# Book 6 Proposition 19

Τὰ ὅμοια τρίγωνα πρὸς ἄλληλα ἐν διπλασίονι λόγῳ ἐστὶ τῶν ὁμολόγων πλευρῶν. Ἔστω ὅμοια τρίγωνα τὰ ΑΒΓ, ΔΕΖ ἴσην ἔχοντα τὴν πρὸς τῷ Β γωνίαν τῇ πρὸς τῷ Ε, ὡς δὲ τὴν ΑΒ πρὸς τὴν ΒΓ, οὕτως τὴν ΔΕ πρὸς τὴν ΕΖ, ὥστε ὁμόλογον εἶναι τὴν ΒΓ τῇ ΕΖ: λέγω, ὅτι τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΔΕΖ τρίγωνον διπλασίονα λόγον ἔχει ἤπερ ἡ ΒΓ πρὸς τὴν ΕΖ. Εἰλήφθω γὰρ τῶν ΒΓ, ΕΖ τρίτη ἀνάλογον ἡ ΒΗ, ὥστε εἶναι ὡς τὴν ΒΓ πρὸς τὴν ΕΖ, οὕτως τὴν ΕΖ πρὸς τὴν ΒΗ: καὶ ἐπεζεύχθω ἡ ΑΗ. Ἐπεὶ οὖν ἐστιν ὡς ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως ἡ ΔΕ πρὸς τὴν ΕΖ, ἐναλλὰξ ἄρα ἐστὶν ὡς ἡ ΑΒ πρὸς τὴν ΔΕ, οὕτως ἡ ΒΓ πρὸς τὴν ΕΖ. ἀλλ' ὡς ἡ ΒΓ πρὸς ΕΖ, οὕτως ἐστὶν ἡ ΕΖ πρὸς ΒΗ. καὶ ὡς ἄρα ἡ ΑΒ πρὸς ΔΕ, οὕτως ἡ ΕΖ πρὸς ΒΗ: τῶν ΑΒΗ, ΔΕΖ ἄρα τριγώνων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας. ὧν δὲ μίαν μιᾷ ἴσην ἐχόντων γωνίαν τριγώνων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας, ἴσα ἐστὶν ἐκεῖνα. ἴσον ἄρα ἐστὶ τὸ ΑΒΗ τρίγωνον τῷ ΔΕΖ τριγώνῳ. καὶ ἐπεί ἐστιν ὡς ἡ ΒΓ πρὸς τὴν ΕΖ, οὕτως ἡ ΕΖ πρὸς τὴν ΒΗ, ἐὰν δὲ τρεῖς εὐθεῖαι ἀνάλογον ὦσιν, ἡ πρώτη πρὸς τὴν τρίτην διπλασίονα λόγον ἔχει ἤπερ πρὸς τὴν δευτέραν, ἡ ΒΓ ἄρα πρὸς τὴν ΒΗ διπλασίονα λόγον ἔχει ἤπερ ἡ ΓΒ πρὸς τὴν ΕΖ. ὡς δὲ ἡ ΓΒ πρὸς τὴν ΒΗ, οὕτως τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΑΒΗ τρίγωνον: καὶ τὸ ΑΒΓ ἄρα τρίγωνον πρὸς τὸ ΑΒΗ διπλασίονα λόγον ἔχει ἤπερ ἡ ΒΓ πρὸς τὴν ΕΖ. ἴσον δὲ τὸ ΑΒΗ τρίγωνον τῷ ΔΕΖ τριγώνῳ: καὶ τὸ ΑΒΓ ἄρα τρίγωνον πρὸς τὸ ΔΕΖ τρίγωνον διπλασίονα λόγον ἔχει ἤπερ ἡ ΒΓ πρὸς τὴν ΕΖ. Τὰ ἄρα ὅμοια τρίγωνα πρὸς ἄλληλα ἐν διπλασίονι λόγῳ ἐστὶ τῶν ὁμολόγων πλευρῶν: [ὅπερ ἔδει δεῖξαι].

Πόρισμα. Ἐκ δὴ τούτου φανερόν, ὅτι, ἐὰν τρεῖς εὐθεῖαι ἀνάλογον ὦσιν, ἔστιν ὡς ἡ πρώτη πρὸς τὴν τρίτην, οὕτως τὸ ἀπὸ τῆς πρώτης εἶδος πρὸς τὸ ἀπὸ τῆς δευτέρας τὸ ὅμοιον καὶ ὁμοίως ἀναγραφόμενον [ἐπείπερ ἐδείχθη, ὡς ἡ ΓΒ πρὸς ΒΗ, οὕτως τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΑΒΗ τρίγωνον, τουτέστι τὸ ΔΕΖ]: ὅπερ ἔδει δεῖξαι.

Similar triangles are to one another in the duplicate ratio of the corresponding sides. Let ABC, DEF be similar triangles having the angle at B equal to the angle at E, and such that, as AB is to BC, so is DE to EF, so that BC corresponds to EF; [V. Def. 11] I say that the triangle ABC has to the triangle DEF a ratio duplicate of that which BC has to EF. For let a third proportional BG be taken to BC, EF, so that, as BC is to EF, so is EF to BG; [VI. 11] and let AG be joined. Since then, as AB is to BC, so is DE to EF, therefore, alternately, as AB is to DE, so is BC to EF. [V. 16] But, as BC is to EF, so is EF to BG; therefore also, as AB is to DE, so is EF to BG. [V. 11] Therefore in the triangles ABG, DEF the sides about the equal angles are reciprocally proportional. But those triangles which have one angle equal to one angle, and in which the sides about the equal angles are reciprocally proportional, are equal; [VI. 15] therefore the triangle ABG is equal to the triangle DEF. Now since, as BC is to EF, so is EF to BG, and, if three straight lines be proportional, the first has to the third a ratio duplicate of that which it has to the second, [V. Def. 9] therefore BC has to BG a ratio duplicate of that which CB has to EF. But, as CB is to BG, so is the triangle ABC to the triangle ABG; [VI. 1] therefore the triangle ABC also has to the triangle ABG a ratio duplicate of that which BC has to EF. But the triangle ABG is equal to the triangle DEF; therefore the triangle ABC also has to the triangle DEF a ratio duplicate of that which BC has to EF. Therefore etc.

Porism. From this it is manifest that, if three straight lines be proportional, then, as the first is to the third, so is the figure described on the first to that which is similar and similarly described on the second [since it was proved that, as CB is to BG, so is the triangle ABC to the triangle ABG, that is DEF].