From a given straight line to cut off a prescribed part.
|Τῆς δοθείσης εὐθείας τὸ προσταχθὲν μέρος ἀφελεῖν. Ἔστω ἡ δοθεῖσα εὐθεῖα ἡ ΑΒ: δεῖ δὴ τῆς ΑΒ τὸ προσταχθὲν μέρος ἀφελεῖν. Ἐπιτετάχθω δὴ τὸ τρίτον. [ καὶ ] διήχθω τις ἀπὸ τοῦ Α εὐθεῖα ἡ ΑΓ γωνίαν περιέχουσα μετὰ τῆς ΑΒ τυχοῦσαν: καὶ εἰλήφθω τυχὸν σημεῖον ἐπὶ τῆς ΑΓ τὸ Δ, καὶ κείσθωσαν τῇ ΑΔ ἴσαι αἱ ΔΕ, ΕΓ. καὶ ἐπεζεύχθω ἡ ΒΓ, καὶ διὰ τοῦ Δ παράλληλος αὐτῇ ἤχθω ἡ ΔΖ. Ἐπεὶ οὖν τριγώνου τοῦ ΑΒΓ παρὰ μίαν τῶν πλευρῶν τὴν ΒΓ ἦκται ἡ ΖΔ, ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΓΔ πρὸς τὴν ΔΑ, οὕτως ἡ ΒΖ πρὸς τὴν ΖΑ. διπλῆ δὲ ἡ ΓΔ τῆς ΔΑ: διπλῆ ἄρα καὶ ἡ ΒΖ τῆς ΖΑ: τριπλῆ ἄρα ἡ ΒΑ τῆς ΑΖ. Τῆς ἄρα δοθείσης εὐθείας τῆς ΑΒ τὸ ἐπιταχθὲν τρίτον μέρος ἀφῄρηται τὸ ΑΖ: ὅπερ ἔδει ποιῆσαι.||From a given straight line to cut off a prescribed part. Let AB be the given straight line; thus it is required to cut off from AB a prescribed part. Let the third part be that prescribed. Let a straight line AC be drawn through from A containing with AB any angle; let a point D be taken at random on AC, and let DE, EC be made equal to AD. [I. 3] Let BC be joined, and through D let DF be drawn parallel to it. [I. 31] Then, since FD has been drawn parallel to BC, one of the sides of the triangle ABC, therefore, proportionally, as CD is to DA, so is BF to FA. [VI. 2] But CD is double of DA; therefore BF is also double of FA; therefore BA is triple of AF.|