§6.prop.3ἐὰν τριγώνου ἡ γωνία δίχα τμηθῇ, ἡ δὲ τέμνουσα τὴν γωνίαν εὐθεῖα τέμνῃ καὶ τὴν βάσιν, τὰ τῆς βάσεως τμήματα τὸν αὐτὸν ἕξει λόγον ταῖς λοιπαῖς τοῦ τριγώνου πλευραῖς· καὶ ἐὰν τὰ τῆς βάσεως τμήματα τὸν αὐτὸν ἔχῃ λόγον ταῖς λοιπαῖς τοῦ τριγώνου πλευραῖς, ἡ ἀπὸ τῆς κορυφῆς ἐπὶ τὴν τομὴν ἐπιζευγνυμένη εὐθεῖα δίχα τεμεῖ τὴν τοῦ τριγώνου γωνίαν.
If an angle of a triangle be bisected, and the straight line cutting the angle cut the base also, the segments of the base will have the same ratio to the remaining sides of the triangle; and, if the segments of the base have the same ratio to the remaining sides of the triangle, the straight line joined from the vertex to the point of section will bisect the angle of the triangle.
ἔστω τρίγωνον τὸ ΑΒΓ, καὶ τετμήσθω ἡ ὑπὸ ΒΑΓ γωνία δίχα ὑπὸ τῆς ΑΔ εὐθείας· λέγω, ὅτι ἐστὶν ὡς ἡ ΒΔ πρὸς τὴν ΓΔ, οὕτως ἡ ΒΑ πρὸς τὴν ΑΓ.
ἤχθω γὰρ διὰ τοῦ Γ τῇ ΔΑ παράλληλος ἡ ΓΕ καὶ διαχθεῖσα ἡ ΒΑ συμπιπτέτω αὐτῇ κατὰ τὸ Ε.
καὶ ἐπεὶ εἰς παραλλήλους τὰς ΑΔ, ΕΓ εὐθεῖα ἐνέπεσεν ἡ ΑΓ, ἡ ἄρα ὑπὸ ΑΓΕ γωνία ἴση ἐστὶ τῇ ὑπὸ ΓΑΔ. ἀλλʼ ἡ ὑπὸ ΓΑΔ τῇ ὑπὸ ΒΑΔ ὑπόκειται ἴση· καὶ ἡ ὑπὸ ΒΑΔ ἄρα τῇ ὑπὸ ΑΓΕ ἐστιν ἴση.
Let ABC be a triangle, and let the angle BAC be bisected by the straight line AD; I say that as BD is to CD, so is BA to AC. For let CE be drawn through C parallel to DA, and let BA produced meet it at E. And since the straight line AC has fallen upon the parallel straight lines AD, EC, therefore the angle ACE is equal to the angle CAD. But the angle CAD is by hypothesis equal to the angle BAD; therefore the angle BAD is also equal to the angle ACE.
πάλιν, ἐπεὶ εἰς παραλλήλους τὰς ΑΔ, ΕΓ εὐθεῖα ἐνέπεσεν ἡ ΒΑΕ, ἡ ἐκτὸς γωνία ἡ ὑπὸ ΒΑΔ ἴση ἐστὶ τῇ ἐντὸς τῇ ὑπὸ ΑΕΓ. ἐδείχθη δὲ καὶ ἡ ὑπὸ ΑΓΕ τῇ ὑπὸ ΒΑΔ ἴση·
Again, since the straight line BAE has fallen upon the parallel straight lines AD, EC, the exterior angle BAD is equal to the interior angle AEC.
καὶ ἡ ὑπὸ ΑΓΕ ἄρα γωνία τῇ ὑπὸ ΑΕΓ ἐστιν ἴση· ὥστε καὶ πλευρὰ ἡ ΑΕ πλευρᾷ τῇ ΑΓ ἐστιν ἴση.
But the angle ACE was also proved equal to the angle BAD; therefore the angle ACE is also equal to the angle AEC, so that the side AE is also equal to the side AC.
καὶ ἐπεὶ τριγώνου τοῦ ΒΓΕ παρὰ μίαν τῶν πλευρῶν τὴν ΕΓ ἦκται ἡ ΑΔ, ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΒΔ πρὸς τὴν ΔΓ, οὕτως ἡ ΒΑ πρὸς τὴν ΑΕ. ἴση δὲ ἡ ΑΕ τῇ ΑΓ· ὡς ἄρα ἡ ΒΔ πρὸς τὴν ΔΓ, οὕτως ἡ ΒΑ πρὸς τὴν ΑΓ.
ἀλλὰ δὴ ἔστω ὡς ἡ ΒΔ πρὸς τὴν ΔΓ, οὕτως ἡ ΒΑ πρὸς τὴν ΑΓ, καὶ ἐπεζεύχθω ἡ ΑΔ· λέγω, ὅτι δίχα τέτμηται ἡ ὑπὸ ΒΑΓ γωνία ὑπὸ τῆς ΑΔ εὐθείας.
And since AD has been drawn parallel to EC, one of the sides of the triangle BCE, therefore, proportionally, as BD is to DC, so is BA to AE. But AE is equal to AC; therefore, as BD is to DC, so is BA to AC. But indeed let as BD is to DC, so BA to AC, and let AD be joined; I say that the angle BAC has been bisected by the straight line AD.
τῶν γὰρ αὐτῶν κατασκευασθέντων, ἐπεί ἐστιν ὡς ἡ ΒΔ πρὸς τὴν ΔΓ, οὕτως ἡ ΒΑ πρὸς τὴν ΑΓ, ἀλλὰ καὶ ὡς ἡ ΒΔ πρὸς τὴν ΔΓ, οὕτως ἐστὶν ἡ ΒΑ πρὸς τὴν ΑΕ· τριγώνου γὰρ τοῦ ΒΓΕ παρὰ μίαν τὴν ΕΓ ἦκται ἡ ΑΔ·
For, with the same construction, since as BD is to DC, so is BA to AC, but also as BD is to DC, so is BA to AE—for AD has been drawn parallel to EC, one of the sides of the triangle BCE—therefore also, as BA is to AC, so is BA to AE.
καὶ ὡς ἄρα ἡ ΒΑ πρὸς τὴν ΑΓ, οὕτως ἡ ΒΑ πρὸς τὴν ΑΕ. ἴση ἄρα ἡ ΑΓ τῇ ΑΕ· ὥστε καὶ γωνία ἡ ὑπὸ ΑΕΓ τῇ ὑπὸ ΑΓΕ ἐστιν ἴση.
Therefore AC is equal to AE; so that the angle AEC is also equal to the angle ACE.
ἀλλʼ ἡ μὲν ὑπὸ ΑΕΓ τῇ ἐκτὸς τῇ ὑπὸ ΒΑΔ ἴση, ἡ δὲ ὑπὸ ΑΓΕ τῇ ἐναλλὰξ τῇ ὑπὸ ΓΑΔ ἐστιν ἴση· καὶ ἡ ὑπὸ ΒΑΔ ἄρα τῇ ὑπὸ ΓΑΔ ἐστιν ἴση.
But the angle AEC is equal to the exterior angle BAD, and the angle ACE is equal to the alternate angle CAD; therefore the angle BAD is also equal to the angle CAD.
ἡ ἄρα ὑπὸ ΒΑΓ γωνία δίχα τέτμηται ὑπὸ τῆς ΑΔ εὐθείας.
Therefore the angle BAC has been bisected by the straight line AD.
ἐὰν ἄρα τριγώνου ἡ γωνία δίχα τμηθῇ, ἡ δὲ τέμνουσα τὴν γωνίαν εὐθεῖα τέμνῃ καὶ τὴν βάσιν, τὰ τῆς βάσεως τμήματα τὸν αὐτὸν ἕξει λόγον ταῖς λοιπαῖς τοῦ τριγώνου πλευραῖς· καὶ ἐὰν τὰ τῆς βάσεως τμήματα τὸν αὐτὸν ἔχῃ λόγον ταῖς λοιπαῖς τοῦ τριγώνου πλευραῖς, ἡ ἀπὸ τῆς κορυφῆς ἐπὶ τὴν τομὴν ἐπιζευγνυμένη εὐθεῖα δίχα τέμνει τὴν τοῦ τριγώνου γωνίαν· ὅπερ ἔδει δεῖξαι.
Therefore, if an angle of a triangle be bisected, and the straight line cutting the angle cut the base also, the segments of the base will have the same ratio to the remaining sides of the triangle; and, if the segments of the base have the same ratio to the remaining sides of the triangle, the straight line joined from the vertex to the point of section will bisect the angle of the triangle; which was to be proved.