§6.prop.2ἐὰν τριγώνου παρὰ μίαν τῶν πλευρῶν ἀχθῇ τις εὐθεῖα, ἀνάλογον τεμεῖ τὰς τοῦ τριγώνου πλευράς· καὶ ἐὰν αἱ τοῦ τριγώνου πλευραὶ ἀνάλογον τμηθῶσιν, ἡ ἐπὶ τὰς τομὰς ἐπιζευγνυμένη εὐθεῖα παρὰ τὴν λοιπὴν ἔσται τοῦ τριγώνου πλευράν.
If a straight line be drawn parallel to one of the sides of a triangle, it will cut the sides of the triangle proportionally; and, if the sides of the triangle be cut proportionally, the straight line joining the points of section will be parallel to the remaining side of the triangle.
τριγώνου γὰρ τοῦ ΑΒΓ παράλληλος μιᾷ τῶν πλευρῶν τῇ ΒΓ ἤχθω ἡ ΔΕ· λέγω, ὅτι ἐστὶν ὡς ἡ ΒΔ πρὸς τὴν ΔΑ, οὕτως ἡ ΓΕ πρὸς τὴν ΕΑ.
ἐπεζεύχθωσαν γὰρ αἱ ΒΕ, ΓΔ.
ἴσον ἄρα ἐστὶ ΒΔΕ τρίγωνον τῷ ΓΔΕ τριγώνῳ· ἐπὶ γὰρ τῆς αὐτῆς βάσεώς ἐστι τῆς ΔΕ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ταῖς ΔΕ, ΒΓ·
For let DE be drawn parallel to one of the sides BC of the triangle ABC; I say that as BD is to DA, so is CE to EA. For let BE, CD be joined. Therefore the triangle BDE is equal to the triangle CDE; for it is on the same base DE and in the same parallels DE, BC.
ἄλλο δέ τι τὸ ΑΔΕ τρίγωνον.
And the triangle ADE is some other (magnitude).
τὰ δὲ ἴσα πρὸς τὸ αὐτὸ τὸν αὐτὸν ἔχει λόγον· ἔστιν ἄρα ὡς τὸ ΒΔΕ τρίγωνον πρὸς τὸ ΑΔΕ, οὕτως τὸ ΓΔΕ τρίγωνον πρὸς τὸ ΑΔΕ τρίγωνον.
And equal (magnitudes) have to the same (magnitude) the same ratio; therefore, as the triangle BDE is to the triangle ADE, so is the triangle CDE to the triangle ADE.
ἀλλʼ ὡς μὲν τὸ ΒΔΕ τρίγωνον πρὸς τὸ ΑΔΕ, οὕτως ἡ ΒΔ πρὸς τὴν ΔΑ· ὑπὸ γὰρ τὸ αὐτὸ ὕψος ὄντα τὴν ἀπὸ τοῦ Ε ἐπὶ τὴν ΑΒ κάθετον ἀγομένην πρὸς ἄλληλά εἰσιν ὡς αἱ βάσεις.
But as the triangle BDE is to ADE, so is BD to DA; for, being under the same height—the perpendicular drawn from E to AB—they are to one another as their bases.
διὰ τὰ αὐτὰ δὴ ὡς τὸ ΓΔΕ τρίγωνον πρὸς τὸ ΑΔΕ, οὕτως ἡ ΓΕ πρὸς τὴν ΕΑ· καὶ ὡς ἄρα ἡ ΒΔ πρὸς τὴν ΔΑ, οὕτως ἡ ΓΕ πρὸς τὴν ΕΑ.
ἀλλὰ δὴ αἱ τοῦ ΑΒΓ τριγώνου πλευραὶ αἱ ΑΒ, ΑΓ ἀνάλογον τετμήσθωσαν, ὡς ἡ ΒΔ πρὸς τὴν ΔΑ, οὕτως ἡ ΓΕ πρὸς τὴν ΕΑ, καὶ ἐπεζεύχθω ἡ ΔΕ· λέγω, ὅτι παράλληλός ἐστιν ἡ ΔΕ τῇ ΒΓ.
τῶν γὰρ αὐτῶν κατασκευασθέντων, ἐπεί ἐστιν ὡς ἡ ΒΔ πρὸς τὴν ΔΑ, οὕτως ἡ ΓΕ πρὸς τὴν ΕΑ, ἀλλʼ ὡς μὲν ἡ ΒΔ πρὸς τὴν ΔΑ, οὕτως τὸ ΒΔΕ τρίγωνον πρὸς τὸ ΑΔΕ τρίγωνον, ὡς δὲ ἡ ΓΕ πρὸς τὴν ΕΑ, οὕτως τὸ ΓΔΕ τρίγωνον πρὸς τὸ ΑΔΕ τρίγωνον, καὶ ὡς ἄρα τὸ ΒΔΕ τρίγωνον πρὸς τὸ ΑΔΕ τρίγωνον, οὕτως τὸ ΓΔΕ τρίγωνον πρὸς τὸ ΑΔΕ τρίγωνον.
For the same reason, as the triangle CDE is to ADE, so is CE to EA; therefore also, as BD is to DA, so is CE to EA. But indeed let the sides AB, AC of the triangle ABC be cut proportionally, as BD to DA, so CE to EA, and let DE be joined; I say that DE is parallel to BC. For, with the same construction, since as BD is to DA, so is CE to EA, but as BD is to DA, so is the triangle BDE to the triangle ADE, and as CE is to EA, so is the triangle CDE to the triangle ADE, therefore also, as the triangle BDE is to the triangle ADE, so is the triangle CDE to the triangle ADE.
ἑκάτερον ἄρα τῶν ΒΔΕ, ΓΔΕ τριγώνων πρὸς τὸ ΑΔΕ τὸν αὐτὸν ἔχει λόγον.
Therefore each of the triangles BDE, CDE has the same ratio to ADE.
ἴσον ἄρα ἐστὶ τὸ ΒΔΕ τρίγωνον τῷ ΓΔΕ τριγώνῳ·
Therefore the triangle BDE is equal to the triangle CDE; and they are on the same base DE.
καί εἰσιν ἐπὶ τῆς αὐτῆς βάσεως τῆς ΔΕ. τὰ δὲ ἴσα τρίγωνα καὶ ἐπὶ τῆς αὐτῆς βάσεως ὄντα καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν.
And equal triangles which are also on the same base are also in the same parallels.
παράλληλος ἄρα ἐστὶν ἡ ΔΕ τῇ ΒΓ.
ἐὰν ἄρα τριγώνου παρὰ μίαν τῶν πλευρῶν ἀχθῇ τις εὐθεῖα, ἀνάλογον τεμεῖ τὰς τοῦ τριγώνου πλευράς· καὶ ἐὰν αἱ τοῦ τριγώνου πλευραὶ ἀνάλογον τμηθῶσιν, ἡ ἐπὶ τὰς τομὰς ἐπιζευγνυμένη εὐθεῖα παρὰ τὴν λοιπὴν ἔσται τοῦ τριγώνου πλευράν· ὅπερ ἔδει δεῖξαι.
Therefore DE is parallel to BC. Therefore, if a straight line be drawn parallel to one of the sides of a triangle, it will cut the sides of the triangle proportionally; and, if the sides of the triangle be cut proportionally, the straight line joining the points of section will be parallel to the remaining side of the triangle; which was to be proved.