§6.prop.7ἐὰν δύο τρίγωνα μίαν γωνίαν μιᾷ γωνίᾳ ἴσην ἔχῃ, περὶ δὲ ἄλλας γωνίας τὰς πλευρὰς ἀνάλογον, τῶν δὲ λοιπῶν ἑκατέραν ἅμα ἤτοι ἐλάσσονα ἢ μὴ ἐλάσσονα ὀρθῆς, ἰσογώνια ἔσται τὰ τρίγωνα καὶ ἴσας ἕξει τὰς γωνίας, περὶ ἃς ἀνάλογόν εἰσιν αἱ πλευραί.
two triangles have one angle equal to one angle, and the sides about other angles proportional, and have each of the remaining angles at the same time either less than or not less than a right angle, the triangles will be equiangular and will have those angles equal about which the sides are proportional.
ἔστω δύο τρίγωνα τὰ ΑΒΓ, ΔΕΖ μίαν γωνίαν μιᾷ γωνίᾳ ἴσην ἔχοντα τὴν ὑπὸ ΒΑΓ τῇ ὑπὸ ΕΔΖ, περὶ δὲ ἄλλας γωνίας τὰς ὑπὸ ΑΒΓ, ΔΕΖ τὰς πλευρὰς ἀνάλογον, ὡς τὴν ΑΒ πρὸς τὴν ΒΓ, οὕτως τὴν ΔΕ πρὸς τὴν ΕΖ, τῶν δὲ λοιπῶν τῶν πρὸς τοῖς Γ, Ζ πρότερον ἑκατέραν ἅμα ἐλάσσονα ὀρθῆς·
Let ABC, DEF be two triangles having one angle BAC equal to one angle EDF, and the sides about other angles ABC, DEF proportional, so that, as AB is to BC, so is DE to EF, and let each of the remaining angles at C, F be first at the same time less than a right angle.
λέγω, ὅτι ἰσογώνιόν ἐστι τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ, καὶ ἴση ἔσται ἡ ὑπὸ ΑΒΓ γωνία τῇ ὑπὸ ΔΕΖ, καὶ λοιπὴ δηλονότι ἡ πρὸς τῷ Γ λοιπῇ τῇ πρὸς τῷ Ζ ἴση.
I say that the triangle ABC is equiangular to the triangle DEF, and the angle ABC will be equal to the angle DEF, and the remaining angle at C will obviously be equal to the remaining angle at F.
εἰ γὰρ ἄνισός ἐστιν ἡ ὑπὸ ΑΒΓ γωνία τῇ ὑπὸ ΔΕΖ, μία αὐτῶν μείζων ἐστίν.
For if the angle ABC is unequal to the angle DEF, one of them is greater.
ἔστω μείζων ἡ ὑπὸ ΑΒΓ. καὶ συνεστάτω πρὸς τῇ ΑΒ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Β τῇ ὑπὸ ΔΕΖ γωνίᾳ ἴση ἡ ὑπὸ ΑΒΗ.
καὶ ἐπεὶ ἴση ἐστὶν ἡ μὲν Α γωνία τῇ Δ, ἡ δὲ ὑπὸ ΑΒΗ τῇ ὑπὸ ΔΕΖ, λοιπὴ ἄρα ἡ ὑπὸ ΑΗΒ λοιπῇ τῇ ὑπὸ ΔΖΕ ἐστιν ἴση.
Let the angle ABC be greater; and let there be constructed on the straight line AB and at the point B on it, the angle ABG equal to the angle DEF. And since the angle A is equal to the angle D, and the angle ABG to the angle DEF, therefore the remaining angle AGB is equal to the remaining angle DFE.
ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΗ τρίγωνον τῷ ΔΕΖ τριγώνῳ.
Therefore the triangle ABG is equiangular to the triangle DEF.
ἔστιν ἄρα ὡς ἡ ΑΒ πρὸς τὴν ΒΗ, οὕτως ἡ ΔΕ πρὸς τὴν ΕΖ. ὡς δὲ ἡ ΔΕ πρὸς τὴν ΕΖ, ὑπόκειται ἡ ΑΒ πρὸς τὴν ΒΓ· ἡ ΑΒ ἄρα πρὸς ἑκατέραν τῶν ΒΓ, ΒΗ τὸν αὐτὸν ἔχει λόγον· ἴση ἄρα ἡ ΒΓ τῇ ΒΗ. ὥστε καὶ γωνία ἡ πρὸς τῷ Γ γωνίᾳ τῇ ὑπὸ ΒΗΓ ἐστιν ἴση.
Therefore, as AB is to BG, so is DE to EF. But, as DE is to EF, so by hypothesis is AB to BC; therefore AB has the same ratio to each of the straight lines BC, BG; therefore BC is equal to BG. Hence the angle at C is also equal to the angle BGC.
ἐλάττων δὲ ὀρθῆς ὑπόκειται ἡ πρὸς τῷ Γ· ἐλάττων ἄρα ἐστὶν ὀρθῆς καὶ ἡ ὑπὸ ΒΗΓ· ὥστε ἡ ἐφεξῆς αὐτῇ γωνία ἡ ὑπὸ ΑΗΒ μείζων ἐστὶν ὀρθῆς.
But the angle at C is by hypothesis less than a right angle; therefore the angle BGC is also less than a right angle; so that the angle AGB adjacent to it is greater than a right angle.
καὶ ἐδείχθη ἴση οὖσα τῇ πρὸς τῷ Ζ· καὶ ἡ πρὸς τῷ Ζ ἄρα μείζων ἐστὶν ὀρθῆς.
And it was proved equal to the angle at F; therefore the angle at F is also greater than a right angle.
ὑπόκειται δὲ ἐλάσσων ὀρθῆς· ὅπερ ἐστὶν ἄτοπον.
But it is by hypothesis less than a right angle; which is absurd.
οὐκ ἄρα ἄνισός ἐστιν ἡ ὑπὸ ΑΒΓ γωνία τῇ ὑπὸ ΔΕΖ· ἴση ἄρα.
Therefore the angle ABC is not unequal to the angle DEF; therefore it is equal.
ἔστι δὲ καὶ ἡ πρὸς τῷ Α ἴση τῇ πρὸς τῷ Δ· καὶ λοιπὴ ἄρα ἡ πρὸς τῷ Γ λοιπῇ τῇ πρὸς τῷ Ζ ἴση ἐστίν.
And the angle at A is also equal to the angle at D; therefore the remaining angle at C is equal to the remaining angle at F.
ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ.
Therefore the triangle ABC is equiangular to the triangle DEF.
ἀλλὰ δὴ πάλιν ὑποκείσθω ἑκατέρα τῶν πρὸς τοῖς Γ, Ζ μὴ ἐλάσσων ὀρθῆς·
But indeed, let again each of the angles at C, F be not less than a right angle.
λέγω πάλιν, ὅτι καὶ οὕτως ἐστὶν ἰσογώνιον τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ.
I say again that, even so, the triangle ABC is equiangular to the triangle DEF.
τῶν γὰρ αὐτῶν κατασκευασθέντων ὁμοίως δείξομεν, ὅτι ἴση ἐστὶν ἡ ΒΓ τῇ ΒΗ· ὥστε καὶ γωνία ἡ πρὸς τῷ Γ τῇ ὑπὸ ΒΗΓ ἴση ἐστίν.
For, with the same construction, we shall prove similarly that BC is equal to BG; so that the angle at C is also equal to the angle BGC.
οὐκ ἐλάττων δὲ ὀρθῆς ἡ πρὸς τῷ Γ·
But the angle at C is not less than a right angle; therefore the angle BGC is also not less than a right angle.
οὐκ ἐλάττων ἄρα ὀρθῆς οὐδὲ ἡ ὑπὸ ΒΗΓ. τριγώνου δὴ τοῦ ΒΗΓ αἱ δύο γωνίαι δύο ὀρθῶν οὔκ εἰσιν ἐλάττονες· ὅπερ ἐστὶν ἀδύνατον.
Thus indeed, in the triangle BGC, the two angles are not less than two right angles; which is impossible.
οὐκ ἄρα πάλιν ἄνισός ἐστιν ἡ ὑπὸ ΑΒΓ γωνία τῇ ὑπὸ ΔΕΖ· ἴση ἄρα.
Therefore again the angle ABC is not unequal to the angle DEF; therefore it is equal.
ἔστι δὲ καὶ ἡ πρὸς τῷ Α τῇ πρὸς τῷ Δ ἴση· λοιπὴ ἄρα ἡ πρὸς τῷ Γ λοιπῇ τῇ πρὸς τῷ Ζ ἴση ἐστίν.
And the angle at A is also equal to the angle at D; therefore the remaining angle at C is equal to the remaining angle at F.
ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ.
Therefore the triangle ABC is equiangular to the triangle DEF.
ἐὰν ἄρα δύο τρίγωνα μίαν γωνίαν μιᾷ γωνίᾳ ἴσην ἔχῃ, περὶ δὲ ἄλλας γωνίας τὰς πλευρὰς ἀνάλογον, τῶν δὲ λοιπῶν ἑκατέραν ἅμα ἐλάττονα ἢ μὴ ἐλάττονα ὀρθῆς, ἰσογώνια ἔσται τὰ τρίγωνα καὶ ἴσας ἕξει τὰς γωνίας, περὶ ἃς ἀνάλογόν εἰσιν αἱ πλευραί· ὅπερ ἔδει δεῖξαι.
Therefore, if two triangles have one angle equal to one angle, and the sides about other angles proportional, and have each of the remaining angles at the same time either less than or not less than a right angle, the triangles will be equiangular and will have those angles equal about which the sides are proportional; which was to be proved.