# Book 1 Proposition 48

Ἐὰν τριγώνου τὸ ἀπὸ μιᾶς τῶν πλευρῶν τετράγωνον ἴσον ᾖ τοῖς ἀπὸ τῶν λοιπῶν τοῦ τριγώνου δύο πλευρῶν τετραγώνοις, ἡ περιεχομένη γωνία ὑπὸ τῶν λοιπῶν τοῦ τριγώνου δύο πλευρῶν ὀρθή ἐστιν. Τριγώνου γὰρ τοῦ ΑΒΓ τὸ ἀπὸ μιᾶς τῆς ΒΓ πλευρᾶς τετράγωνον ἴσον ἔστω τοῖς ἀπὸ τῶν ΒΑ, ΑΓ πλευρῶν τετραγώνοις: λέγω, ὅτι ὀρθή ἐστιν ἡ ὑπὸ ΒΑΓ γωνία. Ἤχθω γὰρ ἀπὸ τοῦ Α σημείου τῇ ΑΓ εὐθείᾳ πρὸς ὀρθὰς ἡ ΑΔ καὶ κείσθω τῇ ΒΑ ἴση ἡ ΑΔ, καὶ ἐπεζεύχθω ἡ ΔΓ. ἐπεὶ ἴση ἐστὶν ἡ ΔΑ τῇ ΑΒ, ἴσον ἐστὶ καὶ τὸ ἀπὸ τῆς ΔΑ τετράγωνον τῷ ἀπὸ τῆς ΑΒ τετραγώνῳ. κοινὸν προσκείσθω τὸ ἀπὸ τῆς ΑΓ τετράγωνον: τὰ ἄρα ἀπὸ τῶν ΔΑ, ΑΓ τετράγωνα ἴσα ἐστὶ τοῖς ἀπὸ τῶν ΒΑ, ΑΓ τετραγώνοις. ἀλλὰ τοῖς μὲν ἀπὸ τῶν ΔΑ, ΑΓ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΔΓ: ὀρθὴ γάρ ἐστιν ἡ ὑπὸ ΔΑΓ γωνία: τοῖς δὲ ἀπὸ τῶν ΒΑ, ΑΓ ἴσον ἐστὶ τὸ ἀπὸ ΒΓ: ὑπόκειται γάρ: τὸ ἄρα ἀπὸ τῆς ΔΓ τετράγωνον ἴσον ἐστὶ τῷ ἀπὸ τῆς ΒΓ τετραγώνῳ: ὥστε καὶ πλευρὰ ἡ ΔΓ τῇ ΒΓ ἐστιν ἴση: καὶ ἐπεὶ ἴση ἐστὶν ἡ ΔΑ τῇ ΑΒ, κοινὴ δὲ ἡ ΑΓ, δύο δὴ αἱ ΔΑ, ΑΓ δύο ταῖς ΒΑ, ΑΓ ἴσαι εἰσίν: καὶ βάσις ἡ ΔΓ βάσει τῇ ΒΓ ἴση: γωνία ἄρα ἡ ὑπὸ ΔΑΓ γωνίᾳ τῇ ὑπὸ ΒΑΓ [ἐστιν] ἴση. ὀρθὴ δὲ ἡ ὑπὸ ΔΑΓ: ὀρθὴ ἄρα καὶ ἡ ὑπὸ ΒΑΓ. Ἐὰν ἄρα τριγώνου τὸ ἀπὸ μιᾶς τῶν πλευρῶν τετράγωνον ἴσον ᾖ τοῖς ἀπὸ τῶν λοιπῶν τοῦ τριγώνου δύο πλευρῶν τετραγώνοις, ἡ περιεχομένη γωνία ὑπὸ τῶν λοιπῶν τοῦ τριγώνου δύο πλευρῶν ὀρθή ἐστιν: ὅπερ ἔδει δεῖξαι.

If in a triangle the square on one of the sides be equal to the squares on the remaining two sides of the triangle, the angle contained by the remaining two sides of the triangle is right. For in the triangle ABC let the square on one side BC be equal to the squares on the sides BA, AC; I say that the angle BAC is right. For let AD be drawn from the point A at right angles to the straight line AC, let AD be made equal to BA, and let DC be joined. Since DA is equal to AB, the square on DA is also equal to the square on AB. Let the square on AC be added to each; therefore the squares on DA, AC are equal to the squares on BA, AC. But the square on DC is equal to the squares on DA, AC, for the angle DAC is right; [I. 47] and the square on BC is equal to the squares on BA, AC, for this is the hypothesis; therefore the square on DC is equal to the square on BC, so that the side DC is also equal to BC. And, since DA is equal to AB, and AC is common, the two sides DA, AC are equal to the two sides BA, AC; and the base DC is equal to the base BC; therefore the angle DAC is equal to the angle BAC. [I. 8] But the angle DAC is right; therefore the angle BAC is also right. Therefore etc.