Ἐὰν πενταγώνου ἰσοπλεύρου αἱ τρεῖς γωνίαι ἤτοι αἱ κατὰ τὸ ἑξῆς ἢ αἱ μὴ κατὰ τὸ ἑξῆς ἴσαι ὦσιν, ἰσογώνιον ἔσται τὸ πεντάγωνον. Πενταγώνου γὰρ ἰσοπλεύρου τοῦ ΑΒΓΔΕ αἱ τρεῖς γωνίαι πρότερον αἱ κατὰ τὸ ἑξῆς αἱ πρὸς τοῖς Α, Β, Γ ἴσαι ἀλλήλαις ἔστωσαν: λέγω, ὅτι ἰσογώνιόν ἐστι τὸ ΑΒΓΔΕ πεντάγωνον. Ἐπεζεύχθωσαν γὰρ αἱ ΑΓ, ΒΕ, ΖΔ. καὶ ἐπεὶ δύο αἱ ΓΒ, ΒΑ δυσὶ ταῖς ΒΑ, ΑΕ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ, καὶ γωνία ἡ ὑπὸ ΓΒΑ γωνίᾳ τῇ ὑπὸ ΒΑΕ ἐστιν ἴση, βάσις ἄρα ἡ ΑΓ βάσει τῇ ΒΕ ἐστιν ἴση, καὶ τὸ ΑΒΓ τρίγωνον τῷ ΑΒΕ τριγώνῳ ἴσον, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι ἔσονται, ὑφ' ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν, ἡ μὲν ὑπὸ ΒΓΑ τῇ ὑπὸ ΒΕΑ, ἡ δὲ ὑπὸ ΑΒΕ τῇ ὑπὸ ΓΑΒ: ὥστε καὶ πλευρὰ ἡ ΑΖ πλευρᾷ τῇ ΒΖ ἐστιν ἴση. ἐδείχθη δὲ καὶ ὅλη ἡ ΑΓ ὅλῃ τῇ ΒΕ ἴση: καὶ λοιπὴ ἄρα ἡ ΖΓ λοιπῇ τῇ ΖΕ ἐστιν ἴση. ἔστι δὲ καὶ ἡ ΓΔ τῇ ΔΕ ἴση. δύο δὴ αἱ ΖΓ, ΓΔ δυσὶ ταῖς ΖΕ, ΕΔ ἴσαι εἰσίν: καὶ βάσις αὐτῶν κοινὴ ἡ ΖΔ: γωνία ἄρα ἡ ὑπὸ ΖΓΔ γωνίᾳ τῇ ὑπὸ ΖΕΔ ἐστιν ἴση. ἐδείχθη δὲ καὶ ἡ ὑπὸ ΒΓΑ τῇ ὑπὸ ΑΕΒ ἴση: καὶ ὅλη ἄρα ἡ ὑπὸ ΒΓΔ ὅλῃ τῇ ὑπὸ ΑΕΔ ἴση. ἀλλ' ἡ ὑπὸ ΒΓΔ ἴση ὑπόκειται ταῖς πρὸς τοῖς Α, Β γωνίαις: καὶ ἡ ὑπὸ ΑΕΔ ἄρα ταῖς πρὸς τοῖς Α, Β γωνίαις ἴση ἐστίν. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἡ ὑπὸ ΓΔΕ γωνία ἴση ἐστὶ ταῖς πρὸς τοῖς Α, Β, Γ γωνίαις: ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΓΔΕ πεντάγωνον. Ἀλλὰ δὴ μὴ ἔστωσαν ἴσαι αἱ κατὰ τὸ ἑξῆς γωνίαι, ἀλλ' ἔστωσαν ἴσαι αἱ πρὸς τοῖς Α, Γ, Δ σημείοις: λέγω, ὅτι καὶ οὕτως ἰσογώνιόν ἐστι τὸ ΑΒΓΔΕ πεντάγωνον. Ἐπεζεύχθω γὰρ ἡ ΒΔ. καὶ ἐπεὶ δύο αἱ ΒΑ, ΑΕ δυσὶ ταῖς ΒΓ, ΓΔ ἴσαι εἰσὶ καὶ γωνίας ἴσας περιέχουσιν, βάσις ἄρα ἡ ΒΕ βάσει τῇ ΒΔ ἴση ἐστίν, καὶ τὸ ΑΒΕ τρίγωνον τῷ ΒΓΔ τριγώνῳ ἴσον ἐστίν, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι ἔσονται, ὑφ' ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν: ἴση ἄρα ἐστὶν ἡ ὑπὸ ΑΕΒ γωνία τῇ ὑπὸ ΓΔΒ. ἔστι δὲ καὶ ἡ ὑπὸ ΒΕΔ γωνία τῇ ὑπὸ ΒΔΕ ἴση, ἐπεὶ καὶ πλευρὰ ἡ ΒΕ πλευρᾷ τῇ ΒΔ ἐστιν ἴση. καὶ ὅλη ἄρα ἡ ὑπὸ ΑΕΔ γωνία ὅλῃ τῇ ὑπὸ ΓΔΕ ἐστιν ἴση. ἀλλὰ ἡ ὑπὸ ΓΔΕ ταῖς πρὸς τοῖς Α, Γ γωνίαις ὑπόκειται ἴση: καὶ ἡ ὑπὸ ΑΕΔ ἄρα γωνία ταῖς πρὸς τοῖς Α, Γ ἴση ἐστίν. διὰ τὰ αὐτὰ δὴ καὶ ἡ ὑπὸ ΑΒΓ ἴση ἐστὶ ταῖς πρὸς τοῖς Α, Γ, Δ γωνίαις. ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΓΔΕ πεντάγωνον: ὅπερ ἔδει δεῖξαι.

If three angles of an equilateral pentagon, taken either in order or not in order, be equal, the pentagon will be equiangular. For in the equilateral pentagon ABCDE let, first, three angles taken in order, those at A, B, C, be equal to one another; I say that the pentagon ABCDE is equiangular. For let AC, BE, FD be joined. Now, since the two sides CB, BA are equal to the two sides BA, AE respectively, and the angle CBA is equal to the angle BAE, therefore the base AC is equal to the base BE, the triangle ABC is equal to the triangle ABE, and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend, [I. 4] that is, the angle BCA to the angle BEA, and the angle ABE to the angle CAB; hence the side AF is also equal to the side BF. [I. 6] But the whole AC was also proved equal to the whole BE; therefore the remainder FC is also equal to the remainder FE. But CD is also equal to DE. Therefore the two sides FC, CD are equal to the two sides FE, ED; and the base FD is common to them; therefore the angle FCD is equal to the angle FED. [I. 8] But the angle BCA was also proved equal to the angle AEB; therefore the whole angle BCD is also equal to the whole angle AED. But, by hypothesis, the angle BCD is equal to the angles at A, B; therefore the angle AED is also equal to the angles at A, B. Similarly we can prove that the angle CDE is also equal to the angles at A, B, C; therefore the pentagon ABCDE is equiangular. Next, let the given equal angles not be angles taken in order, but let the angles at the points A, C, D be equal; I say that in this case too the pentagon ABCDE is equiangular. For let BD be joined. Then, since the two sides BA, AE are equal to the two sides BC, CD, and they contain equal angles, therefore the base BE is equal to the base BD, the triangle ABE is equal to the triangle BCD, and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend; [I. 4] therefore the angle AEB is equal to the angle CDB. But the angle BED is also equal to the angle BDE, since the side BE is also equal to the side BD. [I. 5] Therefore the whole angle AED is equal to the whole angle CDE. But the angle CDE is, by hypothesis, equal to the angles at A, C; therefore the angle AED is also equal to the angles at A, C. For the same reason the angle ABC is also equal to the angles at A, C, D.