§13.prop.8ἐὰν πενταγώνου ἰσοπλεύρου καὶ ἰσογωνίου τὰς κατὰ τὸ ἑξῆς δύο γωνίας ὑποτείνωσιν εὐθεῖαι, ἄκρον καὶ μέσον λόγον τέμνουσιν ἀλλήλας, καὶ τὰ μείζονα αὐτῶν τμήματα ἴσα ἐστὶ τῇ τοῦ πενταγώνου πλευρᾷ.
If in an equilateral and equiangular pentagon straight lines subtend two consecutive angles, they cut one another in extreme and mean ratio, and their greater segments are equal to the side of the pentagon.
πενταγώνου γὰρ ἰσοπλεύρου καὶ ἰσογωνίου τοῦ ΑΒΓ ΔΕ δύο γωνίας τὰς κατὰ τὸ ἑξῆς τὰς πρὸς τοῖς Α, Β ὑποτεινέτωσαν εὐθεῖαι αἱ ΑΓ, ΒΕ τέμνουσαι ἀλλήλας κατὰ τὸ Θ σημεῖον· λέγω, ὅτι ἑκατέρα αὐτῶν ἄκρον καὶ μέσον λόγον τέτμηται κατὰ τὸ Θ σημεῖον, καὶ τὰ μείζονα αὐτῶν τμήματα ἴσα ἐστὶ τῇ τοῦ πενταγώνου πλευρᾷ.
For in the equilateral and equiangular pentagon ABCDE let the straight lines AC, BE subtend two consecutive angles, namely those at A, B, cutting one another at the point Θ; I say that each of them is cut in extreme and mean ratio at the point Θ, and their greater segments are equal to the side of the pentagon.
περιγεγράφθω γὰρ περὶ τὸ ΑΒΓΔΕ πεντάγωνον κύκλος ὁ ΑΒΓΔΕ. καὶ ἐπεὶ δύο εὐθεῖαι αἱ ΕΑ, ΑΒ δυσὶ ταῖς ΑΒ, ΒΓ ἴσαι εἰσὶ καὶ γωνίας ἴσας περιέχουσιν, βάσις ἄρα ἡ ΒΕ βάσει τῇ ΑΓ ἴση ἐστίν, καὶ τὸ ΑΒΕ τρίγωνον τῷ ΑΒΓ τριγώνῳ ἴσον ἐστίν, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι ἔσονται ἑκατέρα ἑκατέρᾳ, ὑφʼ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν.
For let the circle ABCDE be circumscribed about the pentagon ABCDE. And since the two straight lines EA, AB are equal to the two AB, BC, and contain equal angles, therefore the base BE is equal to the base AC, the triangle ABE is equal to the triangle ABC, and the remaining angles will be equal to the remaining angles, each to each, namely those subtended by the equal sides.
ἴση ἄρα ἐστὶν ἡ ὑπὸ ΒΑΓ γωνία τῇ ὑπὸ ΑΒΕ· διπλῆ ἄρα ἡ ὑπὸ ΑΘΕ τῆς ὑπὸ ΒΑΘ. ἔστι δὲ καὶ ἡ ὑπὸ ΕΑΓ τῆς ὑπὸ ΒΑΓ διπλῆ, ἐπειδήπερ καὶ περιφέρεια ἡ ΕΔΓ περιφερείας τῆς ΓΒ ἐστι διπλῆ· ἴση ἄρα ἡ ὑπὸ ΘΑΕ γωνία τῇ ὑπὸ ΑΘΕ· ὥστε καὶ ἡ ΘΕ εὐθεῖα τῇ ΕΑ, τουτέστι τῇ ΑΒ ἐστιν ἴση.
Therefore the angle BAC is equal to the angle ABE; therefore the angle AΘE is double of the angle BAΘ. But the angle EAΓ is also double of the angle BAC, since the circumference EΔC is also double of the circumference ΓB; therefore the angle ΘAE is equal to the angle AΘE; so that the straight line ΘE is also equal to EA, that is, to AB.
καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΑ εὐθεῖα τῇ ΑΕ, ἴση ἐστὶ καὶ γωνία ἡ ὑπὸ ΑΒΕ τῇ ὑπὸ ΑΕΒ. ἀλλὰ ἡ ὑπὸ ΑΒΕ τῇ ὑπὸ ΒΑΘ ἐδείχθη ἴση· καὶ ἡ ὑπὸ ΒΕΑ ἄρα τῇ ὑπὸ ΒΑΘ ἐστιν ἴση.
And since the straight line BA is equal to AE, the angle ABE is also equal to the angle AEB. But the angle ABE was shown to be equal to the angle BAΘ; therefore the angle BEA is also equal to the angle BAΘ.
καὶ κοινὴ τῶν δύο τριγώνων τοῦ τε ΑΒΕ καὶ τοῦ ΑΒΘ ἐστιν ἡ ὑπὸ ΑΒΕ· λοιπὴ ἄρα ἡ ὑπὸ ΒΑΕ γωνία λοιπῇ τῇ ὑπὸ ΑΘΒ ἐστιν ἴση· ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΕ τρίγωνον τῷ ΑΒΘ τριγώνῳ· ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΕΒ πρὸς τὴν ΒΑ, οὕτως ἡ ΑΒ πρὸς τὴν ΒΘ. ἴση δὲ ἡ ΒΑ τῇ ΕΘ· ὡς ἄρα ἡ ΒΕ πρὸς τὴν ΕΘ, οὕτως ἡ ΕΘ πρὸς τὴν ΘΒ. μείζων δὲ ἡ ΒΕ τῆς ΕΘ· μείζων ἄρα καὶ ἡ ΕΘ τῆς ΘΒ. ἡ ΒΕ ἄρα ἄκρον καὶ μέσον λόγον τέτμηται κατὰ τὸ Θ, καὶ τὸ μεῖζον τμῆμα τὸ ΘΕ ἴσον ἐστὶ τῇ τοῦ πενταγώνου πλευρᾷ.
And the angle ABE is common to the two triangles ABE and ABΘ; therefore the remaining angle BAE is equal to the remaining angle AΘB; therefore the triangle ABE is equiangular with the triangle ABΘ; therefore, as EB is to BA, so is AB to BΘ. But BA is equal to EΘ; therefore, as BE is to EΘ, so is EΘ to ΘB. And BE is greater than EΘ; therefore EΘ is also greater than ΘB. Therefore BE is cut in extreme and mean ratio at Θ, and the greater segment ΘE is equal to the side of the pentagon.
ὁμοίως δὴ δείξομεν, ὅτι καὶ ἡ ΑΓ ἄκρον καὶ μέσον λόγον τέτμηται κατὰ τὸ Θ, καὶ τὸ μεῖζον αὐτῆς τμῆμα ἡ ΓΘ ἴσον ἐστὶ τῇ τοῦ πενταγώνου πλευρᾷ· ὅπερ ἔδει δεῖξαι.
Similarly we shall show that AC is also cut in extreme and mean ratio at Θ, and its greater segment ΓΘ is equal to the side of the pentagon; which it was required to prove.