Page 449 - Euclid's Elements of Geometry

ST EW iaþ. G ELEMENTS BOOK 11

A ¨mma B A C B

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Lemma
῝Ον δὲ τρόπον, ᾧ μεῖζόν ἐστι τὸ ἀπὸ τῆς ΑΒ τοῦ ἀπὸ And we can demonstrate, thusly, in which manner to
τῆς ΛΞ, ἐκείνῳ ἴσον λαβεῖν ἔστι τὸ ἀπὸ τῆς ΞΡ, δείξομεν take the (square) on OR equal to that (area) by which
οὕτως. ἐκκείσθωσαν αἱ ΑΒ, ΛΞ εὐθεῖαι, καὶ ἔστω μείζων ἡ the (square) on AB is greater than the (square) on LO.
ΑΒ, καὶ γεγράφθω ἐπ᾿ αὐτῆς ἡμικύκλιον τὸ ΑΒΓ, καὶ εἰς τὸ Let the straight-lines AB and LO be set out, and let AB
ΑΒΓ ἡμικύκλιον ἐνηρμόσθω τῇ ΛΞ εὐθείᾳ μὴ μείζονι οὔσῃ be greater, and let the semicircle ABC have been drawn
τῆς ΑΒ διαμέτρου ἴση ἡ ΑΓ, καὶ ἐπεζεύχθω ἡ ΓΒ. ἐπεὶ οὖν around it. And let AC, equal to the straight-line LO,
ἐν ἡμικυκλίῳ τῷ ΑΓΒ γωνία ἐστὶν ἡ ὑπὸ ΑΓΒ, ὀρθὴ ἄρα which is not greater than the diameter AB, have been
ἐστὶν ἡ ὑπὸ ΑΓΒ. τὸ ἄρα ἀπὸ τῆς ΑΒ ἴσον ἐστὶ τοῖς ἀπὸ inserted into the semicircle ABC [Prop. 4.1]. And let
τῶν ΑΓ, ΓΒ. ὥστε τὸ ἀπὸ τῆς ΑΒ τοῦ ἀπὸ τῆς ΑΓ μεῖζόν CB have been joined. Therefore, since the angle ACB
ἐστι τῷ ἀπὸ τῆς ΓΒ. ἴση δὲ ἡ ΑΓ τῇ ΛΞ. τὸ ἄρα ἀπὸ τῆς is in the semicircle ACB, ACB is thus a right-angle
ΑΒ τοῦ ἀπὸ τῆς ΛΞ μεῖζόν ἐστι τῷ ἀπὸ τῆς ΓΒ. ἐὰν οὖν [Prop. 3.31]. Thus, the (square) on AB is equal to the
kdþ the (square) on AB is greater than the (square) on LO
τῇ ΒΓ ἴσην τὴν ΞΡ ἀπολάβωμεν, ἔσται τὸ ἀπὸ τῆς ΑΒ τοῦ (sum of the) squares on AC and CB [Prop. 1.47]. Hence,
ἀπὸ τῆς ΛΞ μεῖζον τῷ ἀπὸ τῆς ΞΡ· ὅπερ προέκειτο ποιῆσαι. the (square) on AB is greater than the (square) on AC
by the (square) on CB. And AC (is) equal to LO. Thus,

by the (square) on CB. Therefore, if we take OR equal
to BC then the (square) on AB will be greater than the
(square) on LO by the (square) on OR. (Which is) the
very thing it was prescribed to do.

Proposition 24
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B

᾿Εὰν στερεὸν ὑπὸ παραλλήλων ἐπιπέδων περιέχηται, τὰ If a solid (figure) is contained by (six) parallel planes
ἀπεναντίον αὐτοῦ ἐπίπεδα ἴσα τε καὶ παραλληλόγραμμά then its opposite planes are both equal and parallelo-
ἐστιν. grammic. H
A G H Z G
B

D E A C F

E
D
Στερεὸν γὰρ τὸ ΓΔΘΗ ὑπὸ παραλλήλων ἐπιπέδων πε- For let the solid (figure) CDHG have been contained
ριεχέσθω τῶν ΑΓ, ΗΖ, ΑΘ, ΔΖ, ΒΖ, ΑΕ· λέγω, ὅτι τὰ ἀπε- by the parallel planes AC, GF, and AH, DF, and BF,
ναντίον αὐτοῦ ἐπίπεδα ἴσα τε καὶ παραλληλόγραμμά ἐστιν. AE. I say that its opposite planes are both equal and
᾿Επεὶ γὰρ δύο ἐπίπεδα παράλληλα τὰ ΒΗ, ΓΕ ὑπὸ parallelogrammic.
ἐπιπέδου τοῦ ΑΓ τέμνεται, αἱ κοιναὶ αὐτῶν τομαὶ παράλληλοί For since the two parallel planes BG and CE are
εἰσιν. παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΔΓ. πάλιν, ἑπεὶ cut by the plane AC, their common sections are parallel
δύο ἐπίπεδα παράλληλα τὰ ΒΖ, ΑΕ ὑπὸ ἐπιπέδου τοῦ [Prop. 11.16]. Thus, AB is parallel to DC. Again, since
ΑΓ τέμνεται, αἱ κοιναὶ αὐτῶν τομαὶ παράλληλοί εἰσιν. the two parallel planes BF and AE are cut by the plane

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