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. [] 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; [, 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; [] 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.
36. ἐὰν τρεῖς εὐθεῖαι ἀνάλογον ὦσιν, τὸ ἐκ τῶν τριῶν στερεὸν παραλληλεπίπεδον ἴσον ἐστὶ τῷ ἀπὸ τῆς μέσης στερεῷ παραλληλεπιπέδῳ ἰσοπλεύρῳ μέν, ἰσογωνίῳ δὲ τῷ προειρημένῳ. ἔστωσαν τρεῖς εὐθεῖαι ἀνάλογον αἱ Α, Β, Γ, ὡς ἡ Α πρὸς τὴν Β, οὕτως ἡ Β πρὸς τὴν Γ· λέγω, ὅτι τὸ ἐκ τῶν Α, Β, Γ στερεὸν ἴσον ἐστὶ τῷ ἀπὸ τῆς Β στερεῷ ἰσοπλεύρῳ μέν, ἰσογωνίῳ δὲ τῷ προειρημένῳ. Ἐκκείσθω στερεὰ γωνία ἡ πρὸς τῷ Ε περιεχομένη ὑπὸ τῶν ὑπὸ ΔΕΗ, ΗΕΖ, ΖΕΔ, καὶ κείσθω τῇ μὲν Β ἴση ἑκάστη τῶν ΔΕ, ΗΕ, ΕΖ, καὶ συμπεπληρώσθω τὸ ΕΚ στερεὸν παραλληλεπίπεδον, τῇ δὲ Α ἴση ἡ ΛΜ, καὶ συνεστάτω πρὸς τῇ ΛΜ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Λ τῇ πρὸς τῷ Ε στερεᾷ γωνίᾳ ἴση στερεὰ γωνία ἡ περιεχομένη ὑπὸ τῶν ΝΛΞ, ΞΛΜ, ΜΛΝ, καὶ κείσθω τῇ μὲν Β ἴση ἡ ΛΞ, τῇ δὲ Γ ἴση ἡ ΛΝ. καὶ ἐπεί ἐστιν ὡς ἡ Α πρὸς τὴν Β, οὕτως ἡ Β πρὸς τὴν Γ, ἴση δὲ ἡ μὲν Α τῇ ΛΜ, ἡ δὲ Β ἑκατέρᾳ τῶν ΛΞ, ΕΔ, ἡ δὲ Γ τῇ ΛΝ, ἔστιν ἄρα ὡς ἡ ΛΜ πρὸς τὴν ΕΖ, οὕτως ἡ ΔΕ πρὸς τὴν ΛΝ. καὶ περὶ ἴσας γωνίας τὰς ὑπὸ ΝΛΜ, ΔΕΖ αἱ πλευραὶ ἀντιπεπόνθασιν· ἴσον ἄρα ἐστὶ τὸ ΜΝ παραλληλόγραμμον τῷ ΔΖ παραλληλογράμμῳ. καὶ ἐπεὶ δύο γωνίαι ἐπίπεδοι εὐθύγραμμοι ἴσαι εἰσὶν αἱ ὑπὸ ΔΕΖ, ΝΛΜ, καὶ ἐπʼ αὐτῶν μετέωροι εὐθεῖαι ἐφεστᾶσιν αἱ ΛΞ, ΕΗ ἴσαι τε ἀλλήλαις καὶ ἴσας γωνίας περιέχουσαι μετὰ τῶν ἐξ ἀρχῆς εὐθειῶν ἑκατέραν ἑκατέρᾳ, αἱ ἄρα ἀπὸ τῶν Η, Ξ σημείων κάθετοι ἀγόμεναι ἐπὶ τὰ διὰ τῶν ΝΛΜ, ΔΕΖ ἐπίπεδα ἴσαι ἀλλήλαις εἰσίν· ὥστε τὰ ΛΘ, ΕΚ στερεὰ ὑπὸ τὸ αὐτὸ ὕψος ἐστίν. τὰ δὲ ἐπὶ ἴσων βάσεων στερεὰ παραλληλεπίπεδα καὶ ὑπὸ τὸ αὐτὸ ὕψος ἴσα ἀλλήλοις ἐστίν· ἴσον ἄρα ἐστὶ τὸ ΘΛ στερεὸν τῷ ΕΚ στερεῷ. καί ἐστι τὸ μὲν ΛΘ τὸ ἐκ τῶν Α, Β, Γ στερεόν, τὸ δὲ ΕΚ τὸ ἀπὸ τῆς Β στερεόν· τὸ ἄρα ἐκ τῶν Α, Β, Γ στερεὸν παραλληλεπίπεδον ἴσον ἐστὶ τῷ ἀπὸ τῆς Β στερεῷ ἰσοπλεύρῳ μέν, ἰσογωνίῳ δὲ τῷ προειρημένῳ· ὅπερ ἔδει δεῖξαι.
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. [] 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, [] 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.
37. ἐὰν τέσσαρες εὐθεῖαι ἀνάλογον ὦσιν, καὶ τὰ ἀπʼ αὐτῶν στερεὰ παραλληλεπίπεδα ὅμοιά τε καὶ ὁμοίως ἀναγραφόμενα ἀνάλογον ἔσται· καὶ ἐὰν τὰ ἀπʼ αὐτῶν στερεὰ παραλληλεπίπεδα ὅμοιά τε καὶ ὁμοίως ἀναγραφόμενα ἀνάλογον ᾖ, καὶ αὐταὶ αἱ εὐθεῖαι ἀνάλογον ἔσονται. ἔστωσαν τέσσαρες εὐθεῖαι ἀνάλογον αἱ ΑΒ, ΓΔ, ΕΖ, ΗΘ, ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΕΖ πρὸς τὴν ΗΘ, καὶ ἀναγεγράφθωσαν ἀπὸ τῶν ΑΒ, ΓΔ, ΕΖ, ΗΘ ὅμοιά τε καὶ ὁμοίως κείμενα στερεὰ παραλληλεπίπεδα τὰ ΚΑ, ΛΓ, ΜΕ, ΝΗ· λέγω, ὅτι ἐστὶν ὡς τὸ ΚΑ πρὸς τὸ ΛΓ, οὕτως τὸ ΜΕ πρὸς τὸ ΝΗ. ἐπεὶ γὰρ ὅμοιόν ἐστι τὸ ΚΑ στερεὸν παραλληλεπίπεδον τῷ ΛΓ, τὸ ΚΑ ἄρα πρὸς τὸ ΛΓ τριπλασίονα λόγον ἔχει ἤπερ ἡ ΑΒ πρὸς τὴν ΓΔ. διὰ τὰ αὐτὰ δὴ καὶ τὸ ΜΕ πρὸς τὸ ΝΗ τριπλασίονα λόγον ἔχει ἤπερ ἡ ΕΖ πρὸς τὴν ΗΘ. καί ἐστιν ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΕΖ πρὸς τὴν ΗΘ. καὶ ὡς ἄρα τὸ ΑΚ πρὸς τὸ ΛΓ, οὕτως τὸ ΜΕ πρὸς τὸ ΝΗ. ἀλλὰ δὴ ἔστω ὡς τὸ ΑΚ στερεὸν πρὸς τὸ ΛΓ στερεόν, οὕτως τὸ ΜΕ στερεὸν πρὸς τὸ ΝΗ· λέγω, ὅτι ἐστὶν ὡς ἡ ΑΒ εὐθεῖα πρὸς τὴν ΓΔ, οὕτως ἡ ΕΖ πρὸς τὴν ΗΘ. ἐπεὶ γὰρ πάλιν τὸ ΚΑ πρὸς τὸ ΛΓ τριπλασίονα λόγον ἔχει ἤπερ ἡ ΑΒ πρὸς τὴν ΓΔ, ἔχει δὲ καὶ τὸ ΜΕ πρὸς τὸ ΝΗ τριπλασίονα λόγον ἤπερ ἡ ΕΖ πρὸς τὴν ΗΘ, καί ἐστιν ὡς τὸ ΚΑ πρὸς τὸ ΛΓ, οὕτως τὸ ΜΕ πρὸς τὸ ΝΗ, καὶ ὡς ἄρα ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΕΖ πρὸς τὴν ΗΘ. ἐὰν ἄρα τέσσαρες εὐθεῖαι ἀνάλογον ὦσι καὶ τὰ ἑξῆς τῆς προτάσεως· ὅπερ ἔδει δεῖξαι.
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. [] 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; [] therefore the angle OUD is equal to the angle PUE. For this reason DUE is a straight line. [] 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. [] And the straight lines DE, BG join their extremities; therefore DE is parallel to BG. [] Therefore the angle EDT is equal to the angle BGT, for they are alternate; [] and the angle DTU is equal to the angle GTS. [] 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. [] Therefore DT is equal to TG, and UT to TS. Therefore etc. Q. E. D.
38. ἐὰν κύβου τῶν ἀπεναντίον ἐπιπέδων αἱ πλευραὶ δίχα τμηθῶσιν, διὰ δὲ τῶν τομῶν ἐπίπεδα ἐκβληθῇ, ἡ κοινὴ τομὴ τῶν ἐπιπέδων καὶ ἡ τοῦ κύβου διάμετρος δίχα τέμνουσιν ἀλλήλας. κύβου γὰρ τοῦ ΑΖ τῶν ἀπεναντίον ἐπιπέδων τῶν ΓΖ, ΑΘ αἱ πλευραὶ δίχα τετμήσθωσαν κατὰ τὰ Κ, Λ, Μ, Ν, Ξ, Π, Ο, Ρ σημεῖα, διὰ δὲ τῶν τομῶν ἐπίπεδα ἐκβεβλήσθω τὰ ΚΝ, ΞΡ, κοινὴ δὲ τομὴ τῶν ἐπιπέδων ἔστω ἡ ΥΣ, τοῦ δὲ ΑΖ κύβου διαγώνιος ἡ ΔΗ. λέγω, ὅτι ἴση ἐστὶν ἡ μὲν ΥΤ τῇ ΤΣ, ἡ δὲ ΔΤ τῇ ΤΗ. ἐπεζεύχθωσαν γὰρ αἱ ΔΥ, ΥΕ, ΒΣ, ΣΗ. καὶ ἐπεὶ παράλληλός ἐστιν ἡ ΔΞ τῇ ΟΕ, αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΔΞΥ, ΥΟΕ ἴσαι ἀλλήλαις εἰσίν. καὶ ἐπεὶ ἴση ἐστὶν ἡ μὲν ΔΞ τῇ ΟΕ, ἡ δὲ ΞΥ τῇ ΥΟ, καὶ γωνίας ἴσας περιέχουσιν, βάσις ἄρα ἡ ΔΥ τῇ ΥΕ ἐστιν ἴση, καὶ τὸ ΔΞΥ τρίγωνον τῷ ΟΥΕ τριγώνῳ ἐστὶν ἴσον καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι· ἴση ἄρα ἡ ὑπὸ ΞΥΔ γωνία τῇ ὑπὸ ΟΥΕ γωνίᾳ. διὰ δὴ τοῦτο εὐθεῖά ἐστιν ἡ ΔΥΕ. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΒΣΗ εὐθεῖά ἐστιν, καὶ ἴση ἡ ΒΣ τῇ ΣΗ. καὶ ἐπεὶ ἡ ΓΑ τῇ ΔΒ ἴση ἐστὶ καὶ παράλληλος, ἀλλὰ ἡ ΓΑ καὶ τῇ ΕΗ ἴση τέ ἐστι καὶ παράλληλος, καὶ ἡ ΔΒ ἄρα τῇ ΕΗ ἴση τέ ἐστι καὶ παράλληλος. καὶ ἐπιζευγνύουσιν αὐτὰς εὐθεῖαι αἱ ΔΕ, ΒΗ· παράλληλος ἄρα ἐστὶν ἡ ΔΕ τῇ ΒΗ. ἴση ἄρα ἡ μὲν ὑπὸ ΕΔΤ γωνία τῇ ὑπὸ ΒΗΤ· ἐναλλὰξ γάρ· ἡ δὲ ὑπὸ ΔΤΥ τῇ ὑπὸ ΗΤΣ. δύο δὴ τρίγωνά ἐστι τὰ ΔΤΥ, ΗΤΣ τὰς δύο γωνίας ταῖς δυσὶ γωνίαις ἴσας ἔχοντα καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην τὴν ὑποτείνουσαν ὑπὸ μίαν τῶν ἴσων γωνιῶν τὴν ΔΥ τῇ ΗΣ· ἡμίσειαι γάρ εἰσι τῶν ΔΕ, ΒΗ· καὶ τὰς λοιπὰς πλευρὰς ταῖς λοιπαῖς πλευραῖς ἴσας ἕξει. ἴση ἄρα ἡ μὲν ΔΤ τῇ ΤΗ, ἡ δὲ ΥΤ τῇ ΤΣ. ἐὰν ἄρα κύβου τῶν ἀπεναντίον ἐπιπέδων αἱ πλευραὶ δίχα τμηθῶσιν, διὰ δὲ τῶν τομῶν ἐπίπεδα ἐκβληθῇ, ἡ κοινὴ τομὴ τῶν ἐπιπέδων καὶ ἡ τοῦ κύβου διάμετρος δίχα τέμνουσιν ἀλλήλας· ὅπερ ἔδει δεῖξαι.
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.
39. ἐὰν ᾖ δύο πρίσματα ἰσοϋψῆ, καὶ τὸ μὲν ἔχῃ βάσιν παραλληλόγραμμον, τὸ δὲ τρίγωνον, διπλάσιον δὲ ᾖ τὸ παραλληλόγραμμον τοῦ τριγώνου, ἴσα ἔσται τὰ πρίσματα.
Since the parallelogram AF is double of the triangle GHK, while the parallelogram HK is also double of the triangle GHK, [] therefore the parallelogram AF is equal to the parallelogram HK.
ἔστω δύο πρίσματα ἰσοϋψῆ τὰ ΑΒΓΔΕΖ, ΗΘΚΛ ΜΝ, καὶ τὸ μὲν ἐχέτω βάσιν τὸ ΑΖ παραλληλόγραμμον, τὸ δὲ τὸ ΗΘΚ τρίγωνον, διπλάσιον δὲ ἔστω τὸ ΑΖ παραλληλόγραμμον τοῦ ΗΘΚ τριγώνου·
But parallelepipedal solids which are on equal bases and of the same height are equal to one another; [] 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; [] therefore the prism ABCDEF is equal to the prism GHKLMN.
λέγω, ὅτι ἴσον ἐστὶ τὸ ΑΒΓΔΕΖ πρίσμα τῷ ΗΘΚΛΜΝ πρίσματι. συμπεπληρώσθω γὰρ τὰ ΑΞ, ΗΟ στερεά.
Therefore etc. Q. E. D. The predominant feature of Book XII. is the use of the method of exhaustion, which is applied in Propositions,,, and (in a slightly different form) in. 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. ], 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. ]. 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 thelemmaassumed by him and known as theAxiom of Archimedes(see note on 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. ], and that spheres are to one another in the triplicate ratio of their diameters [Eucl. ], and further that every pyramid is one third part of the prism which has the same base with the pyramid and equal height [Eucl. ]; also, that every cone is one third part of the cylinder which has the same base with the cone and equal height [Eucl. ] 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, thatearlier geometersproved these two theorems by means of the lemma known as theAxiom of Archimedesand 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 ( and, and ) is the theorem of Eucl.. As regards the connexion between the twolemmassee note on.
ἐπεὶ διπλάσιόν ἐστι τὸ ΑΖ παραλληλόγραμμον τοῦ ΗΘΚ τριγώνου, ἔστι δὲ καὶ τὸ ΘΚ παραλληλόγραμμον διπλάσιον τοῦ ΗΘΚ τριγώνου, ἴσον ἄρα ἐστὶ τὸ ΑΖ παραλληλόγραμμον τῷ ΘΚ παραλληλογράμμῳ. τὰ δὲ ἐπὶ ἴσων βάσεων ὄντα στερεὰ παραλληλεπίπεδα καὶ ὑπὸ τὸ αὐτὸ ὕψος ἴσα ἀλλήλοις ἐστίν· ἴσον ἄρα ἐστὶ τὸ ΑΞ στερεὸν τῷ ΗΟ στερεῷ. καί ἐστι τοῦ μὲν ΑΞ στερεοῦ ἥμισυ τὸ ΑΒΓΔΕΖ πρίσμα, τοῦ δὲ ΗΟ στερεοῦ ἥμισυ τὸ ΗΘΚΛΜΝ πρίσμα· ἴσον ἄρα ἐστὶ τὸ ΑΒΓΔΕΖ πρίσμα τῷ ΗΘΚΛΜΝ πρίσματι. ἐὰν ἄρα ᾖ δύο πρίσματα ἰσοϋψῆ, καὶ τὸ μὲν ἔχῃ βάσιν παραλληλόγραμμον, τὸ δὲ τρίγωνον, διπλάσιον δὲ ᾖ τὸ παραλληλόγραμμον τοῦ τριγώνου, ἴσα ἐστὶ τὰ πρίσματα· ὅπερ ἔδει δεῖξαι. τὰ ἐν τοῖς κύκλοις ὅμοια πολύγωνα πρὸς ἄλληλά ἐστιν ὡς τὰ ἀπὸ τῶν διαμέτρων τετράγωνα. ἔστωσαν κύκλοι οἱ ΑΒΓ, ΖΗΘ, καὶ ἐν αὐτοῖς ὅμοια πολύγωνα ἔστω τὰ ΑΒΓΔΕ, ΖΗΘΚΛ, διάμετροι δὲ τῶν κύκλων ἔστωσαν αἱ ΒΜ, ΗΝ· λέγω, ὅτι ἐστὶν ὡς τὸ ἀπὸ τῆς ΒΜ τετράγωνον πρὸς τὸ ἀπὸ τῆς ΗΝ τετράγωνον, οὕτως τὸ ΑΒΓΔΕ πολύγωνον πρὸς τὸ ΖΗΘΚΛ πολύγωνον. ἐπεζεύχθωσαν γὰρ αἱ ΒΕ, ΑΜ, ΗΛ, ΖΝ. καὶ ἐπεὶ ὅμοιον τὸ ΑΒΓΔΕ πολύγωνον τῷ ΖΗΘΚΛ πολυγώνῳ, ἴση ἐστὶ καὶ ἡ ὑπὸ ΒΑΕ γωνία τῇ ὑπὸ ΗΖΛ, καί ἐστιν ὡς ἡ ΒΑ πρὸς τὴν ΑΕ, οὕτως ἡ ΗΖ πρὸς τὴν ΖΛ. δύο δὴ τρίγωνά ἐστι τὰ ΒΑΕ, ΗΖΛ μίαν γωνίαν μιᾷ γωνίᾳ ἴσην ἔχοντα τὴν ὑπὸ ΒΑΕ τῇ ὑπὸ ΗΖΛ, περὶ δὲ τὰς ἴσας γωνίας τὰς πλευρὰς ἀνάλογον· ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΕ τρίγωνον τῷ ΖΗΛ τριγώνῳ. ἴση ἄρα ἐστὶν ἡ ὑπὸ ΑΕΒ γωνία τῇ ὑπὸ ΖΛΗ. ἀλλʼ ἡ μὲν ὑπὸ ΑΕΒ τῇ ὑπὸ ΑΜΒ ἐστιν ἴση· ἐπὶ γὰρ τῆς αὐτῆς περιφερείας βεβήκασιν· ἡ δὲ ὑπὸ ΖΛΗ τῇ ὑπὸ ΖΝΗ·
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 and.
καὶ ἡ ὑπὸ ΑΜΒ ἄρα τῇ ὑπὸ ΖΝΗ ἐστιν ἴση. ἔστι δὲ καὶ ὀρθὴ ἡ ὑπὸ ΒΑΜ ὀρθῇ τῇ ὑπὸ ΗΖΝ ἴση· καὶ ἡ λοιπὴ ἄρα τῇ λοιπῇ ἐστιν ἴση.
(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.
ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΜ τρίγωνον τῷ ΖΗΝ τριγώνῳ. ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΒΜ πρὸς τὴν ΗΝ, οὕτως ἡ ΒΑ πρὸς τὴν ΗΖ. ἀλλὰ τοῦ μὲν τῆς ΒΜ πρὸς τὴν ΗΝ λόγου διπλασίων ἐστὶν ὁ τοῦ ἀπὸ τῆς ΒΜ τετραγώνου πρὸς τὸ ἀπὸ τῆς ΗΝ τετράγωνον, τοῦ δὲ τῆς ΒΑ πρὸς τὴν ΗΖ διπλασίων ἐστὶν ὁ τοῦ ΑΒΓΔΕ πολυγώνου πρὸς τὸ ΖΗ ΘΚΛ πολύγωνον·
(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. ThisMethod(orPlan 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 wordswithout 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.
καὶ ὡς ἄρα τὸ ἀπὸ τῆς ΒΜ τετράγωνον πρὸς τὸ ἀπὸ τῆς ΗΝ τετράγωνον, οὕτως τὸ ΑΒΓΔΕ πολύγωνον πρὸς τὸ ΖΗΘΚΛ πολύγωνον. τὰ ἄρα ἐν τοῖς κύκλοις ὅμοια πολύγωνα πρὸς ἄλληλά ἐστιν ὡς τὰ ἀπὸ τῶν διαμέτρων τετράγωνα· ὅπερ ἔδει δεῖξαι. οἱ κύκλοι πρὸς ἀλλήλους εἰσὶν ὡς τὰ ἀπὸ τῶν διαμέτρων τετράγωνα. ἔστωσαν κύκλοι οἱ ΑΒΓΔ, ΕΖΗΘ, διάμετροι δὲ αὐτῶν [ἔστωσαν] αἱ ΒΔ, ΖΘ· λέγω, ὅτι ἐστὶν ὡς ὁ ΑΒΓΔ κύκλος πρὸς τὸν ΕΖΗΘ κύκλον, οὕτως τὸ ἀπὸ τῆς ΒΔ τετράγωνον πρὸς τὸ ἀπὸ τῆς ΖΘ τετράγωνον. εἰ γὰρ μή ἐστιν ὡς ὁ ΑΒΓΔ κύκλος πρὸς τὸν ΕΖΗΘ, οὕτως τὸ ἀπὸ τῆς ΒΔ τετράγωνον πρὸς τὸ ἀπὸ τῆς ΖΘ, ἔσται ὡς τὸ ἀπὸ τῆς ΒΔ πρὸς τὸ ἀπὸ τῆς ΖΘ, οὕτως ὁ ΑΒΓΔ κύκλος ἤτοι πρὸς ἔλασσόν τι τοῦ ΕΖΗΘ κύκλου χωρίον ἢ πρὸς μεῖζον. ἔστω πρότερον πρὸς ἔλασσον τὸ Σ. καὶ ἐγγεγράφθω εἰς τὸν ΕΖΗΘ κύκλον τετράγωνον τὸ ΕΖΗΘ· τὸ δὴ ἐγγεγραμμένον τετράγωνον μεῖζόν ἐστιν ἢ τὸ ἥμισυ τοῦ ΕΖΗΘ κύκλου, ἐπειδήπερ ἐὰν διὰ τῶν ε, Ζ, Η, Θ σημείων ἐφαπτομένας [εὐθείας] τοῦ κύκλου ἀγάγωμεν, τοῦ περιγραφομένου περὶ τὸν κύκλον τετραγώνου ἥμισύ ἐστι τὸ ΕΖΗΘ τετράγωνον, τοῦ δὲ περιγραφέντος τετραγώνου ἐλάττων ἐστὶν ὁ κύκλος· ὥστε τὸ ΕΖΗΘ ἐγγεγραμμένον τετράγωνον μεῖζόν ἐστι τοῦ ἡμίσεως τοῦ ΕΖΗΘ κύκλου. τετμήσθωσαν δίχα αἱ ΕΖ, ΖΗ, ΗΘ, ΘΕ περιφέρειαι κατὰ τὰ Κ, Λ, Μ, Ν σημεῖα, καὶ ἐπεζεύχθωσαν αἱ ΕΚ, ΚΖ, ΖΛ, ΛΗ, ΗΜ, ΜΘ, ΘΝ, ΝΕ· καὶ ἕκαστον ἄρα τῶν ΕΚΖ, ΖΛΗ, ΗΜΘ, ΘΝΕ τριγώνων μεῖζόν ἐστιν ἢ τὸ ἥμισυ τοῦ καθʼ ἑαυτὸ τμήματος τοῦ κύκλου, ἐπειδήπερ ἐὰν διὰ τῶν Κ, Λ, Μ, Ν σημείων ἐφαπτομένας τοῦ κύκλου ἀγάγωμεν καὶ ἀναπληρώσωμεν τὰ ἐπὶ τῶν ΕΖ, ΖΗ, ΗΘ, ΘΕ εὐθειῶν παραλληλόγραμμα, ἕκαστον τῶν ΕΚΖ, ΖΛΗ, ΗΜΘ, ΘΝΕ τριγώνων ἥμισυ ἔσται τοῦ καθʼ ἑαυτὸ παραλληλογράμμου, ἀλλὰ τὸ καθʼ ἑαυτὸ τμῆμα ἔλαττόν ἐστι τοῦ παραλληλογράμμου·
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.
ὥστε ἕκαστον τῶν ΕΚΖ, ΖΛΗ, ΗΜΘ, ΘΝΕ τριγώνων μεῖζόν ἐστι τοῦ ἡμίσεως τοῦ καθʼ ἑαυτὸ τμήματος τοῦ κύκλου.
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.
τέμνοντες δὴ τὰς ὑπολειπομένας περιφερείας δίχα καὶ ἐπιζευγνύντες εὐθείας καὶ τοῦτο ἀεὶ ποιοῦντες καταλείψομέν τινα ἀποτμήματα τοῦ κύκλου, ἃ ἔσται ἐλάσσονα τῆς ὑπεροχῆς, ᾗ ὑπερέχει ὁ ΕΖΗΘ κύκλος τοῦ Σ χωρίου.
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.
ἐδείχθη γὰρ ἐν τῷ πρώτῳ θεωρήματι τοῦ δεκάτου βιβλίου, ὅτι δύο μεγεθῶν ἀνίσων ἐκκειμένων, ἐὰν ἀπὸ τοῦ μείζονος ἀφαιρεθῇ μεῖζον ἢ τὸ ἥμισυ καὶ τοῦ καταλειπομένου μεῖζον ἢ τὸ ἥμισυ, καὶ τοῦτο ἀεὶ γίγνηται, λειφθήσεταί τι μέγεθος, ὃ ἔσται ἔλασσον τοῦ ἐκκειμένου ἐλάσσονος μεγέθους.
The same sort of argument is used for solids, plane sections taking the place of straight lines.
λελείφθω οὖν, καὶ ἔστω τὰ ἐπὶ τῶν ΕΚ, ΚΖ, ΖΛ, ΛΗ, ΗΜ, ΜΘ, ΘΝ, ΝΕ τμήματα τοῦ ΕΖΗΘ κύκλου ἐλάττονα τῆς ὑπεροχῆς, ᾗ ὑπερέχει ὁ ΕΖΗΘ κύκλος τοῦ Σ χωρίου.
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.
λοιπὸν ἄρα τὸ ΕΚΖΛΗ ΜΘΝ πολύγωνον μεῖζόν ἐστι τοῦ Σ χωρίου. ἐγγεγράφθω καὶ εἰς τὸν ΑΒΓΔ κύκλον τῷ ΕΚΖΛΗΜΘΝ πολυγώνῳ ὅμοιον πολύγωνον τὸ ΑΞΒΟΓΠΔΡ·
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 phrasemade 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. ) 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. Similar polygons inscribed in circles are to one another as the squares on the diameters.
ἔστιν ἄρα ὡς τὸ ἀπὸ τῆς ΒΔ τετράγωνον πρὸς τὸ ἀπὸ τῆς ΖΘ τετράγωνον, οὕτως τὸ ΑΞΒΟΓΠΔΡ πολύγωνον πρὸς τὸ ΕΚΖΛ ΗΜΘΝ πολύγωνον. ἀλλὰ καὶ ὡς τὸ ἀπὸ τῆς ΒΔ τετράγωνον πρὸς τὸ ἀπὸ τῆς ΖΘ, οὕτως ὁ ΑΒΓΔ κύκλος πρὸς τὸ Σ χωρίον· καὶ ὡς ἄρα ὁ ΑΒΓΔ κύκλος πρὸς τὸ Σ χωρίον, οὕτως τὸ ΑΞΒΟΓΠΔΡ πολύγωνον πρὸς τὸ ΕΚΖΛΗΜΘΝ πολύγωνον· ἐναλλὰξ ἄρα ὡς ὁ ΑΒΓΔ κύκλος πρὸς τὸ ἐν αὐτῷ πολύγωνον, οὕτως τὸ Σ χωρίον πρὸς τὸ ΕΚΖΛΗΜΘΝ πολύγωνον. μείζων δὲ ὁ ΑΒΓΔ κύκλος τοῦ ἐν αὐτῷ πολυγώνου· μεῖζον ἄρα καὶ τὸ Σ χωρίον τοῦ ΕΚΖΛΗΜΘΝ πολυγώνου. ἀλλὰ καὶ ἔλαττον· ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἐστὶν ὡς τὸ ἀπὸ τῆς ΒΔ τετράγωνον πρὸς τὸ ἀπὸ τῆς ΖΘ, οὕτως ὁ ΑΒΓΔ κύκλος πρὸς ἔλασσόν τι τοῦ ΕΖΗΘ κύκλου χωρίον. ὁμοίως δὴ δείξομεν, ὅτι οὐδὲ ὡς τὸ ἀπὸ ΖΘ πρὸς τὸ ἀπὸ ΒΔ, οὕτως ὁ ΕΖΗΘ κύκλος πρὸς ἔλασσόν τι τοῦ ΑΒΓΔ κύκλου χωρίον. λέγω δή, ὅτι οὐδὲ ὡς τὸ ἀπὸ τῆς ΒΔ πρὸς τὸ ἀπὸ τῆς ΖΘ, οὕτως ὁ ΑΒΓΔ κύκλος πρὸς μεῖζόν τι τοῦ ΕΖΗΘ κύκλου χωρίον. εἰ γὰρ δυνατόν, ἔστω πρὸς μεῖζον τὸ Σ. ἀνάπαλιν ἄρα [ἐστὶν] ὡς τὸ ἀπὸ τῆς ΖΘ τετράγωνον πρὸς τὸ ἀπὸ τῆς ΔΒ, οὕτως τὸ Σ χωρίον πρὸς τὸν ΑΒΓΔ κύκλον. ἀλλʼ ὡς τὸ Σ χωρίον πρὸς τὸν ΑΒΓΔ κύκλον, οὕτως ὁ ΕΖΗΘ κύκλος πρὸς ἔλαττόν τι τοῦ ΑΒΓΔ κύκλου χωρίον· καὶ ὡς ἄρα τὸ ἀπὸ τῆς ΖΘ πρὸς τὸ ἀπὸ τῆς ΒΔ, οὕτως ὁ ΕΖΗΘ κύκλος πρὸς ἔλασσόν τι τοῦ ΑΒΓΔ κύκλου χωρίον·
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.
ὅπερ ἀδύνατον ἐδείχθη.
For let BE, AM, GL, FN be joined.
οὐκ ἄρα ἐστὶν ὡς τὸ ἀπὸ τῆς ΒΔ τετράγωνον πρὸς τὸ ἀπὸ τῆς ΖΘ, οὕτως ὁ ΑΒΓΔ κύκλος πρὸς μεῖζόν τι τοῦ ΕΖΗΘ κύκλου χωρίον.
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]
ἐδείχθη δέ, ὅτι οὐδὲ πρὸς ἔλασσον·
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. []
ἔστιν ἄρα ὡς τὸ ἀπὸ τῆς ΒΔ τετράγωνον πρὸς τὸ ἀπὸ τῆς ΖΘ, οὕτως ὁ ΑΒΓΔ κύκλος πρὸς τὸν ΕΖΗΘ κύκλον.
Therefore the angle AEB is equal to the angle FLG.
οἱ ἄρα κύκλοι πρὸς ἀλλήλους εἰσὶν ὡς τὰ ἀπὸ τῶν διαμέτρων τετράγωνα·
But the angle AEB is equal to the angle AMB, for they stand on the same circumference; [] and the angle FLG to the angle FNG; therefore the angle AMB is also equal to the angle FNG.
ὅπερ ἔδει δεῖξαι.
But the right angle BAM is also equal to the right angle GFN; [] therefore the remaining angle is equal to the remaining angle. [] Therefore the triangle ABM is equiangular with the triangle FGN. Therefore, proportionally, as BM is to GN, so is BA to GF. []
λῆμμα λέγω δή, ὅτι τοῦ Σ χωρίου μείζονος ὄντος τοῦ ΕΖΗΘ κύκλου ἐστὶν ὡς τὸ Σ χωρίον πρὸς τὸν ΑΒΓΔ κύκλον, οὕτως ὁ ΕΖΗΘ κύκλος πρὸς ἔλαττόν τι τοῦ ΑΒΓΔ κύκλου χωρίον.
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; [] 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. Circles are to one another as the squares on the diameters.
γεγονέτω γὰρ ὡς τὸ Σ χωρίον πρὸς τὸν ΑΒΓΔ κύκλον, οὕτως ὁ ΕΖΗΘ κύκλος πρὸς τὸ Τ χωρίον. λέγω, ὅτι ἔλαττόν ἐστι τὸ Τ χωρίον τοῦ ΑΒΓΔ κύκλου.
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.
ἐπεὶ γάρ ἐστιν ὡς τὸ Σ χωρίον πρὸς τὸν ΑΒΓΔ κύκλον, οὕτως ὁ ΕΖΗΘ κύκλος πρὸς τὸ Τ χωρίον, ἐναλλάξ ἐστιν ὡς τὸ Σ χωρίον πρὸς τὸν ΕΖΗΘ κύκλον, οὕτως ὁ ΑΒΓΔ κύκλος πρὸς τὸ Τ χωρίον.
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.
μεῖζον δὲ τὸ Σ χωρίον τοῦ ΕΖΗΘ κύκλου· μείζων ἄρα καὶ ὁ ΑΒΓΔ κύκλος τοῦ Τ χωρίου.
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.
ὥστε ἐστὶν ὡς τὸ Σ χωρίον πρὸς τὸν ΑΒΓΔ κύκλον, οὕτως ὁ ΕΖΗΘ κύκλος πρὸς ἔλαττόν τι τοῦ ΑΒΓΔ κύκλου χωρίον· ὅπερ ἔδει δεῖξαι. πᾶσα πυραμὶς τρίγωνον ἔχουσα βάσιν διαιρεῖται εἰς δύο πυραμίδας ἴσας τε καὶ ὁμοίας ἀλλήλαις καὶ [ὁμοίας] τῇ ὅλῃ τριγώνους ἐχούσας βάσεις καὶ εἰς δύο πρίσματα ἴσα·
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.
καὶ τὰ δύο πρίσματα μείζονά ἐστιν ἢ τὸ ἥμισυ τῆς ὅλης πυραμίδος. ἔστω πυραμίς, ἧς βάσις μέν ἐστι τὸ ΑΒΓ τρίγωνον, κορυφὴ δὲ τὸ Δ σημεῖον· λέγω, ὅτι ἡ ΑΒΓΔ πυραμὶς διαιρεῖται εἰς δύο πυραμίδας ἴσας ἀλλήλαις τριγώνους βάσεις ἐχούσας καὶ ὁμοίας τῇ ὅλῃ καὶ εἰς δύο πρίσματα ἴσα·
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.
καὶ τὰ δύο πρίσματα μείζονά ἐστιν ἢ τὸ ἥμισυ τῆς ὅλης πυραμίδος.
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.
τετμήσθωσαν γὰρ αἱ ΑΒ, ΒΓ, ΓΑ, ΑΔ, ΔΒ, ΔΓ δίχα κατὰ τὰ Ε, Ζ, Η, Θ, Κ, Λ σημεῖα, καὶ ἐπεζεύχθωσαν αἱ ΘΕ, ΕΗ, ΗΘ, ΘΚ, ΚΛ, ΛΘ, ΚΖ, ΖΗ. ἐπεὶ ἴση ἐστὶν ἡ μὲν ΑΕ τῇ ΕΒ, ἡ δὲ ΑΘ τῇ ΔΘ, παράλληλος ἄρα ἐστὶν ἡ ΕΘ τῇ ΔΒ.
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.
διὰ τὰ αὐτὰ δὴ καὶ ἡ ΘΚ τῇ ΑΒ παράλληλός ἐστιν. παραλληλόγραμμον ἄρα ἐστὶ τὸ ΘΕ ΒΚ· ἴση ἄρα ἐστὶν ἡ ΘΚ τῇ ΕΒ.
Therefore the remainder, the polygon EKFLGMHN, is greater than the area S.
ἀλλὰ ἡ ΕΒ τῇ ΕΑ ἐστιν ἴση·
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. []
καὶ ἡ ΑΕ ἄρα τῇ ΘΚ ἐστιν ἴση. ἔστι δὲ καὶ ἡ ΑΘ τῇ ΘΔ ἴση· δύο δὴ αἱ ΕΑ, ΑΘ δυσὶ ταῖς ΚΘ, ΘΔ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ·
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; [] therefore, alternately, as the circle ABCD is to the polygon inscribed in it, so is the area S to the polygon EKFLGMHN. []
καὶ γωνία ἡ ὑπὸ ΕΑΘ γωνίᾳ τῇ ὑπὸ ΚΘΔ ἴση· βάσις ἄρα ἡ ΕΘ βάσει τῇ ΚΔ ἐστιν ἴση. ἴσον ἄρα καὶ ὅμοιόν ἐστι τὸ ΑΕΘ τρίγωνον τῷ ΘΚΔ τριγώνῳ.
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.
διὰ τὰ αὐτὰ δὴ καὶ τὸ ΑΘΗ τρίγωνον τῷ ΘΛΔ τριγώνῳ ἴσον τέ ἐστι καὶ ὅμοιον.
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.
καὶ ἐπεὶ δύο εὐθεῖαι ἁπτόμεναι ἀλλήλων αἱ ΕΘ, ΘΗ παρὰ δύο εὐθείας ἁπτομένας ἀλλήλων τὰς ΚΔ, ΔΛ εἰσιν οὐκ ἐν τῷ αὐτῷ ἐπιπέδῳ οὖσαι, ἴσας γωνίας περιέξουσιν.
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.
ἴση ἄρα ἐστὶν ἡ ὑπὸ ΕΘΗ γωνία τῇ ὑπὸ ΚΔΛ γωνίᾳ.
Therefore, inversely, as the square on FH is to the square on DB, so is the area S to the circle ABCD.
καὶ ἐπεὶ δύο εὐθεῖαι αἱ ΕΘ, ΘΗ δυσὶ ταῖς ΚΔ, ΔΛ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ, καὶ γωνία ἡ ὑπὸ ΕΘΗ γωνίᾳ τῇ ὑπὸ ΚΔΛ ἐστιν ἴση, βάσις ἄρα ἡ ΕΗ βάσει τῇ ΚΛ [ἐστιν] ἴση·
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: [] which was proved impossible.
ἴσον ἄρα καὶ ὅμοιόν ἐστι τὸ ΕΘΗ τρίγωνον τῷ ΚΔΛ τριγώνῳ.
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.
διὰ τὰ αὐτὰ δὴ καὶ τὸ ΑΕΗ τρίγωνον τῷ ΘΚΛ τριγώνῳ ἴσον τε καὶ ὅμοιόν ἐστιν.
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.
ἡ ἄρα πυραμίς, ἧς βάσις μέν ἐστι τὸ ΑΕΗ τρίγωνον, κορυφὴ δὲ τὸ Θ σημεῖον, ἴση καὶ ὁμοία ἐστὶ πυραμίδι, ἧς βάσις μέν ἐστι τὸ ΘΚΛ τρίγωνον, κορυφὴ δὲ τὸ Δ σημεῖον.
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.
καὶ ἐπεὶ τριγώνου τοῦ ΑΔΒ παρὰ μίαν τῶν πλευρῶν τὴν ΑΒ ἦκται ἡ ΘΚ, ἰσογώνιόν ἐστι τὸ ΑΔΒ τρίγωνον τῷ ΔΘΚ τριγώνῳ, καὶ τὰς πλευρὰς ἀνάλογον ἔχουσιν·
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. []
ὅμοιον ἄρα ἐστὶ τὸ ΑΔΒ τρίγωνον τῷ ΔΘΚ τριγώνῳ.
But the area S is greater than the circle EFGH; therefore the circle ABCD is also greater than the area T.
διὰ τὰ αὐτὰ δὴ καὶ τὸ μὲν ΔΒΓ τρίγωνον τῷ ΔΚΛ τριγώνῳ ὅμοιόν ἐστιν, τὸ δὲ ΑΔΓ τῷ ΔΛΘ.
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. 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.
καὶ ἐπεὶ δύο εὐθεῖαι ἁπτόμεναι ἀλλήλων αἱ ΒΑ, ΑΓ παρὰ δύο εὐθείας ἁπτομένας ἀλλήλων τὰς ΚΘ, ΘΛ εἰσιν οὐκ ἐν τῷ αὐτῷ ἐπιπέδῳ, ἴσας γωνίας περιέξουσιν.
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.
ἴση ἄρα ἐστὶν ἡ ὑπὸ ΒΑΓ γωνία τῇ ὑπὸ ΚΘΛ. καί ἐστιν ὡς ἡ ΒΑ πρὸς τὴν ΑΓ, οὕτως ἡ ΚΘ πρὸς τὴν ΘΛ· ὅμοιον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΘΚΛ τριγώνῳ.
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. [] For the same reason HK is also parallel to AB. Therefore HEBK is a parallelogram; therefore HK is equal to EB. [] But EB is equal to EA; therefore AE is also equal to HK.
καὶ πυραμὶς ἄρα, ἧς βάσις μέν ἐστι τὸ ΑΒΓ τρίγωνον, κορυφὴ δὲ τὸ Δ σημεῖον, ὁμοία ἐστὶ πυραμίδι, ἧς βάσις μέν ἐστι τὸ ΘΚΛ τρίγωνον, κορυφὴ δὲ τὸ Δ σημεῖον.
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. [] 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. []
ἀλλὰ πυραμίς, ἧς βάσις μὲν [ἐστι] τὸ ΘΚΛ τρίγωνον, κορυφὴ δὲ τὸ Δ σημεῖον, ὁμοία ἐδείχθη πυραμίδι, ἧς βάσις μέν ἐστι τὸ ΑΕΗ τρίγωνον, κορυφὴ δὲ τὸ Θ σημεῖον [ὥστε καὶ πυραμίς, ἧς βάσις μὲν τὸ ΑΒΓ τρίγωνον, κορυφὴ δὲ τὸ Δ σημεῖον, ὁμοία ἐστὶ πυραμίδι, ἧς βάσις μὲν τὸ ΑΕΗ τρίγωνον, κορυφὴ δὲ τὸ Θ σημεῖον].
Therefore the angle EHG is equal to the angle KDL.
ἑκατέρα ἄρα τῶν ΑΕΗΘ, ΘΚΛΔ πυραμίδων ὁμοία ἐστὶ τῇ ὅλῃ τῇ ΑΒΓΔ πυραμίδι.
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; [] 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. []
¯Καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΖ τῇ ΖΓ, διπλάσιόν ἐστι τὸ ΕΒΖΗ παραλληλόγραμμον τοῦ ΗΖΓ τριγώνου.
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, [] and they have their sides proportional; therefore the triangle ADB is similar to the triangle DHK. [] 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. [] 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.
καὶ ἐπεί, ἐὰν ᾖ δύο πρίσματα ἰσοϋψῆ, καὶ τὸ μὲν ἔχῃ βάσιν παραλληλόγραμμον, τὸ δὲ τρίγωνον, διπλάσιον δὲ ᾖ τὸ παραλληλόγραμμον τοῦ τριγώνου, ἴσα ἐστὶ τὰ πρίσματα, ἴσον ἄρα ἐστὶ τὸ πρίσμα τὸ περιεχόμενον ὑπὸ δύο μὲν τριγώνων τῶν ΒΚΖ, ΕΘΗ, τριῶν δὲ παραλληλογράμμων τῶν ΕΒΖΗ, ΕΒΚΘ, ΘΚΖΗ τῷ πρίσματι τῷ περιεχομένῳ ὑπὸ δύο μὲν τριγώνων τῶν ΗΖΓ, ΘΚΛ, τριῶν δὲ παραλληλογράμμων τῶν ΚΖΓΛ, ΛΓΗΘ, ΘΚΖΗ.
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.
καὶ φανερόν, ὅτι ἑκάτερον τῶν πρισμάτων, οὗ τε βάσις τὸ ΕΒΖΗ παραλληλόγραμμον, ἀπεναντίον δὲ ἡ ΘΚ εὐθεῖα, καὶ οὗ βάσις τὸ ΗΖΓ τρίγωνον, ἀπεναντίον δὲ τὸ ΘΚΛ τρίγωνον, μεῖζόν ἐστιν ἑκατέρας τῶν πυραμίδων, ὧν βάσεις μὲν τὰ ΑΕΗ, ΘΚΛ τρίγωνα, κορυφαὶ δὲ τὰ Θ, Δ σημεῖα, ἐπειδήπερ [καὶ] ἐὰν ἐπιζεύξωμεν τὰς ΕΖ, ΕΚ εὐθείας, τὸ μὲν πρίσμα, οὗ βάσις τὸ ΕΒΖΗ παραλληλόγραμμον, ἀπεναντίον δὲ ἡ ΘΚ εὐθεῖα, μεῖζόν ἐστι τῆς πυραμίδος, ἧς βάσις τὸ ΕΒΖ τρίγωνον, κορυφὴ δὲ τὸ Κ σημεῖον.
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, [] 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.
ἀλλʼ ἡ πυραμίς, ἧς βάσις τὸ ΕΒΖ τρίγωνον, κορυφὴ δὲ τὸ Κ σημεῖον, ἴση ἐστὶ πυραμίδι, ἧς βάσις τὸ ΑΕΗ τρίγωνον, κορυφὴ δὲ τὸ Θ σημεῖον·
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.
ὑπὸ γὰρ ἴσων καὶ ὁμοίων ἐπιπέδων περιέχονται. ὥστε καὶ τὸ πρίσμα, οὗ βάσις μὲν τὸ ΕΒΖΗ παραλληλόγραμμον, ἀπεναντίον δὲ ἡ ΘΚ εὐθεῖα, μεῖζόν ἐστι πυραμίδος, ἧς βάσις μὲν τὸ ΑΕΗ τρίγωνον, κορυφὴ δὲ τὸ Θ σημεῖον.
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.
ἴσον δὲ τὸ μὲν πρίσμα, οὗ βάσις τὸ ΕΒΖΗ παραλληλόγραμμον, ἀπεναντίον δὲ ἡ ΘΚ εὐθεῖα, τῷ πρίσματι, οὗ βάσις μὲν τὸ ΗΖΓ τρίγωνον, ἀπεναντίον δὲ τὸ ΘΚΛ τρίγωνον·
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.
ἡ δὲ πυραμίς, ἧς βάσις τὸ ΑΕΗ τρίγωνον, κορυφὴ δὲ τὸ Θ σημεῖον, ἴση ἐστὶ πυραμίδι, ἧς βάσις τὸ ΘΚΛ τρίγωνον, κορυφὴ δὲ τὸ Δ σημεῖον.
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.
τὰ ἄρα εἰρημένα δύο πρίσματα μείζονά ἐστι τῶν εἰρημένων δύο πυραμίδων, ὧν βάσεις μὲν τὰ ΑΕΗ, ΘΚΛ τρίγωνα, κορυφαὶ δὲ τὰ Θ, Δ σημεῖα.
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.
ἡ ἄρα ὅλη πυραμίς, ἧς βάσις τὸ ΑΒΓ τρίγωνον, κορυφὴ δὲ τὸ Δ σημεῖον, διῄρηται εἴς τε δύο πυραμίδας ἴσας ἀλλήλαις [καὶ ὁμοίας τῇ ὅλῃ] καὶ εἰς δύο πρίσματα ἴσα, καὶ τὰ δύο πρίσματα μείζονά ἐστιν ἢ τὸ ἥμισυ τῆς ὅλης πυραμίδος· ὅπερ ἔδει δεῖξαι.