§6.prop.20#2ἐπεζεύχθωσαν γὰρ αἱ ΑΓ, ΖΘ. καὶ ἐπεὶ διὰ τὴν ὁμοιότητα τῶν πολυγώνων ἴση ἐστὶν ἡ ὑπὸ ΑΒΓ γωνία τῇ ὑπὸ ΖΗΘ, καί ἐστιν ὡς ἡ ΑΒ πρὸς ΒΓ, οὕτως ἡ ΖΗ πρὸς ΗΘ, ἰσογώνιόν ἐστι τὸ ΑΒΓ τρίγωνον τῷ ΖΗΘ τριγώνῳ· ἴση ἄρα ἐστὶν ἡ μὲν ὑπὸ ΒΑΓ γωνία τῇ ὑπὸ ΗΖΘ, ἡ δὲ ὑπὸ ΒΓΑ τῇ ὑπὸ ΗΘΖ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ὑπὸ ΒΑΜ γωνία τῇ ὑπὸ ΗΖΝ, ἔστι δὲ καὶ ἡ ὑπὸ ΑΒΜ τῇ ὑπὸ ΖΗΝ ἴση, καὶ λοιπὴ ἄρα ἡ ὑπὸ ΑΜΒ λοιπῇ τῇ ὑπὸ ΖΝΗ ἴση ἐστίν· ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΜ τρίγωνον τῷ ΖΗΝ τριγώνῳ.
For let AΓ, ΖΘ be joined. And since, because of the similarity of the polygons, the angle ABΓ is equal to the angle ZHΘ, and as AB is to BΓ, so is ZH to HΘ, the triangle ABΓ is equiangular with the triangle ΖΗΘ; therefore the angle BAΓ is equal to the angle HΖΘ, and the angle BΓA to the angle HΘΖ. And since the angle BAM is equal to the angle HΖN, and the angle ABM is also equal to the angle ΖHN, therefore the remaining angle AMB is also equal to the remaining angle ΖNH; therefore the triangle ABM is equiangular with the triangle ΖHN.
ὁμοίως δὴ δείξομεν, ὅτι καὶ τὸ ΒΜΓ τρίγωνον ἰσογώνιόν ἐστι τῷ ΗΝΘ τριγώνῳ.
Similarly indeed we shall show that also the triangle BMΓ is equiangular with the triangle HNΘ.
ἀνάλογον ἄρα ἐστίν, ὡς μὲν ἡ ΑΜ πρὸς ΜΒ, οὕτως ἡ ΖΝ πρὸς ΝΗ, ὡς δὲ ἡ ΒΜ πρὸς ΜΓ, οὕτως ἡ ΗΝ πρὸς ΝΘ· ὥστε καὶ διʼ ἴσου, ὡς ἡ ΑΜ πρὸς ΜΓ, οὕτως ἡ ΖΝ πρὸς ΝΘ. ἀλλʼ ὡς ἡ ΑΜ πρὸς ΜΓ, οὕτως τὸ ΑΒΜ πρὸς τὸ ΜΒΓ, καὶ τὸ ΑΜΕ πρὸς τὸ ΕΜΓ· πρὸς ἄλληλα γάρ εἰσιν ὡς αἱ βάσεις.
Therefore, proportionally, as AM is to MB, so is ΖN to NH, and as BM is to MΓ, so is HN to NΘ; so that also, ex aequali, as AM is to MΓ, so is ΖN to NΘ. But as AM is to MΓ, so is ABM to MBΓ, and AME to EMΓ; for they are to one another as their bases.
καὶ ὡς ἄρα ἓν τῶν ἡγουμένων πρὸς ἓν τῶν ἑπομένων, οὕτως ἅπαντα τὰ ἡγούμενα πρὸς ἅπαντα τὰ ἑπόμενα· ὡς ἄρα τὸ ΑΜΒ τρίγωνον πρὸς τὸ ΒΜΓ, οὕτως τὸ ΑΒΕ πρὸς τὸ ΓΒΕ. ἀλλʼ ὡς τὸ ΑΜΒ πρὸς τὸ ΒΜΓ, οὕτως ἡ ΑΜ πρὸς ΜΓ· καὶ ὡς ἄρα ἡ ΑΜ πρὸς ΜΓ, οὕτως τὸ ΑΒΕ τρίγωνον πρὸς τὸ ΕΒΓ τρίγωνον.
And as, therefore, one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents; as therefore the triangle AMB is to the triangle BMΓ, so is the triangle ABE to the triangle ΓBE. But as the triangle AMB is to the triangle BMΓ, so is AM to MΓ; and as therefore AM is to MΓ, so is the triangle ABE to the triangle EBΓ.
διὰ τὰ αὐτὰ δὴ καὶ ὡς ἡ ΖΝ πρὸς ΝΘ, οὕτως τὸ ΖΗΛ τρίγωνον πρὸς τὸ ΗΛΘ τρίγωνον.
For the same reasons indeed, as ΖN is to NΘ, so is the triangle ΖHΛ to the triangle HΛΘ.
καί ἐστιν ὡς ἡ ΑΜ πρὸς ΜΓ, οὕτως ἡ ΖΝ πρὸς ΝΘ· καὶ ὡς ἄρα τὸ ΑΒΕ τρίγωνον πρὸς τὸ ΒΕΓ τρίγωνον, οὕτως τὸ ΖΗΛ τρίγωνον πρὸς τὸ ΗΛΘ τρίγωνον, καὶ ἐναλλὰξ, ὡς τὸ ΑΒΕ τρίγωνον πρὸς τὸ ΖΗΛ τρίγωνον, οὕτως τὸ ΒΕΓ τρίγωνον πρὸς τὸ ΗΛΘ τρίγωνον.
And as AM is to MΓ, so is ΖN to NΘ; and as therefore the triangle ABE is to the triangle BEΓ, so is the triangle ΖHΛ to the triangle HΛΘ, and, alternando, as the triangle ABE is to the triangle ΖHΛ, so is the triangle BEΓ to the triangle HΛΘ.
ὁμοίως δὴ δείξομεν ἐπιζευχθεισῶν τῶν ΒΔ, ΗΚ, ὅτι καὶ ὡς τὸ ΒΕΓ τρίγωνον πρὸς τὸ ΛΗΘ τρίγωνον, οὕτως τὸ ΕΓΔ τρίγωνον πρὸς τὸ ΛΘΚ τρίγωνον.
Similarly indeed we shall show, BΔ, HK being joined, that also as the triangle BEΓ is to the triangle ΛHΘ, so is the triangle EΓΔ to the triangle ΛΘK.
καὶ ἐπεί ἐστιν ὡς τὸ ΑΒΕ τρίγωνον πρὸς τὸ ΖΗΛ τρίγωνον, οὕτως τὸ ΕΒΓ πρὸς τὸ ΛΗΘ, καὶ ἔτι τὸ ΕΓΔ πρὸς τὸ ΛΘΚ, καὶ ὡς ἄρα ἓν τῶν ἡγουμένων πρὸς ἓν τῶν ἑπομένων, οὕτως ἅπαντα τὰ ἡγούμενα πρὸς ἅπαντα τὰ ἑπόμενα· ἔστιν ἄρα ὡς τὸ ΑΒΕ τρίγωνον πρὸς τὸ ΖΗΛ τρίγωνον, οὕτως τὸ ΑΒΓΔΕ πολύγωνον πρὸς τὸ ΖΗΘΚΛ πολύγωνον.
And since, as the triangle ABE is to the triangle ΖHΛ, so is the triangle EBΓ to the triangle ΛHΘ, and further the triangle EΓΔ to the triangle ΛΘK, therefore also, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents; therefore, as the triangle ABE is to the triangle ΖHΛ, so is the polygon ABΓΔE to the polygon ZHΘKΛ.
ἀλλὰ τὸ ΑΒΕ τρίγωνον πρὸς τὸ ΖΗΛ τρίγωνον διπλασίονα λόγον ἔχει ἤπερ ἡ ΑΒ ὁμόλογος πλευρὰ πρὸς τὴν ΖΗ ὁμόλογον πλευράν· τὰ γὰρ ὅμοια τρίγωνα ἐν διπλασίονι λόγῳ ἐστὶ τῶν ὁμολόγων πλευρῶν.
But the triangle ABE has to the triangle ΖHΛ a duplicate ratio of that which the corresponding side AB has to the corresponding side ΖH; for similar triangles are in the duplicate ratio of their corresponding sides.
καὶ τὸ ΑΒΓΔΕ ἄρα πολύγωνον πρὸς τὸ ΖΗΘΚΛ πολύγωνον διπλασίονα λόγον ἔχει ἤπερ ἡ ΑΒ ὁμόλογος πλευρὰ πρὸς τὴν ΖΗ ὁμόλογον πλευράν.
Therefore also the polygon ABΓΔE has to the polygon ZHΘKΛ a duplicate ratio of that which the corresponding side AB has to the corresponding side ΖH.
τὰ ἄρα ὅμοια πολύγωνα εἴς τε ὅμοια τρίγωνα διαιρεῖται καὶ εἰς ἴσα τὸ πλῆθος καὶ ὁμόλογα τοῖς ὅλοις, καὶ τὸ πολύγωνον πρὸς τὸ πολύγωνον διπλασίονα λόγον ἔχει ἤπερ ἡ ὁμόλογος πλευρὰ πρὸς τὴν ὁμόλογον πλευράν· .
Therefore similar polygons are divided into similar triangles, and equal in multitude and corresponding to the wholes, and the polygon has to the polygon a duplicate ratio of that which the corresponding side has to the corresponding side.
Πόρισμα
ὡσαύτως δὲ καὶ ἐπὶ τῶν τετραπλεύρων δειχθήσεται, ὅτι ἐν διπλασίονι λόγῳ εἰσὶ τῶν ὁμολόγων πλευρῶν.
Porism And in the same manner it will also be shown in the case of quadrilaterals that they are in the duplicate ratio of their corresponding sides.
ἐδείχθη δὲ καὶ ἐπὶ τῶν τριγώνων· ὥστε καὶ καθόλου τὰ ὅμοια εὐθύγραμμα σχήματα πρὸς ἄλληλα ἐν διπλασίονι λόγῳ εἰσὶ τῶν ὁμολόγων πλευρῶν.
And it was also shown in the case of triangles; so that also generally, similar rectilinear figures are to one another in the duplicate ratio of their corresponding sides.
ὅπερ ἔδει δεῖξαι.
Which was to be demonstrated.