§6.17ιζ΄.
17.
Ἀλλὰ δὴ μὴ ἔστω παράλληλος ἡ ΑΒ τῇ Γ∠, ἀλλὰ συμπιπτέτω κατὰ τὸ Ν.
ἐπεὶ οὖν ἀπὸ τοῦ αὐτοῦ σημείου τοῦ ∠ εἰς τρεῖς εὐθείας τὰς ΒΝ, ΒΓ, ΒΖ δύο εὐθεῖαι διηγμέναι εἰσὶν αἱ ∠Ε, ∠Ν, ἔστιν, ὡς τὸ ὑπὸ Ν∠, ΓΖ πρὸς τὸ ὑπὸ ΝΓ, ∠Ζ, οὕτως τὸ ὑπὸ ∠Ε, ΚΛ πρὸς τὸ ὑπὸ ΕΛ, Κ∠.
But indeed, let ΑΒ not be parallel to Γ∠, but let them meet at point Ν. Since therefore, from the same point ∠ to three straight lines ΒΝ, ΒΓ, ΒΖ, two straight lines ∠Ε, ∠Ν have been drawn, it follows that, as the rectangle contained by Ν∠, ΓΖ is to the rectangle contained by ΝΓ, ∠Ζ, so is the rectangle contained by ∠Ε, ΚΛ to the rectangle contained by ΕΛ, Κ∠.
ὡς δὲ τὸ ὑπὸ Ε∠, ΚΛ πρὸς τὸ ὑπὸ ΕΛ, Κ∠, οὕτως ἐστὶν τὸ ὑπὸ ΕΘ, ΓΗ πρὸς τὸ ὑπὸ ΕΓ, ΘΗ· πάλιν γὰρ εἰς τρεῖς τὰς ΓΛ, ∠Θ, ΗΚ ἀπὸ τοῦ αὐτοῦ σημείου τοῦ Ε δύο ἠγμέναι εἰσὶν αἱ ΕΓ, Ε∠·
And as the rectangle contained by Ε∠, ΚΛ is to the rectangle contained by ΕΛ, Κ∠, so is the rectangle contained by ΕΘ, ΓΗ to the rectangle contained by ΕΓ, ΘΗ; for again to three straight lines ΓΛ, ∠Θ, ΗΚ, from the same point Ε, two straight lines ΕΓ, Ε∠ have been drawn.
καὶ ὡς ἄρα τὸ ὑπὸ ΕΘ, ΓΗ πρὸς τὸ ὑπὸ ΕΓ, ΘΗ, οὕτως τὸ ὑπὸ Ν∠, ΓΖ πρὸς τὸ ὑπὸ ΝΓ, Ζ∠.
Therefore also, as the rectangle contained by ΕΘ, ΓΗ is to the rectangle contained by ΕΓ, ΘΗ, so is the rectangle contained by Ν∠, ΓΖ to the rectangle contained by ΝΓ, Ζ∠.
διὰ δὴ τὸ προγεγραμμένον εὐθεῖά ἐστιν ἡ διὰ τῶν Α, Θ, ∠· καὶ ἡ διὰ τῶν Α, Μ, Δ ἄρα εὐθεῖά ἐστιν.
Therefore, by what was written before, the line through Α, Θ, ∠ is a straight line; therefore also, the line through Α, Μ, Δ is a straight line.
§6.18ιη΄.
18.
Τρίγωνον τὸ ΑΒΓ, καὶ τῇ ΒΓ παράλληλος ἤχθω ἡ Α∠, καὶ διήχθωσαν αἱ ∠Ε, ΖΗ, ἔστω δέ, ὡς τὸ ἀπὸ ΕΒ πρὸς τὸ ὑπὸ ΕΓΒ, οὕτως ἡ ΒΗ πρὸς τὴν ΗΓ· ὅτι, ἐὰν ἐπιζευχθῇ ἡ Β∠, γίνεται εὐθεῖα ἡ διὰ τῶν Θ, Κ, Γ.
ἐπεί ἔστιν, ὡς τὸ ἀπὸ τῆς ΕΒ πρὸς τὸ ὑπὸ ΕΓΒ, οὕτως ἡ ΒH πρὸς ΗΓ, κοινὸς ἄρα προσκείσθω ὁ τῆς ΓΕ πρὸς ΕΒ λόγος ὁ αὐτὸς ὢν τῷ τοῦ ὑπὸ ΕΓΒ πρὸς τὸ ὑπὸ ΚΒΓ· δι᾿ ἴσου ἄρα ὁ τοῦ ἀπὸ ΕΒ πρὸς τὸ ὑπὸ ΕΒΓ λόγος, τουτέστιν ὁ τῆς ΕΒ πρὸς τὴν ΒΓ, ὁ αὐτός ἐστιν τῷ συνημμένῳ ἔκ τε τοῦ τῆς ΒΗ πρὸς ΗΓ καὶ τοῦ τοῦ ὑπὸ ΕΓΒ πρὸς τὸ ὑπὸ ΕΒΓ, ὅς ἔστιν ὁ αὐτὸς τῷ τῆς ΕΓ πρὸς ΕΒ·
Let there be a triangle ΑΒΓ, and let Α∠ be drawn parallel to ΒΓ, and let ∠Ε, ΖΗ be drawn, and let it be that, as the square on ΕΒ is to the rectangle contained by ΕΓ, ΓΒ, so is ΒΗ to ΗΓ; (to prove) that, if Β∠ is joined, the line through Θ, Κ, Γ becomes a straight line. Since, as the square on ΕΒ is to the rectangle contained by ΕΓ, ΓΒ, so is ΒΗ to ΗΓ, let the common ratio of ΓΕ to ΕΒ be added, which is the same as the ratio of the rectangle contained by ΕΓ, ΓΒ to the rectangle contained by ΚΒ, ΒΓ; therefore, by equality (δι' ἴσου), the ratio of the square on ΕΒ to the rectangle contained by ΕΒ, ΒΓ, that is, the ratio of ΕΒ to ΒΓ, is the same as the ratio compounded of that of ΒΗ to ΗΓ and that of the rectangle contained by ΕΓ, ΓΒ to the rectangle contained by ΕΒ, ΒΓ, which is the same as the ratio of ΕΓ to ΕΒ.
ὥστε ὁ τοῦ ἀπὸ ΕΒ πρὸς τὸ ὑπὸ ΕΒΓ συνῆπται ἔκ τε τοῦ ὃν ἔχει ἡ ΒΗ πρὸς ΗΓ καὶ τοῦ ὃν ἔχει ἡ ΕΓ πρὸς ΕΒ, ὅς ἐστιν ὁ αὐτὸς τῷ τοῦ ὑπὸ ΕΓ, ΒΗ πρὸς τὸ ὑπὸ ΕΒ, ΓΗ. ὡς δὲ ἡ ΕΒ πρὸς τὴν ΒΓ, οὕτως ἐστὶν διὰ τὸ προγεγραμμένον λῆμμα τὸ ὑπὸ ∠Ε, ΖΘ πρὸς τὸ ὑπὸ ∠Ζ, ΘΕ·
Consequently, the ratio of the square on ΕΒ to the rectangle contained by ΕΒ, ΒΓ is compounded of that which ΒΗ has to ΗΓ and of that which ΕΓ has to ΕΒ, which is the same as the ratio of the rectangle contained by ΕΓ, ΒΗ to the rectangle contained by ΕΒ, ΓΗ. And as ΕΒ is to ΒΓ, so is, by the lemma written before, the rectangle contained by ∠Ε, ΖΘ to the rectangle contained by ∠Ζ, ΘΕ.
καὶ ὡς ἄρα τὸ ὑπὸ ΓΕ, ΒΗ πρὸς τὸ ὑπὸ ΓΗ, ΕΒ, οὕτως ἐστὶν τὸ ὑπὸ ∠Ε, ΖΘ πρὸς τὸ ὑπὸ ∠Ζ, ΘΕ. εὐθεῖα ἄρα ἐστὶν ἡ διὰ τῶν Θ, Κ, Γ· τοῦτο γὰρ ἐν τοῖς πτωτικοῖς τῶν ἀναστροφίων.
Therefore also, as the rectangle contained by ΓΕ, ΒΗ is to the rectangle contained by ΓΗ, ΕΒ, so is the rectangle contained by ∠Ε, ΖΘ to the rectangle contained by ∠Ζ, ΘΕ. Therefore, the line through Θ, Κ, Γ is a straight line; for this is among the cases of the conversions.
§6.19ιθ΄.
19.
Εἰς τρεῖς εὐθείας τὰς ΑΒ, ΑΓ, Α∠ ἀπό τινος σημείου τοῦ Ε δύο διήχθωσαν αἱ ΕΖ, ΕΒ, ἔστω δέ, ὡς ἡ ΕΖ πρὸς τὴν ΖΗ, οὕτως ἡ ΘΕ πρὸς τὴν ΘΗ ὅτι γίνεται καί, ὡς ἡ ΒΕ πρὸς τὴν ΒΓ, οὕτως ἡ Ε∠ πρὸς τὴν ∠Γ.
ἤχθω διὰ τοῦ Η τῇ ΒΕ παράλληλος ἡ ΛΚ.
ἐπεὶ οὖν ἐστιν, ὡς ἡ ΕΖ πρὸς τὴν ΖΗ, οὕτως ἡ ΕΘ πρὸς τὴν ΘΗ, ἀλλ᾿ ὡς μὲν ἡ ΕΖ πρὸς τὴν ΖΗ, οὕτως ἡ ΕΒ πρὸς τὴν ΗΚ, ὡς δὲ ἡ ΕΘ πρὸς τὴν ΘΗ, οὕτως ἐστὶν ἡ ∠Ε πρὸς τὴν ΗΛ, καὶ ὡς ἄρα ἡ ΒΕ πρὸς τὴν ΗΚ, οὕτως ἐστὶν ἡ ∠Ε πρὸς τὴν ΗΛ. ἐναλλάξ ἐστιν, ὡς ἡ ΕΒ πρὸς τὴν Ε∠, οὕτως ἡ Κ πρὸς τὴν ΗΛ. ὡς δὲ ἡ ΚΗ πρὸς τὴν ΗΛ, οὕτως ἐστὶν ἡ ΒΓ πρὸς τὴν Γ∠ καὶ ὡς ἄρα ἡ ΒΕ πρὸς τὴν Ε∠, οὕτως ἡ ΒΓ πρὸς τὴν Γ∠.
To three straight lines ΑΒ, ΑΓ, Α∠, from some point Ε, let two straight lines ΕΖ, ΕΒ be drawn, and let it be that, as ΕΖ is to ΖΗ, so is ΘΕ to ΘΗ; (to prove) that it also follows that, as ΒΕ is to ΒΓ, so is Ε∠ to ∠Γ. Let ΛΚ be drawn through Η parallel to ΒΕ. Since therefore, as ΕΖ is to ΖΗ, so is ΕΘ to ΘΗ, but as ΕΖ is to ΖΗ, so is ΕΒ to ΗΚ, and as ΕΘ is to ΘΗ, so is ∠Ε to ΗΛ, therefore also, as ΒΕ is to ΗΚ, so is ∠Ε to ΗΛ. Alternately, as ΕΒ is to Ε∠, so is Κ to ΗΛ. And as ΚΗ is to ΗΛ, so is ΒΓ to Γ∠; therefore also, as ΒΕ is to Ε∠, so is ΒΓ to Γ∠.
ἐναλλάξ ἐστιν, ὡς ἡ ΕΒ πρὸς τὴν ΒΓ, οὕτως ἡ Ε∠ πρὸς τὴν ∠Γ.
τὰ δὲ πτωτικὰ ὁμοίως.
Alternately, as ΒΕ is to ΒΓ, so is Ε∠ to ∠Γ. And the other cases likewise.