contained by equilateral and equiangular figures equal to one another. +

+

For a solid angle cannot be constructed with two triangles, or indeed planes.

+

With three triangles the angle of the pyramid is constructed, with four the angle of the octahedron, and with five the angle of the icosahedron; but a solid angle cannot be formed by six equilateral and equiangular triangles placed together at one point, for, the angle of the equilateral triangle being two-thirds of a right angle, the six will be equal to four right angles: which is impossible, for any solid angle is contained by angles less than four right angles. [XI. 21]

+

For the same reason, neither can a solid angle be constructed by more than six plane angles.

+

By three squares the angle of the cube is contained, but by four it is impossible for a solid angle to be contained, for they will again be four right angles.

+

By three equilateral and equiangular pentagons the angle of the dodecahedron is contained; but by four such it is impossible for any solid angle to be contained, for, the angle of the equilateral pentagon being a right angle and a fifth, the four angles will be greater than four right angles: which is impossible.

+

Neither again will a solid angle be contained by other polygonal figures by reason of the same absurdity.

+

Therefore etc. Q. E. D.

+ +

Lemma. But that the angle of the equilateral and equiangular pentagon is a right angle and a fifth we must prove thus.

+

Let ABCDE be an equilateral and equiangular pentagon, let the circle ABCDE be circumscribed about it, let its centre F be taken, and let FA, FB, FC, FD, FE be joined.

+

Therefore they bisect the angles of the pentagon at A, B, C, D, E.

+

+

And, since the angles at F are equal to four right angles and are equal, therefore one of them, as the angle AFB, is one right angle less a fifth; therefore the remaining angles FAB, ABF consist of one right angle and a fifth.

+

But the angle FAB is equal to the angle FBC; therefore the whole angle ABC of the pentagon consists of one right angle and a fifth. Q. E. D.

+ +
+
+
+
+ + + diff --git a/data/tlg1799/tlg001/tlg1799.tlg001.perseus-grc2.xml b/data/tlg1799/tlg001/tlg1799.tlg001.perseus-grc2.xml index 77772b168..83ec819a4 100644 --- a/data/tlg1799/tlg001/tlg1799.tlg001.perseus-grc2.xml +++ b/data/tlg1799/tlg001/tlg1799.tlg001.perseus-grc2.xml @@ -1,36 +1,37 @@ - + -Elementa + Στοιχεῖα Euclid -J. L. Heiberg + Johan Ludvig Heiberg Perseus Project, Tufts University Gregory Crane Prepared under the supervision of Lisa Cerrato -William Merrill -Elli Mylonas David Smith The National Science Foundation -Trustees of Tufts University -Medford, MA -Perseus Project - + Trustees of Tufts University + Medford, MA + Perseus Digital Library Project + Perseus 3.0 + tlg1799.tlg001.perseus-grc2.xml + + Available under a Creative Commons Attribution-ShareAlike 4.0 International License + Euclid Euclidis Opera Omnia - Euclidis Elementa -J. L. Heiberg + Euclidis Elementa + Johan Ludvig Heiberg Leipzig Teubner @@ -38,11 +39,11 @@ schematypens="http://relaxng.org/ns/structure/1.0"?> 1-5 - http://www.archive.org/stream/euclidisoperaomn01eucluoft - http://www.archive.org/stream/euclidisoperaomn02eucluoft - https://hdl.handle.net/2027/coo.31924101918617 - https://hdl.handle.net/2027/coo.31924102028861 - https://hdl.handle.net/2027/coo.31924102028929 + Internet Archive + Internet Archive + HathiTrust + HathiTrust + HathiTrust @@ -55,19 +56,19 @@ schematypens="http://relaxng.org/ns/structure/1.0"?> --> -

This pointer pattern extracts book and type and number

+ matchPattern="(\w+).(\w+).(\w+)" + replacementPattern="#xpath(/tei:TEI/tei:text/tei:body/tei:div/tei:div[@n='$1']/tei:div[@n='$2']/tei:div[@n='$3'])"> +

This pointer pattern extracts book and type and number.

 -

This pointer pattern extracts book and type

+ replacementPattern="#xpath(/tei:TEI/tei:text/tei:body/tei:div/tei:div[@n='$1']/tei:div[@n='$2'])"> +

This pointer pattern extracts book and type.

 -

This pointer pattern extracts book

+ replacementPattern="#xpath(/tei:TEI/tei:text/tei:body/tei:div/tei:div[@n='$1'])"> +

This pointer pattern extracts book.

 @@ -77,77 +78,34 @@ schematypens="http://relaxng.org/ns/structure/1.0"?> - - - + Epidoc and CTS conversion and header review for best practices + Epidoc and CTS conversion + adding rend=displayNum to lb tags + base sdl file + moved more xml files around based on copyright status + fixed some more entities and chunking + lots of cleanup to chunking schemes, entities, head tags, etc + removed stray item tags + Converted to XML + missing div for 7.20; added without checking source + Made DIV types explicit on all elements, since the inheritance behavior didn't survive XML conversion. + changed TEI header to reflect correct edition + Updated texts to TEI P4 and Perseus P4 extensions; minor cleanup (esp. character encodings and typos.) + + fixed numeration problem at book 7 definitions + Added separate funder entity to TEI header. + Type all DIVs properly. Write refsDecl that works for whole document. + Added refsDecl for definitions, etc. + Tagged points and line segments as NUMs. + New header - + +
-
+

σημεῖόν ἐστιν, οὗ μέρος οὐθέν.

@@ -253,7 +211,7 @@ New header. μέρη ἐπὶ μηδέτερα συμπίπτουσιν ἀλλήλαις.

-
+

Ἠιτήσθω ἀπὸ παντὸς σημείου ἐπὶ πᾶν σημεῖον εὐθεῖαν γραμμὴν ἀγαγεῖν.

@@ -275,7 +233,7 @@ New header. ἐφʼ ἃ μέρη εἰσὶν αἱ τῶν δύο ὀρθῶν ἐλάσσονες.

-
+

τὰ τῷ αὐτῷ ἴσα καὶ ἀλλήλοις ἐστὶν ἴσα.

@@ -311,7 +269,7 @@ New header.

καὶ δύο εὐθεῖαι χωρίον οὐ περιέχουσιν.

-
+

ἐπὶ τῆς δοθείσης εὐθείας πεπερασμένης τρίγωνον ἰσόπλευρον συστήσασθαι.

@@ -1790,7 +1748,7 @@ New header.
-
+

πᾶν παραλληλόγραμμον ὀρθογώνιον περιέχεσθαι λέγεται ὑπὸ δύο τῶν τὴν ὀρθὴν γωνίαν περιεχουσῶν @@ -1802,7 +1760,7 @@ New header. σὺν τοῖς δυσὶ παραπληρώμασι γνώμων καλείσθω.

-
+

ἐὰν ὦσι δύο εὐθεῖαι, τμηθῇ δὲ ἡ ἑτέρα αὐτῶν εἰς ὁσαδηποτοῦν τμήματα, τὸ περιεχόμενον ὀρθογώνιον ὑπὸ τῶν @@ -2420,7 +2378,7 @@ New header.

-
+

ἴσοι κύκλοι εἰσίν, ὧν αἱ διάμετροι ἴσαι εἰσίν, ἢ ὧν αἱ ἐκ τῶν κέντρων ἴσαι εἰσίν.

@@ -2473,7 +2431,7 @@ New header. ἴσας, ἢ ἐν οἷς αἱ γωνίαι ἴσαι ἀλλήλαις εἰσίν.

-
+

τοῦ δοθέντος κύκλου τὸ κέντρον εὑρεῖν.

ἔστω ὁ δοθεὶς κύκλος ὁ ΑΒΓ· δεῖ δὴ τοῦ ΑΒΓ κύκλου @@ -3869,7 +3827,7 @@ New header.

-
+

σχῆμα εὐθύγραμμον εἰς σχῆμα εὐθύγραμμον ἐγγράφεσθαι λέγεται, ὅταν ἑκάστη τῶν τοῦ ἐγγραφομένου @@ -3906,7 +3864,7 @@ New header. πέρατα αὐτῆς ἐπὶ τῆς περιφερείας ᾖ τοῦ κύκλου.

-
+

εἰς τὸν δοθέντα κύκλον τῇ δοθείσῃ εὐθείᾳ μὴ μείζονι οὔσῃ τῆς τοῦ κύκλου διαμέτρου ἴσην εὐθεῖαν ἐναρμόσαι.

@@ -4555,7 +4513,7 @@ New header.
-
+

μέρος ἐστὶ μέγεθος μεγέθους τὸ ἔλασσον τοῦ μείζονος, ὅταν καταμετρῇ τὸ μεῖζον.

@@ -4648,7 +4606,7 @@ New header. ἄλλο τι, οὕτως ἐν τοῖς δευτέροις ἄλλο τι πρὸς ἡγούμενον.

-
+

ἐὰν ᾖ ὁποσαοῦν μεγέθη ὁποσωνοῦν μεγεθῶν ἴσων τὸ πλῆθος ἕκαστον ἑκάστου ἰσάκις πολλαπλάσιον, ὁσαπλάσιόν @@ -5546,7 +5504,7 @@ New header.

-
+

ὅμοια σχήματα εὐθύγραμμά ἐστιν, ὅσα τάς τε γωνίας ἴσας ἔχει κατὰ μίαν καὶ τὰς περὶ τὰς ἴσας γωνίας @@ -5575,7 +5533,7 @@ New header.

-
+

τὰ τρίγωνα καὶ τὰ παραλληλόγραμμα, τὰ ὑπὸ τὸ αὐτὸ ὕψος ὄντα πρὸς ἄλληλά ἐστιν ὡς αἱ βάσεις.

@@ -7097,7 +7055,7 @@ New header.
-
+

μονάς ἐστιν, καθʼ ἣν ἕκαστον τῶν ὄντων ἓν λέγεται.

@@ -7189,7 +7147,7 @@ New header. ἴσος ὤν.

-
+

δύο ἀριθμῶν ἀνίσων ἐκκειμένων, ἀνθυφαιρουμένου δὲ ἀεὶ τοῦ ἐλάσσονος ἀπὸ τοῦ μείζονος, ἐὰν ὁ λειπόμενος @@ -8249,7 +8207,7 @@ New header.

-
+

ἐὰν ὦσιν ὁσοιδηποτοῦν ἀριθμοὶ ἑξῆς ἀνάλογον, οἱ δὲ ἄκροι αὐτῶν πρῶτοι πρὸς ἀλλήλους ὦσιν, ἐλάχιστοί εἰσι @@ -9149,7 +9107,7 @@ New header.

-
+

ἐὰν δύο ὅμοιοι ἐπίπεδοι ἀριθμοὶ πολλαπλασιάσαντες ἀλλήλους ποιῶσί τινα, ὁ γενόμενος τετράγωνος ἔσται.

@@ -10118,7 +10076,7 @@ New header.
-
+

σύμμετρα μεγέθη λέγεται τὰ τῷ αὐτῷ μέτρῳ μετρούμενα, ἀσύμμετρα δέ, ὧν μηδὲν ἐνδέχεται κοινὸν @@ -10146,7 +10104,7 @@ New header. εὐθύγραμμα, αἱ ἴσα αὐτοῖς τετράγωνα ἀναγράφουσαι.

-
+

δύο μεγεθῶν ἀνίσων ἐκκειμένων, ἐὰν ἀπὸ τοῦ μείζονος ἀφαιρεθῇ μεῖζον ἢ τὸ ἥμισυ καὶ τοῦ καταλειπομένου @@ -11945,7 +11903,7 @@ New header. διαιρεῖται· καθʼ ἓν ἄρα μόνον σημεῖον διαιρεῖται.

-
+

ὑποκειμένης ῥητῆς καὶ τῆς ἐκ δύο ὀνομάτων διῃρημένης εἰς τὰ ὀνόματα, ἧς τὸ μεῖζον ὄνομα τοῦ @@ -11975,7 +11933,7 @@ New header.

ἐὰν δὲ μηδέτερον, ἕκτη.

-
+

εὑρεῖν τὴν ἐκ δύο ὀνομάτων πρώτην.

Ἐκκείσθωσαν δύο ἀριθμοὶ οἱ ΑΓ, ΓΒ, ὥστε τὸν @@ -13549,7 +13507,7 @@ New header. δὶς ὑπʼ αὐτῶν· ὅπερ ἔδει δεῖξαι.

-
+

ὑποκειμένης ῥητῆς καὶ ἀποτομῆς, ἐὰν μὲν ἡ ὅλη τῆς προσαρμοζούσης μεῖζον δύνηται τῷ ἀπὸ συμμέτρου @@ -13580,7 +13538,7 @@ New header.

ἐὰν δὲ μηδετέρα, ἕκτη.

-
+

εὑρεῖν τὴν πρώτην ἀποτομήν.

Ἐκκείσθω ῥητὴ ἡ Α, καὶ τῇ Α μήκει σύμμετρος ἔστω @@ -15207,7 +15165,7 @@ New header.

-
+

στερεόν ἐστι τὸ μῆκος καὶ πλάτος καὶ βάθος ἔχον.

@@ -15340,7 +15298,7 @@ New header. πενταγώνων ἴσων καὶ ἰσοπλεύρων καὶ ἰσογωνίων περιεχόμενον.

-
+

εὐθείας γραμμῆς μέρος μέν τι οὐκ ἔστιν ἐν τῷ ὑποκειμένῳ, ἐπιπέδῳ, μέρος δέ τι ἐν μετεωροτέρῳ.

@@ -17009,7 +16967,7 @@ New header.
-
+

τὰ ἐν τοῖς κύκλοις ὅμοια πολύγωνα πρὸς ἄλληλά ἐστιν ὡς τὰ ἀπὸ τῶν διαμέτρων τετράγωνα.

@@ -18434,7 +18392,7 @@ New header.
-
+

ἐὰν εὐθεῖα γραμμὴ ἄκρον καὶ μέσον λόγον τμηθῇ, τὸ μεῖζον τμῆμα προσλαβὸν τὴν ἡμίσειαν τῆς ὅλης πενταπλάσιον diff --git a/manifest.txt b/manifest.txt index e2286012c..a5a7df259 100644 --- a/manifest.txt +++ b/manifest.txt @@ -2464,6 +2464,7 @@ /data/tlg1622/tlg001/tlg1622.tlg001.perseus-eng1.xml /data/tlg1799/__cts__.xml /data/tlg1799/tlg001/__cts__.xml +/data/tlg1799/tlg001/tlg1799.tlg001.perseus-eng2.xml /data/tlg1799/tlg001/tlg1799.tlg001.perseus-grc2.xml /data/tlg2003/__cts__.xml /data/tlg2003/tlg001/__cts__.xml