Page 469 - Euclid's Elements of Geometry

ST EW iaþ.

D U Z E D O K U ELEMENTS BOOK 11

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B S T R H A B Q C M S T L H R E

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Κύβου γὰρ τοῦ ΑΖ τῶν ἀπεναντίον ἐπιπέδων τῶν ΓΖ, For let the opposite planes CF and AH of the cube
ΑΘ αἱ πλευραὶ δίχα τετμήσθωσαν κατὰ τὰ Κ, Λ, Μ, Ν, Ξ, AF have been cut in half at the points K, L, M, N, O,
Π, Ο, Ρ σημεῖα, διὰ δὲ τῶν τομῶν ἐπίπεδα ἐκβεβλήσθω τὰ Q, P, and R. And let the planes KN and OR have been
ΚΝ, ΞΡ, κοινὴ δὲ τομὴ τῶν ἐπιπέδων ἔστω ἡ ΥΣ, τοῦ δὲ produced through the pieces. And let US be the common
ΑΖ κύβου διαγώνιος ἡ ΔΗ. λέγω, ὅτι ἴση ἐστὶν ἡ μὲν ΥΤ section of the planes, and DG the diameter of cube AF.
τῇ ΤΣ, ἡ δὲ ΔΤ τῇ ΤΗ. I say that UT is equal to T S, and DT to T G.
᾿Επεζεύχθωσαν γὰρ αἱ ΔΥ, ΥΕ, ΒΣ, ΣΗ. καὶ ἐπεὶ For let DU, UE, BS, and SG have been joined. And
παράλληλός ἐστιν ἡ ΔΞ τῇ ΟΕ, αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ since DO is parallel to PE, the alternate angles DOU and
ΔΞΥ, ΥΟΕ ἴσαι ἀλλήλαις εἰσίν. καὶ ἐπεὶ ἴση ἐστὶν ἡ μὲν UPE are equal to one another [Prop. 1.29]. And since
ΔΞ τῇ ΟΕ, ἡ δὲ ΞΥ τῇ ΥΟ, καὶ γωνίας ἴσας περιέχουσιν, DO is equal to PE, and OU to UP, and they contain
βάσις ἄρα ἡ ΔΥ τῇ ΥΕ ἐστιν ἴση, καὶ τὸ ΔΞΥ τρίγωνον τῷ equal angles, base DU is thus equal to base UE, and tri-
ΟΥΕ τριγώνῳ ἐστὶν ἴσον καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς angle DOU is equal to triangle PUE, and the remaining
γωνίαις ἴσαι· ἴση ἄρα ἡ ὑπὸ ΞΥΔ γωνία τῇ ὑπὸ ΟΥΕ γωνίᾳ. angles (are) equal to the remaining angles [Prop. 1.4].
διὰ δὴ τοῦτο εὐθεῖά ἐστιν ἡ ΔΥΕ. διὰ τὰ αὐτὰ δὴ καὶ ΒΣΗ Thus, angle OUD (is) equal to angle PUE. So, for this
εὐθεῖά ἐστιν, καὶ ἴση ἡ ΒΣ τῇ ΣΗ. καὶ ἐπεὶ ἡ ΓΑ τῇ ΔΒ (reason), DUE is a straight-line [Prop. 1.14]. So, for
ἴση ἐστὶ καὶ παράλληλος, ἀλλὰ ἡ ΓΑ καὶ τῇ ΕΗ ἴση τέ the same (reason), BSG is also a straight-line, and BS
ἐστι καὶ παράλληλος, καὶ ἡ ΔΒ ἄρα τῇ ΕΗ ἴση τέ ἐστι equal to SG. And since CA is equal and parallel to DB,
καὶ παράλληλος. καὶ ἐπιζευγνύουσιν αὐτὰς εὐθεῖαι αἱ ΔΕ, but CA is also equal and parallel to EG, DB is thus also
ΒΗ· παράλληλος ἄρα ἐστὶν ἡ ΔΕ τῇ ΒΗ. ἴση ἄρα ἡ μὲν ὑπὸ equal and parallel to EG [Prop. 11.9]. And the straight-
ΕΔΤ γωνία τῇ ὑπὸ ΒΗΤ· ἐναλλὰξ γάρ· ἡ δὲ ὑπὸ ΔΤΥ τῇ lines DE and BG join them. DE is thus parallel to BG
ὑπὸ ΗΤΣ. δύο δὴ τρίγωνά ἐστι τὰ ΔΤΥ, ΗΤΣ τὰς δύο [Prop. 1.33]. Thus, angle EDT (is) equal to BGT . For
γωνίας ταῖς δυσὶ γωνίαις ἴσας ἔχοντα καὶ μίαν πλευρὰν μιᾷ (they are) alternate [Prop. 1.29]. And (angle) DT U (is
πλευρᾷ ἴσην τὴν ὑποτείνουσαν ὑπὸ μίαν τῶν ἴσων γωνιῶν equal) to GT S [Prop. 1.15]. So, DT U and GT S are two
τὴν ΔΥ τῇ ΗΣ· ἡμίσειαι γάρ εἰσι τῶν ΔΕ, ΒΗ· καὶ τὰς triangles having two angles equal to two angles, and one
λοιπὰς πλευρὰς ταῖς λοιπαῖς πλευραῖς ἴσας ἕξει. ἴση ἄρα ἡ side equal to one side—(namely), that subtended by one
μὲν ΔΤ τῇ ΤΗ, ἡ δὲ ΥΤ τῇ ΤΣ. of the equal angles—(that is), DU (equal) to GS. For
᾿Εὰν ἄρα κύβου τῶν ἀπεναντίον ἐπιπέδων αἱ πλευραὶ δίχα they are halves of DE and BG (respectively). (Thus),
τμηθῶσιν, διὰ δὲ τῶν τομῶν ἐπίπεδα ἐκβληθῇ, ἡ κοινὴ τομὴ they will also have the remaining sides equal to the re-
τῶν ἐπιπέδων καὶ ἡ τοῦ κύβου διάμετρος δίχα τέμνουσιν maining sides [Prop. 1.26]. Thus, DT (is) equal to T G,
ἀλλήλας· ὅπερ ἔδει δεῖξαι. and UT to T S.
Thus, if the sides of the opposite planes of a cube are
cut in half, and planes are produced through the pieces,
then the common section of the (latter) planes and the
diameter of the cube cut one another in half. (Which is)
the very thing it was required to show.

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