§13.prop.11#1ἐὰν εἰς κύκλον ῥητὴν ἔχοντα τὴν διάμετρον πεντάγωνον ἰσόπλευρον ἐγγραφῇ, ἡ τοῦ πενταγώνου πλευρὰ ἄλογός ἐστιν ἡ καλουμένη ἐλάσσων.
If an equilateral pentagon is inscribed in a circle having a rational diameter, the side of the pentagon is the irrational straight line called minor.
εἰς γὰρ κύκλον τὸν ΑΒΓΔΕ ῥητὴν ἔχοντα τὴν διάμετρον πεντάγωνον ἰσόπλευρον ἐγγεγράφθω τὸ ΑΒΓΔΕ· λέγω, ὅτι ἡ τοῦ πενταγώνου πλευρὰ ἄλογός ἐστιν ἡ καλουμένη ἐλάσσων.
For let the equilateral pentagon ABCDE be inscribed in the circle ABCDE having a rational diameter; I say that the side of the pentagon is the irrational straight line called minor.
εἰλήφθω γὰρ τὸ κέντρον τοῦ κύκλου τὸ Ζ σημεῖον, καὶ ἐπεζεύχθωσαν αἱ ΑΖ, ΖΒ καὶ διήχθωσαν ἐπὶ τὰ Η, Θ σημεῖα, καὶ ἐπεζεύχθω ἡ ΑΓ, καὶ κείσθω τῆς ΑΖ τέταρτον μέρος ἡ ΖΚ. ῥητὴ δὲ ἡ ΑΖ·
For let the center of the circle be taken as the point F, and let AF, FB be joined and drawn through to the points H, Th, and let AG be joined, and let FK be laid down as a fourth part of AF.
ῥητὴ ἄρα καὶ ἡ ΖΚ. ἔστι δὲ καὶ ἡ ΒΖ ῥητή· ὅλη ἄρα ἡ ΒΚ ῥητή ἐστιν.
And AF is rational; therefore FK is also rational. And BF is also rational; therefore the whole BK is rational.
καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΓΗ περιφέρεια τῇ ΑΔΗ περιφερείᾳ, ὧν ἡ ΑΒΓ τῇ ΑΕΔ ἐστιν ἴση, λοιπὴ ἄρα ἡ ΓΗ λοιπῇ τῇ ΗΔ ἐστιν ἴση.
And since the circumference AGH is equal to the circumference ADH, of which ABG is equal to AED, therefore the remaining GH is equal to the remaining HD.
καὶ ἐὰν ἐπιζεύξωμεν τὴν ΑΔ, συνάγονται ὀρθαὶ αἱ πρὸς τῷ Λ γωνίαι, καὶ διπλῆ ἡ ΓΔ τῆς ΓΛ. διὰ τὰ αὐτὰ δὴ καὶ αἱ πρὸς τῷ Μ ὀρθαί εἰσιν, καὶ διπλῆ ἡ ΑΓ τῆς ΓΜ. ἐπεὶ οὖν ἴση ἐστὶν ἡ ὑπὸ ΑΛΓ γωνία τῇ ὑπὸ ΑΜΖ, κοινὴ δὲ τῶν δύο τριγώνων τοῦ τε ΑΓΛ καὶ τοῦ ΑΜΖ ἡ ὑπὸ ΛΑΓ, λοιπὴ ἄρα ἡ ὑπὸ ΑΓΛ λοιπῇ τῇ ὑπὸ ΜΖΑ ἐστιν ἴση· ἰσογώνιον ἄρα ἐστὶ τὸ ΑΓΛ τρίγωνον τῷ ΑΜΖ τριγώνῳ·
And if we join AD, the angles at L are right, and GD is double of GL. For the same reason indeed, the angles at M are also right, and AG is double of GM. Since therefore the angle ALG is equal to AMF, and the angle LAG is common to the two triangles AGL and AMF, therefore the remaining angle AGL is equal to the remaining MFA; therefore the triangle AGL is equiangular with the triangle AMF.
ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΛΓ πρὸς ΓΑ, οὕτως ἡ ΜΖ πρὸς ΖΑ· καὶ τῶν ἡγουμένων τὰ διπλάσια· ὡς ἄρα ἡ τῆς ΛΓ διπλῆ πρὸς τὴν ΓΑ, οὕτως ἡ τῆς ΜΖ διπλῆ πρὸς τὴν ΖΑ. ὡς δὲ ἡ τῆς ΜΖ διπλῆ πρὸς τὴν ΖΑ, οὕτως ἡ ΜΖ πρὸς τὴν ἡμίσειαν τῆς ΖΑ· καὶ ὡς ἄρα ἡ τῆς ΛΓ διπλῆ πρὸς τὴν ΓΑ, οὕτως ἡ ΜΖ πρὸς τὴν ἡμίσειαν τῆς ΖΑ. καὶ τῶν ἑπομένων τὰ ἡμίσεα· ὡς ἄρα ἡ τῆς ΛΓ διπλῆ πρὸς τὴν ἡμίσειαν τῆς ΓΑ, οὕτως ἡ ΜΖ πρὸς τὸ τέταρτον τῆς ΖΑ. καί ἐστι τῆς μὲν ΛΓ διπλῆ ἡ ΔΓ, τῆς δὲ ΓΑ ἡμίσεια ἡ ΓΜ, τῆς δὲ ΖΑ τέταρτον μέρος ἡ ΖΚ·
Therefore, proportionally, as LG is to GA, so is MF to FA; and of the antecedents, their doubles: as therefore the double of LG is to GA, so is the double of MF to FA. But as the double of MF is to FA, so is MF to the half of FA; therefore also as the double of LG is to GA, so is MF to the half of FA. And of the consequents, their halves: as therefore the double of LG is to the half of GA, so is MF to the fourth part of FA. And DG is the double of LG, GM is the half of GA, and FK is the fourth part of FA; therefore, as DG is to GM, so is MF to FK.
ἔστιν ἄρα ὡς ἡ ΔΓ πρὸς τὴν ΓΜ, οὕτως ἡ ΜΖ πρὸς τὴν ΖΚ. συνθέντι καὶ ὡς συναμφότερος ἡ ΔΓΜ πρὸς τὴν ΓΜ, οὕτως ἡ ΜΚ πρὸς ΚΖ· καὶ ὡς ἄρα τὸ ἀπὸ συναμφοτέρου τῆς ΔΓΜ πρὸς τὸ ἀπὸ ΓΜ, οὕτως τὸ ἀπὸ ΜΚ πρὸς τὸ ἀπὸ ΚΖ. καὶ ἐπεὶ τῆς ὑπὸ δύο πλευρὰς τοῦ πενταγώνου ὑποτεινούσης, οἷον τῆς ΑΓ, ἄκρον καὶ μέσον λόγον τεμνομένης τὸ μεῖζον τμῆμα ἴσον ἐστὶ τῇ τοῦ πενταγώνου πλευρᾷ, τουτέστι τῇ ΔΓ, τὸ δὲ μεῖζον τμῆμα προσλαβὸν τὴν ἡμίσειαν τῆς ὅλης πενταπλάσιον δύναται τοῦ ἀπὸ τῆς ἡμισείας τῆς ὅλης, καί ἐστιν ὅλης τῆς ΑΓ ἡμίσεια ἡ ΓΜ, τὸ ἄρα ἀπὸ τῆς ΔΓΜ ὡς μιᾶς πενταπλάσιόν ἐστι τοῦ ἀπὸ τῆς ΓΜ.
By addition, as the sum of DG, GM is to GM, so is MK to KF; and therefore as the square on the sum of DG, GM is to the square on GM, so is the square on MK to the square on KF. And since, when the straight line subtending two sides of the pentagon, such as AG, is cut in extreme and mean ratio, the greater segment is equal to the side of the pentagon, that is, to DG, and the greater segment plus the half of the whole is in square five times the square on the half of the whole, and GM is the half of the whole AG, therefore the square on DG, GM as one straight line is five times the square on GM.