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## Translations

Ἐὰν εὐθεῖα γραμμὴ τμηθῇ, ὡς ἔτυχεν, τὸ ἀπὸ τῆς ὅλης καὶ τὸ ἀφ' ἑνὸς τῶν τμημάτων τὰ συναμφότερα τετράγωνα ἴσα ἐστὶ τῷ τε δὶς ὑπὸ τῆς ὅλης καὶ τοῦ εἰρημένου τμήματος περιεχομένῳ ὀρθογωνίῳ καὶ τῷ ἀπὸ τοῦ λοιποῦ τμήματος τετραγώνῳ. Εὐθεῖα γάρ τις ἡ ΑΒ τετμήσθω, ὡς ἔτυχεν, κατὰ τὸ Γ σημεῖον: λέγω, ὅτι τὰ ἀπὸ τῶν ΑΒ, ΒΓ τετράγωνα ἴσα ἐστὶ τῷ τε δὶς ὑπὸ τῶν ΑΒ, ΒΓ περιεχομένῳ ὀρθογωνίῳ καὶ τῷ ἀπὸ τῆς ΓΑ τετραγώνῳ. Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΑΒ τετράγωνον τὸ ΑΔΕΒ: καὶ καταγεγράφθω τὸ σχῆμα. Ἐπεὶ οὖν ἴσον ἐστὶ τὸ ΑΗ τῷ ΗΕ, κοινὸν προσκείσθω τὸ ΓΖ: ὅλον ἄρα τὸ ΑΖ ὅλῳ τῷ ΓΕ ἴσον ἐστίν: τὰ ἄρα ΑΖ, ΓΕ διπλάσιά ἐστι τοῦ ΑΖ. ἀλλὰ τὰ ΑΖ, ΓΕ ὁ ΚΛΜ ἐστι γνώμων καὶ τὸ ΓΖ τετράγωνον: ὁ ΚΛΜ ἄρα γνώμων καὶ τὸ ΓΖ διπλάσιά ἐστι τοῦ ΑΖ. ἔστι δὲ τοῦ ΑΖ διπλάσιον καὶ τὸ δὶς ὑπὸ τῶν ΑΒ, ΒΓ: ἴση γὰρ ἡ ΒΖ τῇ ΒΓ: ὁ ἄρα ΚΛΜ γνώμων καὶ τὸ ΓΖ τετράγωνον ἴσον ἐστὶ τῷ δὶς ὑπὸ τῶν ΑΒ, ΒΓ. κοινὸν προσκείσθω τὸ ΔΗ, ὅ ἐστιν ἀπὸ τῆς ΑΓ τετράγωνον: ὁ ἄρα ΚΛΜ γνώμων καὶ τὰ ΒΗ, ΗΔ τετράγωνα ἴσα ἐστὶ τῷ τε δὶς ὑπὸ τῶν ΑΒ, ΒΓ περιεχομένῳ ὀρθογωνίῳ καὶ τῷ ἀπὸ τῆς ΑΓ τετραγώνῳ. ἀλλὰ ὁ ΚΛΜ γνώμων καὶ τὰ ΒΗ, ΗΔ τετράγωνα ὅλον ἐστὶ τὸ ΑΔΕΒ καὶ τὸ ΓΖ, ἅ ἐστιν ἀπὸ τῶν ΑΒ, ΒΓ τετράγωνα: τὰ ἄρα ἀπὸ τῶν ΑΒ, ΒΓ τετράγωνα ἴσα ἐστὶ τῷ [ τε ] δὶς ὑπὸ τῶν ΑΒ, ΒΓ περιεχομένῳ ὀρθογωνίῳ μετὰ τοῦ ἀπὸ τῆς ΑΓ τετραγώνου. Ἐὰν ἄρα εὐθεῖα γραμμὴ τμηθῇ, ὡς ἔτυχεν, τὸ ἀπὸ τῆς ὅλης καὶ τὸ ἀφ' ἑνὸς τῶν τμημάτων τὰ συναμφότερα τετράγωνα ἴσα ἐστὶ τῷ τε δὶς ὑπὸ τῆς ὅλης καὶ τοῦ εἰρημένου τμήματος περιεχομένῳ ὀρθογωνίῳ καὶ τῷ ἀπὸ τοῦ λοιποῦ τμήματος τετραγώνῳ: ὅπερ ἔδει δεῖξαι.

If a straight line be cut at random, the square on the whole and that on one of the segments both together are equal to twice the rectangle contained by the whole and the said segment and the square on the remaining segment. For let a straight line AB be cut at random at the point C; I say that the squares on AB, BC are equal to twice the rectangle contained by AB, BC and the square on CA. For let the square ADEB be described on AB, [I. 46] and let the figure be drawn. Then, since AG is equal to GE, [I. 43] let CF be added to each; therefore the whole AF is equal to the whole CE. Therefore AF, CE are double of AF. But AF, CE are the gnomon KLM and the square CF; therefore the gnomon KLM and the square CF are double of AF. But twice the rectangle AB, BC is also double of AF; for BF is equal to BC; therefore the gnomon KLM and the square CF are equal to twice the rectangle AB, BC. Let DG, which is the square on AC, be added to each; therefore the gnomon KLM and the squares BG, GD are equal to twice the rectangle contained by AB, BC and the square on AC. But the gnomon KLM and the squares BG, GD are the whole ADEB and CF, which are squares described on AB, BC; therefore the squares on AB, BC are equal to twice the rectangle contained by AB, BC together with the square on AC.