ἐπεὶ οὖν ἴση ἐστὶν ἡ μὲν ΑΓ τῇ ΔΖ, ἡ δὲ ΑΒ τῇ ΔΕ, δύο δὴ αἱ ΓΑ, ΑΒ δυσὶ ταῖς ΖΔ, ΔΕ ἴσαι εἰσίν.
Since therefore AG is equal to DZ, and AB to DE, the two straight lines GA, AB are equal to the two straight lines ZD, DE.
ἀλλὰ καὶ γωνία ἡ ὑπὸ ΓΑΒ γωνίᾳ τῇ ὑπὸ ΖΔΕ ἐστιν ἴση· βάσις ἄρα ἡ ΒΓ βάσει τῇ ΕΖ ἴση ἐστὶ καὶ τὸ τρίγωνον τῷ τριγώνῳ καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις· ἴση ἄρα ἡ ὑπὸ ΑΓΒ γωνία τῇ ὑπὸ ΔΖΕ. ἔστι δὲ καὶ ὀρθὴ ἡ ὑπὸ ΑΓΚ ὀρθῇ τῇ ὑπὸ ΔΖΝ ἴση· καὶ λοιπὴ ἄρα ἡ ὑπὸ ΒΓΚ λοιπῇ τῇ ὑπὸ ΕΖΝ ἐστιν ἴση.
But the angle GAB is also equal to the angle ZDE; therefore the base BG is equal to the base EZ, and the triangle to the triangle, and the remaining angles to the remaining angles; therefore the angle AGB is equal to the angle DZE. But the right angle AGK is also equal to the right angle DZN; therefore the remaining angle BGK is also equal to the remaining angle EZN.
διὰ τὰ αὐτὰ δὴ καὶ ἡ ὑπὸ ΓΒΚ τῇ ὑπὸ ΖΕΝ ἐστιν ἴση.
For the same reasons indeed the angle GBK is also equal to the angle ZEN.
δύο δὴ τρίγωνά ἐστι τὰ ΒΓΚ, ΕΖΝ δύο γωνίας δυσὶ γωνίαις ἴσας ἔχοντα ἑκατέραν ἑκατέρᾳ καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην τὴν πρὸς ταῖς ἴσαις γωνίαις τὴν ΒΓ τῇ ΕΖ· καὶ τὰς λοιπὰς ἄρα πλευρὰς ταῖς λοιπαῖς πλευραῖς ἴσας ἕξουσιν.
Thus there are two triangles BGK, EZN having two angles equal to two angles, each to each, and one side equal to one side, namely that adjacent to the equal angles, BG to EZ; therefore they will also have the remaining sides equal to the remaining sides.
ἴση ἄρα ἐστὶν ἡ ΓΚ τῇ ΖΝ. ἔστι δὲ καὶ ἡ ΑΓ τῇ ΔΖ ἴση· δύο δὴ αἱ ΑΓ, ΓΚ δυσὶ ταῖς ΔΖ, ΖΝ ἴσαι εἰσίν· καὶ ὀρθὰς γωνίας περιέχουσιν.
Therefore GK is equal to ZN. And AG is also equal to DZ; therefore the two straight lines AG, GK are equal to the two straight lines DZ, ZN; and they contain right angles.
βάσις ἄρα ἡ ΑΚ βάσει τῇ ΔΝ ἴση ἐστίν.
Therefore the base AK is equal to the base DN.
καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΘ τῇ ΔΜ, ἴσον ἐστὶ καὶ τὸ ἀπὸ τῆς ΑΘ τῷ ἀπὸ τῆς ΔΜ. ἀλλὰ τῷ μὲν ἀπὸ τῆς ΑΘ ἴσα ἐστὶ τὰ ἀπὸ τῶν ΑΚ, ΚΘ·
And since AQ is equal to DM, the square on AQ is also equal to the square on DM.
ὀρθὴ γὰρ ἡ ὑπὸ ΑΚΘ· τῷ δὲ ἀπὸ τῆς ΔΜ ἴσα τὰ ἀπὸ τῶν ΔΝ, ΝΜ· ὀρθὴ γὰρ ἡ ὑπὸ ΔΝΜ· τὰ ἄρα ἀπὸ τῶν ΑΚ, ΚΘ ἴσα ἐστὶ τοῖς ἀπὸ τῶν ΔΝ, ΝΜ, ὧν τὸ ἀπὸ τῆς ΑΚ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΔΝ· λοιπὸν ἄρα τὸ ἀπὸ τῆς ΚΘ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΝΜ·
But the squares on AK, KQ are equal to the square on AQ; for the angle AKQ is right; and the squares on DN, NM are equal to the square on DM; for the angle DNM is right; therefore the squares on AK, KQ are equal to the squares on DN, NM, of which the square on AK is equal to the square on DN; therefore the remaining square on KQ is equal to the remaining square on NM; therefore QK is equal to MN.
ἴση ἄρα ἡ ΘΚ τῇ ΜΝ. καὶ ἐπεὶ δύο αἱ ΘΑ, ΑΚ δυσὶ ταῖς ΜΔ, ΔΝ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ, καὶ βάσις ἡ ΘΚ βάσει τῇ ΜΝ ἐδείχθη ἴση, γωνία ἄρα ἡ ὑπὸ ΘΑΚ γωνίᾳ τῇ ὑπὸ ΜΔΝ ἐστιν ἴση.
And since the two QA, AK are equal to the two MD, DN, each to each, and the base QK was proved equal to the base MN, therefore the angle QAK is equal to the angle MDN.
ἐὰν ἄρα ὦσι δύο γωνίαι ἐπίπεδοι ἴσαι καὶ τὰ ἑξῆς τῆς προτάσεως.
Therefore, if there be two equal plane angles, and so on of the proposition.
Πόρισμα
ἐκ δὴ τούτου φανερόν, ὅτι, ἐὰν ὦσι δύο γωνίαι ἐπίπεδοι ἴσαι, ἐπισταθῶσι δὲ ἐπʼ αὐτῶν μετέωροι εὐθεῖαι ἴσαι ἴσας γωνίας περιέχουσαι μετὰ τῶν ἐξ ἀρχῆς εὐθειῶν ἑκατέραν ἑκατέρᾳ, αἱ ἀπʼ αὐτῶν κάθετοι ἀγόμεναι ἐπὶ τὰ ἐπίπεδα, ἐν οἷς εἰσιν αἱ ἐξ ἀρχῆς γωνίαι, ἴσαι ἀλλήλαις εἰσίν.
Porism From this indeed it is manifest that, if there be two equal plane angles, and there be set up on them equal elevated straight lines containing equal angles with the original straight lines, each to each, the perpendiculars drawn from them to the planes in which the original angles are, are equal to one another.
ὅπερ ἔδει δεῖξαι.
Which it was required to prove.