Page 465 - Euclid's Elements of Geometry

ST EW iaþ. ELEMENTS BOOK 11

the planes at points L and N (respectively). And let LA
H BAC and EDF (respectively). And let them have joined
and ND have been joined. I say that angle GAL is equal
to angle MDN.
G

A B G D E A B C K L E M
H

Z

D
N
F

Κείσθω τῇ ΔΜ ἴση ἡ ΑΘ, καὶ ἤχθω διὰ τοῦ Θ σημείου Let AH be made equal to DM. And let HK have been
τῇ ΗΛ παράλληλος ἡ ΘΚ. ἡ δὲ ΗΛ κάθετός ἐστιν ἐπὶ τὸ drawn through point H parallel to GL. And GL is per-
διὰ τῶν ΒΑΓ ἐπίπεδον· καὶ ἡ ΘΚ ἄρα κάθετός ἐστιν ἐπὶ τὸ pendicular to the plane through BAC. Thus, HK is also
διὰ τῶν ΒΑΓ ἐπίπεδον. ἤχθωσαν ἀπὸ τῶν Κ, Ν σημείων perpendicular to the plane through BAC [Prop. 11.8].
ἐπὶ τὰς ΑΓ, ΔΖ, ΑΒ, ΔΕ εὐθείας κάθετοι αἱ ΚΓ, ΝΖ, ΚΒ, And let KC, NF, KB, and NE have been drawn from
ΝΕ, καὶ ἐπεζεύχθωσαν αἱ ΘΓ, ΓΒ, ΜΖ, ΖΕ. ἐπεὶ τὸ ἀπὸ points K and N perpendicular to the straight-lines AC,
τῆς ΘΑ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΘΚ, ΚΑ, τῷ δὲ ἀπὸ τῆς ΚΑ DF, AB, and DE. And let HC, CB, MF, and FE have
ἴσα ἐστὶ τὰ ἀπὸ τῶν ΚΓ, ΓΑ, καὶ τὸ ἀπὸ τῆς ΘΑ ἄρα ἴσον been joined. Since the (square) on HA is equal to the
ἐστὶ τοῖς ἀπὸ τῶν ΘΚ, ΚΓ, ΓΑ. τοῖς δὲ ἀπὸ τῶν ΘΚ, ΚΓ (sum of the squares) on HK and KA [Prop. 1.47], and
ἴσον ἐστὶ τὸ ἀπὸ τῆς ΘΓ· τὸ ἄρα ἀπὸ τῆς ΘΑ ἴσον ἐστὶ τοῖς the (sum of the squares) on KC and CA is equal to the
ἀπὸ τῶν ΘΓ, ΓΑ. ὀρθὴ ἄρα ἐστὶν ἡ ὑπὸ ΘΓΑ γωνία. διὰ τὰ (square) on KA [Prop. 1.47], thus the (square) on HA
αὐτὰ δὴ καὶ ἡ ὑπὸ ΔΖΜ γωνία ὀρθή ἐστιν. ἴση ἄρα ἐστὶν is equal to the (sum of the squares) on HK, KC, and
ἡ ὑπὸ ΑΓΘ γωνία τῇ ὑπὸ ΔΖΜ. ἔστι δὲ καὶ ἡ ὑπὸ ΘΑΓ CA. And the (square) on HC is equal to the (sum of
τῇ ὑπὸ ΜΔΖ ἴση. δύο δὴ τρίγωνά ἐστι τὰ ΜΔΖ, ΘΑΓ δύο the squares) on HK and KC [Prop. 1.47]. Thus, the
γωνίας δυσὶ γωνίαις ἴσας ἔχοντα ἑκατέραν ἑκατέρᾳ καὶ μίαν (square) on HA is equal to the (sum of the squares)
πλευρὰν μιᾷ πλευρᾷ ἴσην τὴν ὑποτείνουσαν ὑπὸ μίαν τῶν on HC and CA. Thus, angle HCA is a right-angle
ἴσων γωνιῶν τὴν ΘΑ τῇ ΜΔ· καὶ τὰς λοιπὰς ἄρα πλευρὰς [Prop. 1.48]. So, for the same (reasons), angle DFM
ταῖς λοιπαῖς πλευραῖς ἴσας ἕξει ἑκατέραν ἑκαρέρᾳ. ἴση ἄρα is also a right-angle. Thus, angle ACH is equal to (an-
ἐστὶν ἡ ΑΓ τῇ ΔΖ. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἡ ΑΒ τῇ gle) DFM. And HAC is also equal to MDF. So, MDF
ΔΕ ἐστιν ἴση. ἐπεὶ οὖν ἴση ἐστὶν ἡ μὲν ΑΓ τῇ ΔΖ, ἡ δὲ and HAC are two triangles having two angles equal to
ΑΒ τῇ ΔΕ, δύο δὴ αἱ ΓΑ, ΑΒ δυσὶ ταῖς ΖΔ, ΔΕ ἴσαι εἰσίν. two angles, respectively, and one side equal to one side—
ἀλλὰ καὶ γωνία ἡ ὑπὸ ΓΑΒ γωνίᾳ τῇ ὑπὸ ΖΔΕ ἐστιν ἴση· (namely), that subtending one of the equal angles —(that
βάσις ἄρα ἡ ΒΓ βάσει τῇ ΕΖ ἴση ἐστὶ καὶ τὸ τρίγωνον τῷ is), HA (equal) to MD. Thus, they will also have the re-
τριγώνῳ καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις· ἴση ἄρα ἡ maining sides equal to the remaining sides, respectively
ὑπὸ ΑΓΒ γωνία τῇ ὑπὸ ΔΖΕ. ἔστι δὲ καὶ ὀρθὴ ἡ ὑπὸ ΑΓΚ [Prop. 1.26]. Thus, AC is equal to DF. So, similarly, we
ὀρθῇ τῇ ὑπὸ ΔΖΝ ἴση· καὶ λοιπὴ ἄρα ἡ ὑπὸ ΒΓΚ λοιπῇ τῇ can show that AB is also equal to DE. Therefore, since
ὑπὸ ΕΖΝ ἐστιν ἴση. διὰ τὰ αὐτὰ δὴ καὶ ἡ ὑπὸ ΓΒΚ τῇ ὑπὸ AC is equal to DF, and AB to DE, so the two (straight-
ΖΕΝ ἐστιν ἴση. δύο δὴ τρίγωνά ἐστι τὰ ΒΓΚ, ΕΖΝ [τὰς] lines) CA and AB are equal to the two (straight-lines)
δύο γωνίας δυσὶ γωνίαις ἴσας ἔχοντα ἑκατέραν ἑκατέρᾳ καὶ FD and DE (respectively). But, angle CAB is also equal
μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην τὴν πρὸς ταῖς ἴσαις γωνίαις to angle FDE. Thus, base BC is equal to base EF, and
τὴν ΒΓ τῇ ΕΖ· καὶ τὰς λοιπὰς ἄρα πλευρὰς ταῖς λοιπαῖς triangle (ACB) to triangle (DFE), and the remaining
πλευραῖς ἴσας ἕξουσιν. ἴση ἄρα ἐστὶν ἡ ΓΚ τῇ ΖΝ. ἔστι δὲ angles to the remaining angles (respectively) [Prop. 1.4].

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