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To three given straight lines to find a fourth proportional.
| Τριῶν δοθεισῶν εὐθειῶν τετάρτην ἀνάλογον προσευρεῖν. Ἔστωσαν αἱ δοθεῖσαι τρεῖς εὐθεῖαι αἱ Α, Β, Γ: δεῖ δὴ τῶν Α, Β, Γ τετάρτην ἀνάλογον προσευρεῖν. Ἐκκείσθωσαν δύο εὐθεῖαι αἱ ΔΕ, ΔΖ γωνίαν περιέχουσαι [ τυχοῦσαν ] τὴν ὑπὸ ΕΔΖ: καὶ κείσθω τῇ μὲν Α ἴση ἡ ΔΗ, τῇ δὲ Β ἴση ἡ ΗΕ, καὶ ἔτι τῇ Γ ἴση ἡ ΔΘ: καὶ ἐπιζευχθείσης τῆς ΗΘ παράλληλος αὐτῇ ἤχθω διὰ τοῦ Ε ἡ ΕΖ. Ἐπεὶ οὖν τριγώνου τοῦ ΔΕΖ παρὰ μίαν τὴν ΕΖ ἦκται ἡ ΗΘ, ἔστιν ἄρα ὡς ἡ ΔΗ πρὸς τὴν ΗΕ, οὕτως ἡ ΔΘ πρὸς τὴν ΘΖ. ἴση δὲ ἡ μὲν ΔΗ τῇ Α, ἡ δὲ ΗΕ τῇ Β, ἡ δὲ ΔΘ τῇ Γ: ἔστιν ἄρα ὡς ἡ Α πρὸς τὴν Β, οὕτως ἡ Γ πρὸς τὴν ΘΖ. Τριῶν ἄρα δοθεισῶν εὐθειῶν τῶν Α, Β, Γ τετάρτη ἀνάλογον προσεύρηται ἡ ΘΖ: ὅπερ ἔδει ποιῆσαι. | To three given straight lines to find a fourth proportional. Let A, B, C be the three given straight lines; thus it is required to find a fourth proportional to A, B, C. Let two straight lines DE, DF be set out containing any angle EDF; let DG be made equal to A, GE equal to B, and further DH equal to C; let GH be joined, and let EF be drawn through E parallel to it. [I. 31] Since, then, GH has been drawn parallel to EF, one of the sides of the triangle DEF, therefore, as DG is to GE, so is DH to HF. [VI. 2] But DG is equal to A, GE to B, and DH to C; therefore, as A is to B, so is C to HF. |