§6.12ιβ΄.
12.
Ἀποδεδειγμένων οὖν τούτων ἔσται δεῖξαι, ὅτι, ἐὰν παράλληλοι ὦσιν αἱ ΑΒ, Γ∠, καὶ εἰς αὐτὰς ἐμπίπτωσιν εὐθεῖαί τινες αἱ Α∠, Α Ζ, ΒΓ ΒΖ, καὶ ἐπιζευχθῶσιν αἱ Ε∠, ΕΓ, γίνεται εὐθεῖα ἡ διὰ τῶν Η, Μ, Κ.
ἐπεὶ γὰρ τρίγωνον τὸ ∠ΑΖ, καὶ τῇ ∠ Ζ παράλληλος ἡ ΑΕ, καὶ διῆκται ἡ ΕΓ συμπίπτουσα τῇ ∠ κατὰ τὸ Γ, διὰ τὸ προγεγραμμένον γίνεται, ὡς ἡ ∠ Ζ πρὸς τὴν ΖΓ, οὕτως τὸ ὑπὸ ΓΕ, ΗΘ πρὸς τὸ ὑπὸ ΓΗ, ΘΕ πάλιν, ἐπεὶ τρίγωνόν ἐστιν τὸ Γ ΒΖ, καὶ τῇ Γ∠ παράλληλος ἦκται ἡ ΒΕ, καὶ διῆκται ἡ ∠Ε συμπίπτουσα τῇ ΓΖ∠ κατὰ τὸ ∠, γίνεται, ὡς ἡ ΓΖ πρὸς τὴν Ζ∠, οὕτως τὸ ὑπὸ ∠Ε, ΛΚ πρὸς τὸ ὑπὸ ∠Κ, ΛΕ·
Therefore, these things having been proved, it will be possible to prove that, if ΑΒ, Γ∠ are parallel, and some straight lines Α∠, ΑΖ, ΒΓ, ΒΖ fall upon them, and Ε∠, ΕΓ are joined, the line through Η, Μ, Κ becomes a straight line. For since ∠ΑΖ is a triangle, and ΑΕ is drawn parallel to ∠Ζ, and ΕΓ is drawn meeting ∠Ζ at Γ, by what was written before it follows that, as ∠Ζ is to ΖΓ, so is the rectangle contained by ΓΕ, ΗΘ to the rectangle contained by ΓΗ, ΘΕ. Again, since ΓΒΖ is a triangle, and ΒΕ is drawn parallel to Γ∠, and ∠Ε is drawn meeting ΓΖ at ∠, it follows that, as ΓΖ is to Ζ∠, so is the rectangle contained by ∠Ε, ΛΚ to the rectangle contained by ∠Κ, ΛΕ.
ἀνάπαλιν ἄρα γίνεται, ὡς ἡ ∠ Ζ πρὸς τὴν ΖΓ, οὕτως τὸ ὑπὸ ∠Κ, ΛΕ πρὸς τὸ ὑπὸ ∠Ε, ΛΚ. ἦν δὲ καί, ὡς ἡ ∠Ζ πρὸς τὴν Ζ οὕτως τὸ ὑπὸ ΓΕ, ΗΘ πρὸς τὸ ὑπὸ ΓΗ, ΘΕ·
Therefore, inversely, as ∠Ζ is to ΖΓ, so is the rectangle contained by ∠Κ, ΛΕ to the rectangle contained by ∠Ε, ΛΚ. But also, as ∠Ζ was to ΖΓ, so was the rectangle contained by ΓΕ, ΗΘ to the rectangle contained by ΓΗ, ΘΕ.
καὶ ὡς ἄρα τὸ ὑπὸ ΓΕ, ΗΘ πρὸς τὸ ὑπὸ ΓΗ, ΘΕ ὕτως ἐστὶν τὸ ὑπὸ ∠Κ, ΛΕ πρὸς τὸ ὑπὸ ∠Ε, ΚΛ. πεὶ οὖν εἰς δύο εὐθείας τὰς Γ ΜΛ, ΘΜ∠ δύο εὐθεῖαι διηγμέναι εἰσὶν αἱ ΕΓ, Ε∠, καί ἐστιν, ὡς τὸ ὑπὸ ΓΕ, ΗΘ πρὸς τὸ ὑπὸ ΓΗ, ΘΕ, οὕτωσ τὸ ὑπὸ ∠Κ, ΕΛ πρὸς τὸ ὑπὸ ∠Ε, ΛΚ, εὐθεῖα ἄρα ἐστὶν ἡ διὰ τῶν Η, Μ, Κ τοῦτο γὰρ προδέδεικται.
Therefore also, as the rectangle contained by ΓΕ, ΗΘ is to the rectangle contained by ΓΗ, ΘΕ, so is the rectangle contained by ∠Κ, ΛΕ to the rectangle contained by ∠Ε, ΚΛ. Since therefore to two straight lines ΓΜΛ, ΘΜ∠, two straight lines ΕΓ, Ε∠ have been drawn from Ε, and as the rectangle ΓΕ, ΗΘ is to the rectangle ΓΗ, ΘΕ, so is the rectangle ∠Κ, ΕΛ to the rectangle ∠Ε, ΛΚ, therefore the line through Η, Μ, Κ is a straight line; for this has been proved before.
§6.13ιγ΄.
13.
Ἀλλὰ δὴ μὴ ἔστωσαν αἱ ΑΒ, Γ∠ παράλληλοι, ἀλλὰ συμπιπτέτωσαν κατὰ τὸ Ν· ὅτι πάλιν εὐθεῖά ἐστιν ἡ διὰ τῶν Η, Μ, Κ.
ἐπεὶ εἰς τρεῖς εὐθείας τὰς ΑΝ, ΑΖ, Α∠ ἀπὸ τοῦ αὐτοῦ σημείου τοῦ Γ δύο διηγμέναι εἰσὶν αἱ ΓΕ Γ∠, γίνεται, ὡς τὸ ὑπὸ ΓΕ, ΗΘ πρὸς τὸ ὑπὸ ΓΗ, ΘΕ, οὕτως τὸ ὑπὸ τῶν ΓΝ, Ζ∠ πρὸς τὸ ὑπὸ τῶν Ν∠, Γ Ζ. πάλιν ἐπεὶ ἀπὸ τοῦ αὐτοῦ σημείου τοῦ ∠ εἰς τρεῖς εὐθείας τὰς ΒΝ, ΒΓ, ΒΖ δύο εἰσὶν διηγμέναι αἱ ∠Ε, ∠Ν, ἔστιν, ὡς τὸ ὑπὸ ΝΓ, Ζ∠ πρὸς τὸ ὑπὸ Ν∠, ΖΓ οὕτως τὸ ὑπὸ ∠Κ, ΕΛ πρὸς τὸ ὑπὸ ∠Ε, ΚΛ. ἀλλʼ ὡς τὸ ὑπὸ ΝΓ, Ζ∠ πρὸς τὸ ὑπὸ Ν∠, Γ Ζ, οὕτως ἐδείχθη τὸ ὑπὸ ΓΕ, πρὸς τὸ ὑπὸ ΓΗ, ΘΕ καὶ ὡς ἄρα τὸ ὑπὸ ΓΕ ΘΗ πρὸς τὸ ὑπὸ ΓΗ, ΘΕ, οὕτως ἐστὶν τὸ ὑπὸ ∠Κ, ΕΛ πρὸς τὸ ὑπὸ ∠Ε, ΚΛ. διὰ δὴ τὸ προγεγραμμένον εὐθεῖά ἐστιν ἡ διὰ τῶν Η, Μ, Κ.
But now let ΑΒ, Γ∠ not be parallel, but let them meet at Ν; (to prove) that again the line through Η, Μ, Κ is a straight line. For since from the same point Γ, to three straight lines ΑΝ, ΑΖ, Α∠, two straight lines ΓΕ, Γ∠ have been drawn, it follows that, as the rectangle ΓΕ, ΗΘ is to the rectangle ΓΗ, ΘΕ, so is the rectangle contained by ΓΝ, Ζ∠ to the rectangle contained by Ν∠, ΓΖ. Again, since from the same point ∠, to three straight lines ΒΝ, ΒΓ, ΒΖ, two straight lines ∠Ε, ∠Ν have been drawn, as the rectangle ΝΓ, Ζ∠ is to the rectangle Ν∠, ΖΓ, so is the rectangle ∠Κ, ΕΛ to the rectangle ∠Ε, ΚΛ. But as the rectangle ΝΓ, Ζ∠ is to the rectangle Ν∠, ΓΖ, so was shown the rectangle ΓΕ, ΗΘ to the rectangle ΓΗ, ΘΕ. Therefore also, as the rectangle ΓΕ, ΘΗ is to the rectangle ΓΗ, ΘΕ, so is the rectangle ∠Κ, ΕΛ to the rectangle ∠Ε, ΚΛ. Therefore, by what was written before, the line through Η, Μ, Κ is a straight line.
§6.14ιδʹ.
14.
Ἔστω παράλληλος ἡ ΑΒ τῇ Γ∠, καὶ διήχθωσαν αἱ ΑΕ, ΓΒ, καὶ σημεῖον ἐπὶ τῆς ΒΗ τὸ Ζ, ὥστε εἶναι, ὡς τὴν ∠Ε πρὸς τὴν ΕΓ, οὕτως τὸ ὑπὸ ΓΒ, ΗΖ πρὸς τὸ ὑπὸ ΖΒ, ΓΗ ὅτι εὐθεῖά ἐστιν ἡ διὰ τῶν Α, Ζ, ∠.
Let ΑΒ be parallel to Γ∠, and let ΑΕ, ΓΒ be drawn, and let there be a point Ζ on ΒΗ, so that as ∠Ε is to ΕΓ, so is the rectangle contained by ΓΒ, ΗΖ to the rectangle contained by ΖΒ, ΓΗ; (to prove) that the line through Α, Ζ, ∠ is a straight line.
ἤχθω διὰ μὲν τοῦ ∠ τῇ ΒΓ παράλληλος ἡ ∠Θ, καὶ ἐκβεβλήσθω ἡ ΑΕ ἐπὶ τὸ Θ, διὰ δὲ τοῦ Θ τῇ Γ∠ παράλληλος ἡ Θ Κ, καὶ ἐκβεβλήσθω ἡ Β ἐπὶ τὸ Κ.
ἐπεὶ οὖν ἐστιν, ὡς ἡ ∠Ε πρὸς τὴν ΕΓ. οὕτως τὸ ὑπὸ ΓΒ, Ζ πρὸς τὸ ὑπὸ ΒΖ, ΓΗ, ὡς δὲ ἡ ∠Ε πρὸς τὴν ΕΓ, οὕτως ἐστὶν ἥ τε ∠Θ πρὸς τὴν Γ καὶ τὸ ὑπὸ ∠Θ, ΒΖ πρὸς τὸ ὑπὸ τῶν ΓΗ, ἴσον ἄρα ἐστὶν τὸ ὑπὸ τῶν ΒΓ ΖΗ τῷ ὑπὸ ∠Θ, ΒΖ ἀνάλογον ἄρα ἐστίν, ὡς ἡ ΓΒ πρὸς τήν ΒΖ, οὕτως ἡ ∠Θ, τουτέστιν ἡ Γ Κ, πρὸς τὴν ΗΖ καὶ ὅλη ἄρα ἡ ΚΒ πρὸς ὅλην τὴν Β ἐστιν, ὡς ἡ ΚΓ πρὸς ΖΗ τουτέστιν ὡς ἡ ∠Θ πρὸς ΖΗ. ἀλλʼ ὡς ἡ ΚΒ πρὸς Β ἐν παραλλήλῳ, οὕτως ἐστὶν ἡ ΘΑ πρὸς ΑΗ καὶ ἡ ∠Θ πρὸς ΖΗ. καί εἰσιν παράλληλοι αἱ ∠Θ, ΖΗ· εὐθεῖα ἄρα ἐστὶν ἡ διὰ τῶν Α, Ζ, ∠ σημείων.
Let ∠Θ be drawn through ∠ parallel to ΒΓ, and let ΑΕ be produced to Θ, and through Θ let ΘΚ be drawn parallel to Γ∠, and let ΒΓ be produced to Κ. Since therefore, as ∠Ε is to ΕΓ, so is the rectangle contained by ΓΒ, ΗΖ to the rectangle contained by ΒΖ, ΓΗ, and as ∠Ε is to ΕΓ, so is ∠Θ to ΓΗ, and the rectangle contained by ∠Θ, ΒΖ to the rectangle contained by ΓΗ, ΒΖ, therefore the rectangle contained by ΒΓ, ΖΗ is equal to the rectangle contained by ∠Θ, ΒΖ. Therefore, proportionally, as ΓΒ is to ΒΖ, so is ∠Θ, that is ΓΚ, to ΗΖ; and therefore the whole ΚΒ is to the whole ΒΖ as ΚΓ is to ΖΗ, that is, as ∠Θ is to ΖΗ. But as ΚΒ is to ΒΖ, between parallels, so is ΘΑ to ΑΗ, and therefore ΘΑ to ΑΗ is as ∠Θ to ΖΗ. And ∠Θ, ΖΗ are parallel; therefore the line through the points Α, Ζ, ∠ is a straight line.