§6.prop.31ἐν τοῖς ὀρθογωνίοις τριγώνοις τὸ ἀπὸ τῆς τὴν ὀρθὴν γωνίαν ὑποτεινούσης πλευρᾶς εἶδος ἴσον ἐστὶ τοῖς ἀπὸ τῶν τὴν ὀρθὴν γωνίαν περιεχουσῶν πλευρῶν εἴδεσι τοῖς ὁμοίοις τε καὶ ὁμοίως ἀναγραφομένοις.
In right-angled triangles the figure described on the side subtending the right angle is equal to the similar and similarly described figures on the sides containing the right angle.
ἔστω τρίγωνον ὀρθογώνιον τὸ ΑΒΓ ὀρθὴν ἔχον τὴν ὑπὸ ΒΑΓ γωνίαν· λέγω, ὅτι τὸ ἀπὸ τῆς ΒΓ εἶδος ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΒΑ, ΑΓ εἴδεσι τοῖς ὁμοίοις τε καὶ ὁμοίως ἀναγραφομένοις.
Let ΑΒΓ be a right-angled triangle having the angle ΒΑΓ right; I say that the figure on ΒΓ is equal to the similar and similarly described figures on ΒΑ, ΑΓ.
ἤχθω κάθετος ἡ ΑΔ.
ἐπεὶ οὖν ἐν ὀρθογωνίῳ τριγώνῳ τῷ ΑΒΓ ἀπὸ τῆς πρὸς τῷ Α ὀρθῆς γωνίας ἐπὶ τὴν ΒΓ βάσιν κάθετος ἦκται ἡ ΑΔ, τὰ ΑΒΔ, ΑΔΓ πρὸς τῇ καθέτῳ τρίγωνα ὅμοιά ἐστι τῷ τε ὅλῳ τῷ ΑΒΓ καὶ ἀλλήλοις.
Let the perpendicular ΑΔ be drawn. Since then, in a right-angled triangle ΑΒΓ, the perpendicular ΑΔ has been drawn from the right angle at Α to the base ΒΓ, the triangles ΑΒΔ, ΑΔΓ about the perpendicular are similar to the whole ΑΒΓ and to one another.
καὶ ἐπεὶ ὅμοιόν ἐστι τὸ ΑΒΓ τῷ ΑΒΔ, ἔστιν ἄρα ὡς ἡ ΓΒ πρὸς τὴν ΒΑ, οὕτως ἡ ΑΒ πρὸς τὴν ΒΔ. καὶ ἐπεὶ τρεῖς εὐθεῖαι ἀνάλογόν εἰσιν, ἔστιν ὡς ἡ πρώτη πρὸς τὴν τρίτην, οὕτως τὸ ἀπὸ τῆς πρώτης εἶδος πρὸς τὸ ἀπὸ τῆς δευτέρας τὸ ὅμοιον καὶ ὁμοίως ἀναγραφόμενον.
And since ΑΒΓ is similar to ΑΒΔ, therefore, as ΓΒ is to ΒΑ, so is ΑΒ to ΒΔ. And since three straight lines are proportional, as the first is to the third, so is the figure described on the first to the similar and similarly described figure on the second.
ὡς ἄρα ἡ ΓΒ πρὸς τὴν ΒΔ, οὕτως τὸ ἀπὸ τῆς ΓΒ εἶδος πρὸς τὸ ἀπὸ τῆς ΒΑ τὸ ὅμοιον καὶ ὁμοίως ἀναγραφόμενον.
Therefore, as ΓΒ is to ΒΔ, so is the figure on ΓΒ to the similar and similarly described figure on ΒΑ.
διὰ τὰ αὐτὰ δὴ καὶ ὡς ἡ ΒΓ πρὸς τὴν ΓΔ, οὕτως τὸ ἀπὸ τῆς ΒΓ εἶδος πρὸς τὸ ἀπὸ τῆς ΓΑ. ὥστε καὶ ὡς ἡ ΒΓ πρὸς τὰς ΒΔ, ΔΓ, οὕτως τὸ ἀπὸ τῆς ΒΓ εἶδος πρὸς τὰ ἀπὸ τῶν ΒΑ, ΑΓ τὰ ὅμοια καὶ ὁμοίως ἀναγραφόμενα.
For the same reasons also, as ΒΓ is to ΓΔ, so is the figure on ΒΓ to the figure on ΓΑ. And so also, as ΒΓ is to the sum of ΒΔ, ΔΓ, so is the figure on ΒΓ to the similar and similarly described figures on ΒΑ, ΑΓ.
ἴση δὲ ἡ ΒΓ ταῖς ΒΔ, ΔΓ· ἴσον ἄρα καὶ τὸ ἀπὸ τῆς ΒΓ εἶδος τοῖς ἀπὸ τῶν ΒΑ, ΑΓ εἴδεσι τοῖς ὁμοίοις τε καὶ ὁμοίως ἀναγραφομένοις.
And ΒΓ is equal to the sum of ΒΔ, ΔΓ; therefore the figure on ΒΓ is also equal to the similar and similarly described figures on the sides ΒΑ, ΑΓ.
ἐν ἄρα τοῖς ὀρθογωνίοις τριγώνοις τὸ ἀπὸ τῆς τὴν ὀρθὴν γωνίαν ὑποτεινούσης πλευρᾶς εἶδος ἴσον ἐστὶ τοῖς ἀπὸ τῶν τὴν ὀρθὴν γωνίαν περιεχουσῶν πλευρῶν εἴδεσι τοῖς ὁμοίοις τε καὶ ὁμοίως ἀναγραφομένοις· ὅπερ ἔδει δεῖξαι.
Therefore, in right-angled triangles the figure described on the side subtending the right angle is equal to the similar and similarly described figures on the sides containing the right angle; which was to be proved.
§6.prop.32ἐὰν δύο τρίγωνα συντεθῇ κατὰ μίαν γωνίαν τὰς δύο πλευρὰς ταῖς δυσὶ πλευραῖς ἀνάλογον ἔχοντα ὥστε τὰς ὁμολόγους αὐτῶν πλευρὰς καὶ παραλλήλους εἶναι, αἱ λοιπαὶ τῶν τριγώνων πλευραὶ ἐπʼ εὐθείας ἔσονται.
If two triangles having two sides proportional to two sides be placed together at one angle so that their corresponding sides are also parallel, the remaining sides of the triangles will be in a straight line.
ἔστω δύο τρίγωνα τὰ ΑΒΓ, ΔΓΕ τὰς δύο πλευρὰς τὰς ΒΑ, ΑΓ ταῖς δυσὶ πλευραῖς ταῖς ΔΓ, ΔΕ ἀνάλογον ἔχοντα, ὡς μὲν τὴν ΑΒ πρὸς τὴν ΑΓ, οὕτως τὴν ΔΓ πρὸς τὴν ΔΕ, παράλληλον δὲ τὴν μὲν ΑΒ τῇ ΔΓ, τὴν δὲ ΑΓ τῇ ΔΕ· λέγω, ὅτι ἐπʼ εὐθείας ἐστὶν ἡ ΒΓ τῇ ΓΕ.
ἐπεὶ γὰρ παράλληλός ἐστιν ἡ ΑΒ τῇ ΔΓ, καὶ εἰς αὐτὰς ἐμπέπτωκεν εὐθεῖα ἡ ΑΓ, αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΒΑΓ, ΑΓΔ ἴσαι ἀλλήλαις εἰσίν.
Let there be two triangles ΑΒΓ, ΔΓΕ having the two sides ΒΑ, ΑΓ proportional to the two sides ΔΓ, ΔΕ, so that, as ΑΒ is to ΑΓ, so is ΔΓ to ΔΕ, and let ΑΒ be parallel to ΔΓ, and ΑΓ to ΔΕ; I say that ΒΓ is in a straight line with ΓΕ. For since ΑΒ is parallel to ΔΓ, and the straight line ΑΓ has fallen upon them, the alternate angles ΒΑΓ, ΑΓΔ are equal to one another.
διὰ τὰ αὐτὰ δὴ καὶ ἡ ὑπὸ ΓΔΕ τῇ ὑπὸ ΑΓΔ ἴση ἐστίν.
For the same reasons also, the angle ΓΔΕ is equal to the angle ΑΓΔ.
ὥστε καὶ ἡ ὑπὸ ΒΑΓ τῇ ὑπὸ ΓΔΕ ἐστιν ἴση.
And so also the angle ΒΑΓ is equal to the angle ΓΔΕ.
καὶ ἐπεὶ δύο τρίγωνά ἐστι τὰ ΑΒΓ, ΔΓΕ μίαν γωνίαν τὴν πρὸς τῷ Α μιᾷ γωνίᾳ τῇ πρὸς τῷ Δ ἴσην ἔχοντα, περὶ δὲ τὰς ἴσας γωνίας τὰς πλευρὰς ἀνάλογον, ὡς τὴν ΒΑ πρὸς τὴν ΑΓ, οὕτως τὴν ΓΔ πρὸς τὴν ΔΕ, ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΓΕ τριγώνῳ· ἴση ἄρα ἡ ὑπὸ ΑΒΓ γωνία τῇ ὑπὸ ΔΓΕ. ἐδείχθη δὲ καὶ ἡ ὑπὸ ΑΓΔ τῇ ὑπὸ ΒΑΓ ἴση· ὅλη ἄρα ἡ ὑπὸ ΑΓΕ δυσὶ ταῖς ὑπὸ ΑΒΓ, ΒΑΓ ἴση ἐστίν.
And since there are two triangles ΑΒΓ, ΔΓΕ having one angle at Α equal to one angle at Δ, and the sides about the equal angles proportional, namely, as ΒΑ is to ΑΓ, so is ΓΔ to ΔΕ, therefore the triangle ΑΒΓ is equiangular with the triangle ΔΓΕ; therefore the angle ΑΒΓ is equal to the angle ΔΓΕ. But the angle ΑΓΔ was also proved equal to the angle ΒΑΓ; therefore the whole angle ΑΓΕ is equal to the two angles ΑΒΓ, ΒΑΓ.
κοινὴ προσκείσθω ἡ ὑπὸ ΑΓΒ· αἱ ἄρα ὑπὸ ΑΓΕ, ΑΓΒ ταῖς ὑπὸ ΒΑΓ, ΑΓΒ, ΓΒΑ ἴσαι εἰσίν.
Let the angle ΑΓΒ be added as common; therefore the angles ΑΓΕ, ΑΓΒ are equal to the angles ΒΑΓ, ΑΓΒ, ΓΒΑ.
ἀλλʼ αἱ ὑπὸ ΒΑΓ, ΑΒΓ, ΑΓΒ δυσὶν ὀρθαῖς ἴσαι εἰσίν· καὶ αἱ ὑπὸ ΑΓΕ, ΑΓΒ ἄρα δυσὶν ὀρθαῖς ἴσαι εἰσίν.
But the angles ΒΑΓ, ΑΒΓ, ΑΓΒ are equal to two right angles; therefore the angles ΑΓΕ, ΑΓΒ are also equal to two right angles.
πρὸς δή τινι εὐθείᾳ τῇ ΑΓ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Γ δύο εὐθεῖαι αἱ ΒΓ, ΓΕ μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι τὰς ἐφεξῆς γωνίας τὰς ὑπὸ ΑΓΕ, ΑΓΒ δυσὶν ὀρθαῖς ἴσας ποιοῦσιν· ἐπʼ εὐθείας ἄρα ἐστὶν ἡ ΒΓ τῇ ΓΕ.
ἐὰν ἄρα δύο τρίγωνα συντεθῇ κατὰ μίαν γωνίαν τὰς δύο πλευρὰς ταῖς δυσὶ πλευραῖς ἀνάλογον ἔχοντα ὥστε τὰς ὁμολόγους αὐτῶν πλευρὰς καὶ παραλλήλους εἶναι, αἱ λοιπαὶ τῶν τριγώνων πλευραὶ ἐπʼ εὐθείας ἔσονται· ὅπερ ἔδει δεῖξαι.
Since then, at some straight line ΑΓ and at the point Γ on it, two straight lines ΒΓ, ΓΕ not lying on the same side make the adjacent angles ΑΓΕ, ΑΓΒ equal to two right angles; therefore ΒΓ is in a straight line with ΓΕ. If therefore two triangles having two sides proportional to two sides be placed together at one angle so that their corresponding sides are also parallel, the remaining sides of the triangles will be in a straight line; which was to be proved.