Book 4 Proposition 12
Περὶ τὸν δοθέντα κύκλον πεντάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον περιγράψαι. Ἔστω ὁ δοθεὶς κύκλος ὁ ΑΒΓΔΕ: δεῖ <δὴ> περὶ τὸν ΑΒΓΔΕ κύκλον πεντάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον περιγράψαι. Νενοήσθω τοῦ ἐγγεγραμμένου πενταγώνου τῶν γωνιῶν σημεῖα τὰ Α, Β, Γ, Δ, Ε, ὥστε ἴσας εἶναι τὰς ΑΒ, ΒΓ, ΓΔ, ΔΕ, ΕΑ περιφερείας: καὶ διὰ τῶν Α, Β, Γ, Δ, Ε ἤχθωσαν τοῦ κύκλου ἐφαπτόμεναι αἱ ΗΘ, ΘΚ, ΚΛ, ΛΜ, ΜΗ, καὶ εἰλήφθω τοῦ ΑΒΓΔΕ κύκλου κέντρον τὸ Ζ, καὶ ἐπεζεύχθωσαν αἱ ΖΒ, ΖΚ, ΖΓ, ΖΛ, ΖΔ. Καὶ ἐπεὶ ἡ μὲν ΚΛ εὐθεῖα ἐφάπτεται τοῦ ΑΒΓΔΕ κατὰ τὸ Γ, ἀπὸ δὲ τοῦ Ζ κέντρου ἐπὶ τὴν κατὰ τὸ Γ ἐπαφὴν ἐπέζευκται ἡ ΖΓ, ἡ ΖΓ ἄρα κάθετός ἐστιν ἐπὶ τὴν ΚΛ: ὀρθὴ ἄρα ἐστὶν ἑκατέρα τῶν πρὸς τῷ Γ γωνιῶν. διὰ τὰ αὐτὰ δὴ καὶ αἱ πρὸς τοῖς Β, Δ σημείοις γωνίαι ὀρθαί εἰσιν. καὶ ἐπεὶ ὀρθή ἐστιν ἡ ὑπὸ ΖΓΚ γωνία, τὸ ἄρα ἀπὸ τῆς ΖΚ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΖΓ, ΓΚ. διὰ τὰ αὐτὰ δὴ καὶ τοῖς ἀπὸ τῶν ΖΒ, ΒΚ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΖΚ: ὥστε τὰ ἀπὸ τῶν ΖΓ, ΓΚ τοῖς ἀπὸ τῶν ΖΒ, ΒΚ ἐστιν ἴσα, ὧν τὸ ἀπὸ τῆς ΖΓ τῷ ἀπὸ τῆς ΖΒ ἐστιν ἴσον: λοιπὸν ἄρα τὸ ἀπὸ τῆς ΓΚ τῷ ἀπὸ τῆς ΒΚ ἐστιν ἴσον. ἴση ἄρα ἡ ΒΚ τῇ ΓΚ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΖΒ τῇ ΖΓ, καὶ κοινὴ ἡ ΖΚ, δύο δὴ αἱ ΒΖ, ΖΚ δυσὶ ταῖς ΓΖ, ΖΚ ἴσαι εἰσίν: καὶ βάσις ἡ ΒΚ βάσει τῇ ΓΚ [ἐστιν ] ἴση: γωνία ἄρα ἡ μὲν ὑπὸ ΒΖΚ [γωνίᾳ] τῇ ὑπὸ ΚΖΓ ἐστιν ἴση: ἡ δὲ ὑπὸ ΒΚΖ τῇ ὑπὸ ΖΚΓ: διπλῆ ἄρα ἡ μὲν ὑπὸ ΒΖΓ τῆς ὑπὸ ΚΖΓ, ἡ δὲ ὑπὸ ΒΚΓ τῆς ὑπὸ ΖΚΓ. διὰ τὰ αὐτὰ δὴ καὶ ἡ μὲν ὑπὸ ΓΖΔ τῆς ὑπὸ ΓΖΛ ἐστι διπλῆ, ἡ δὲ ὑπὸ ΔΛΓ τῆς ὑπὸ ΖΛΓ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΓ περιφέρεια τῇ ΓΔ, ἴση ἐστὶ καὶ γωνία ἡ ὑπὸ ΒΖΓ τῇ ὑπὸ ΓΖΔ. καί ἐστιν ἡ μὲν ὑπὸ ΒΖΓ τῆς ὑπὸ ΚΖΓ διπλῆ, ἡ δὲ ὑπὸ ΔΖΓ τῆς ὑπὸ ΛΖΓ: ἴση ἄρα καὶ ἡ ὑπὸ ΚΖΓ τῇ ὑπὸ ΛΖΓ: ἐστὶ δὲ καὶ ἡ ὑπὸ ΖΓΚ γωνία τῇ ὑπὸ ΖΓΛ ἴση. δύο δὴ τρίγωνά ἐστι τὰ ΖΚΓ, ΖΛΓ τὰς δύο γωνίας ταῖς δυσὶ γωνίαις ἴσας ἔχοντα καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην κοινὴν αὐτῶν τὴν ΖΓ: καὶ τὰς λοιπὰς ἄρα πλευρὰς ταῖς λοιπαῖς πλευραῖς ἴσας ἕξει καὶ τὴν λοιπὴν γωνίαν τῇ λοιπῇ γωνίᾳ: ἴση ἄρα ἡ μὲν ΚΓ εὐθεῖα τῇ ΓΛ, ἡ δὲ ὑπὸ ΖΚΓ γωνία τῇ ὑπὸ ΖΛΓ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΚΓ τῇ ΓΛ, διπλῆ ἄρα ἡ ΚΛ τῆς ΚΓ. διὰ τὰ αὐτὰ δὴ δειχθήσεται καὶ ἡ ΘΚ τῆς ΒΚ διπλῆ. καί ἐστιν ἡ ΒΚ τῇ ΚΓ ἴση: καὶ ἡ ΘΚ ἄρα τῇ ΚΛ ἐστιν ἴση. ὁμοίως δὴ δειχθήσεται καὶ ἑκάστη τῶν ΘΗ, ΗΜ, ΜΛ ἑκατέρᾳ τῶν ΘΚ, ΚΛ ἴση: ἰσόπλευρον ἄρα ἐστὶ τὸ ΗΘΚΛΜ πεντάγωνον. λέγω δή, ὅτι καὶ ἰσογώνιον. ἐπεὶ γὰρ ἴση ἐστὶν ἡ ὑπὸ ΖΚΓ γωνία τῇ ὑπὸ ΖΛΓ, καὶ ἐδείχθη τῆς μὲν ὑπὸ ΖΚΓ διπλῆ ἡ ὑπὸ ΘΚΛ, τῆς δὲ ὑπὸ ΖΛΓ διπλῆ ἡ ὑπὸ ΚΛΜ, καὶ ἡ ὑπὸ ΘΚΛ ἄρα τῇ ὑπὸ ΚΛΜ ἐστιν ἴση. ὁμοίως δὴ δειχθήσεται καὶ ἑκάστη τῶν ὑπὸ ΚΘΗ, ΘΗΜ, ΗΜΛ ἑκατέρᾳ τῶν ὑπὸ ΘΚΛ, ΚΛΜ ἴση: αἱ πέντε ἄρα γωνίαι αἱ ὑπὸ ΗΘΚ, ΘΚΛ, ΚΛΜ, ΛΜΗ, ΜΗΘ ἴσαι ἀλλήλαις εἰσίν. ἰσογώνιον ἄρα ἐστὶ τὸ ΗΘΚΛΜ πεντάγωνον. ἐδείχθη δὲ καὶ ἰσόπλευρον, καὶ περιγέγραπται περὶ τὸν ΑΒΓΔΕ κύκλον. [Περὶ τὸν δοθέντα ἄρα κύκλον πεντάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον περιγέγραπται]: ὅπερ ἔδει ποιῆσαι.
About a given circle to circumscribe an equilateral and equiangular pentagon. Let ABCDE be the given circle; thus it is required to circumscribe an equilateral and equiangular pentagon about the circle ABCDE. Let A, B, C, D, E be conceived to be the angular points of the inscribed pentagon, so that the circumferences AB, BC, CD, DE, EA are equal; [IV. 11] through A, B, C, D, E let GH, HK, KL, LM, MG be drawn touching the circle; [III. 16, Por.] let the centre F of the circle ABCDE be taken [III. 1], and let FB, FK, FC, FL, FD be joined. Then, since the straight line KL touches the circle ABCDE at C, and FC has been joined from the centre F to the point of contact at C, therefore FC is perpendicular to KL; [III. 18] therefore each of the angles at C is right. For the same reason the angles at the points B, D are also right. And, since the angle FCK is right, therefore the square on FK is equal to the squares on FC, CK. For the same reason [I. 47] the square on FK is also equal to the squares on FB, BK; so that the squares on FC, CK are equal to the squares on FB, BK, of which the square on FC is equal to the square on FB; therefore the square on CK which remains is equal to the square on BK. Therefore BK is equal to CK. And, since FB is equal to FC, and FK common, the two sides BF, FK are equal to the two sides CF, FK; and the base BK equal to the base CK; therefore the angle BFK is equal to the angle KFC, [I. 8] and the angle BKF to the angle FKC. Therefore the angle BFC is double of the angle KFC, and the angle BKC of the angle FKC. For the same reason the angle CFD is also double of the angle CFL, and the angle DLC of the angle FLC. Now, since the circumference BC is equal to CD, the angle BFC is also equal to the angle CFD. [III. 27] And the angle BFC is double of the angle KFC, and the angle DFC of the angle LFC; therefore the angle KFC is also equal to the angle LFC. But the angle FCK is also equal to the angle FCL; therefore FKC, FLC are two triangles having two angles equal to two angles and one side equal to one side, namely FC which is common to them; therefore they will also have the remaining sides equal to the remaining sides, and the remaining angle to the remaining angle; [I. 26] therefore the straight line KC is equal to CL, and the angle FKC to the angle FLC. And, since KC is equal to CL, therefore KL is double of KC. For the same reason it can be proved that HK is also double of BK. And BK is equal to KC; therefore HK is also equal to KL. Similarly each of the straight lines HG, GM, ML can also be proved equal to each of the straight lines HK, KL; therefore the pentagon GHKLM is equilateral. I say next that it is also equiangular. For, since the angle FKC is equal to the angle FLC, and the angle HKL was proved double of the angle FKC, and the angle KLM double of the angle FLC, therefore the angle HKL is also equal to the angle KLM. Similarly each of the angles KHG, HGM, GML can also be proved equal to each of the angles HKL, KLM; therefore the five angles GHK, HKL, KLM, LMG, MGH are equal to one another. Therefore the pentagon GHKLM is equiangular.