# Book 6 Proposition 10

Τὴν δοθεῖσαν εὐθεῖαν ἄτμητον τῇ δοθείσῃ τετμημένῃ ὁμοίως τεμεῖν. Ἔστω ἡ μὲν δοθεῖσα εὐθεῖα ἄτμητος ἡ ΑΒ, ἡ δὲ τετμημένη ἡ ΑΓ κατὰ τὰ Δ, Ε σημεῖα, καὶ κείσθωσαν ὥστε γωνίαν τυχοῦσαν περιέχειν, καὶ ἐπεζεύχθω ἡ ΓΒ, καὶ διὰ τῶν Δ, Ε τῇ ΒΓ παράλληλοι ἤχθωσαν αἱ ΔΖ, ΕΗ, διὰ δὲ τοῦ Δ τῇ ΑΒ παράλληλος ἤχθω ἡ ΔΘΚ. Παραλληλόγραμμον ἄρα ἐστὶν ἑκάτερον τῶν ΖΘ, ΘΒ: ἴση ἄρα ἡ μὲν ΔΘ τῇ ΖΗ, ἡ δὲ ΘΚ τῇ ΗΒ. καὶ ἐπεὶ τριγώνου τοῦ ΔΚΓ παρὰ μίαν τῶν πλευρῶν τὴν ΚΓ εὐθεῖα ἦκται ἡ ΘΕ, ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΓΕ πρὸς τὴν ΕΔ, οὕτως ἡ ΚΘ πρὸς τὴν ΘΔ. ἴση δὲ ἡ μὲν ΚΘ τῇ ΒΗ, ἡ δὲ ΘΔ τῇ ΗΖ. ἔστιν ἄρα ὡς ἡ ΓΕ πρὸς τὴν ΕΔ, οὕτως ἡ ΒΗ πρὸς τὴν ΗΖ. πάλιν, ἐπεὶ τριγώνου τοῦ ΑΗΕ παρὰ μίαν τῶν πλευρῶν τὴν ΗΕ ἦκται ἡ ΖΔ, ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΕΔ πρὸς τὴν ΔΑ, οὕτως ἡ ΗΖ πρὸς τὴν ΖΑ. ἐδείχθη δὲ καὶ ὡς ἡ ΓΕ πρὸς τὴν ΕΔ, οὕτως ἡ ΒΗ πρὸς τὴν ΗΖ: ἔστιν ἄρα ὡς μὲν ἡ ΓΕ πρὸς τὴν ΕΔ, οὕτως ἡ ΒΗ πρὸς τὴν ΗΖ, ὡς δὲ ἡ ΕΔ πρὸς τὴν ΔΑ, οὕτως ἡ ΗΖ πρὸς τὴν ΖΑ. Ἡ ἄρα δοθεῖσα εὐθεῖα ἄτμητος ἡ ΑΒ τῇ δοθείσῃ εὐθείᾳ τετμημένῃ τῇ ΑΓ ὁμοίως τέτμηται: ὅπερ ἔδει ποιῆσαι.

To cut a given uncut straight line similarly to a given cut straight line. Let AB be the given uncut straight line, and AC the straight line cut at the points D, E; and let them be so placed as to contain any angle; let CB be joined, and through D, E let DF, EG be drawn parallel to BC, and through D let DHK be drawn parallel to AB. [I. 31] Therefore each of the figures FH, HB is a parallelogram; therefore DH is equal to FG and HK to GB. [I. 34] Now, since the straight line HE has been drawn parallel to KC, one of the sides of the triangle DKC, therefore, proportionally, as CE is to ED, so is KH to HD. [VI. 2] But KH is equal to BG, and HD to GF; therefore, as CE is to ED, so is BG to GF. Again, since FD has been drawn parallel to GE, one of the sides of the triangle AGE, therefore, proportionally, as ED is to DA, so is GF to FA. [VI. 2] But it was also proved that, as CE is to ED, so is BG to GF; therefore, as CE is to ED, so is BG to GF, and, as ED is to DA, so is GF to FA.