§15## ιε΄.
## ιε΄.
Ἔστω ἁμικύκλιον τὸ ΑΒ καὶ ἁ ΑΓ πλευρὰ τοῦ ἐγγεγραμμένου ἰσοπλεύρου τε καὶ ἰσογωνίου πενταγώνου, τετμάσθω δὲ περιφέρεια ἁ ΑΓ δίχα κατὰ τὸ △, ἐπιζευχθεῖσα δὲ ἁ Γ△ ἐκβεβλήσθω ἐπὶ τὸ Ε, καὶ ἀπὸ τοῦ △ σαμείου διάχθω ἁ △Β τέμνουσα πλευρὰν τὰν ΑΓ κατὰ τὸ Ζ, καὶ ἀπὸ τοῦ Ζ ἄχθω τᾷ ΑΒ ποτʼ ὀρθὰς ἁ ΖΗ φαμὶ δή, εὐθεῖα ἁ ΕΗ τᾷ ἐκ τοῦ κέντρου τοῦ κύκλου ἴσα ἐστίν.
Let there be a semicircle AB, and let AΓ be a side of the inscribed equilateral and equiangular pentagon, and let the arc AΓ be bisected at △, and let the joined Γ△ be produced to E, and from the point △ let △B be drawn cutting the side AΓ at Z, and from Z let ZH be drawn at right angles to AB; I say indeed that the straight line EH is equal to the radius (the straight line from the center) of the circle.
Ἐπεζεύχθω γὰρ ἁ ΓΒ, καὶ ἔστω κέντρον τοῦ κύκλου τὸ Θ σαμεῖον, καὶ ἄχθωσαν αἱ Θ△, △Η, Α△ εὐθεῖαι.
For let ΓB be joined, and let the center of the circle be the point Θ, and let the straight lines Θ△, △H, A△ be drawn.
Ἐπεὶ οὖν γωνία ἁ ὑπὸ ΑΒΓ δύο πέμπτα ὀρθᾶς ἐστιν, γωνία ἁ ὑπὸ ΓΒ△, τουτέστι ἁ ὑπὸ △ΒΑ, ἓν πεμπταμόριον ὀρθᾶς ἐστιν· γωνία ἄρα ἁ ὑπὸ △ΘΑ δύο πέμπτα ὀρθᾶς ἐστιν.
Since, then, the angle ABΓ is two-fifths of a right angle, the angle ΓB△, that is, △BA, is one-fifth of a right angle; therefore, the angle △ΘA is two-fifths of a right angle.
Καὶ ἐπεὶ ἐν δυσὶ τριγώνοις τοῖς ΓΒΖ, ΗΒΖ δύο γωνίαι αἱ ποτὶ τῷ Β ἴσαι ἀλλάλαις ἐντί, ὀρθαὶ δὲ αἱ ποτὶ τὰ Η, Γ σαμεῖα, κοινὰ δὲ πλευρὰ ἁ ΒΖ, ἐσσεῖται ἄρα καὶ βάσις ἁ ΒΓ βάσει τᾷ ΒΗ ἴσα.
And since in the two triangles ΓBZ, HBZ two angles at B are equal to one another, and those at the points H, Γ are right angles, and the side BZ is common, therefore the base BΓ will also be equal to the base BH.
Πάλιν ἐπεὶ ἐν δυσὶ τριγώνοις τοῖς ΓΒ△, ΗΒ△ δύο πλευραὶ αἱ ΓΒ, ΒΗ ἴσαι ἀλλάλαις ἐντί, γωνίαι δὲ αἱ ποτὶ τῷ Β ἴσαι, κοινὰ δὲ πλευρὰ ἁ Β△, ἐσσεῖται ἄρα γωνία ἁ ὑπὸ ΒΓ△ γωνίᾳ τᾷ ὑπὸ ΒΗ△, τουτέστιν ἐπιπέμπτῳ ὀρθᾶς, ἴσα·
Again, since in the two triangles ΓB△, HB△ the two sides ΓB, BH are equal to one another, and the angles at B are equal, and the side B△ is common, therefore the angle BΓ△ will be equal to the angle BH△, that is, one and one-fifth of a right angle.
ἔστι BD δὲ ἑκατέρα τῶν ὑπὸ ΒΓ△, ΒΗ△ γωνιῶν γωνίᾳ τᾷ ἐκτὸς τοῦ ἐν τῷ κύκλῳ τετραπλεύτοῦ ῥοῦ ΒΑ△Γ, τουτέστι τᾷ △ΑΕ, ἴσα γωνία ἄρα ἁ ὑπὸ △ΑΒ γωνίᾳ τᾷ ὑπὸ △ΗΑ ἔστιν ἴσα, καὶ πλευρὰ ἁ △Α τᾷ △Η. Καὶ ἐπεὶ γωνία ἁ ὑπὸ △ΘΗ βε΄ ὀρθᾶς ἐστι καὶ ἁ ὑπὸ △ΗΘ ἐπίπεμπτος ὀρθᾶς, γωνία ἄρα ἁ ὑπὸ Θ△Η βέ ὀρθᾶς ἐστιν πλευρὰ ἄρα ἁ △ Η πλευρᾷ τᾷ ΗΘ ἐστὶν ἴσα.
And each of the angles BΓ△, BH△ is equal to the exterior angle of the cyclic quadrilateral BA△Γ, that is, the angle △AE; therefore, the angle △AB is equal to the angle △HA, and the side △A to △H. And since the angle △ΘH is two-fifths of a right angle, and △HΘ is one and one-fifth of a right angle, therefore the angle Θ△H is two-fifths of a right angle; therefore, the side △H is equal to the side HΘ.
Πάλιν, ἐπεὶ γωνία ἁ ὑπὸ Α△Ε τοῦ ἐν τῷ κύκλῳ τετραπλεύρου τοῦ Α△ΓΒ ἐκτός ἐστιν, ἐσσεῖται ἄρα γωνία ἁ ὑπὸ Α△Ε γωνίᾳ τᾷ ὑπὸ ΑΒΓ ἴσα· ἔστι δὲ γωνία ἁ ὑπὸ ΑΒΓ βγ΄ ὀρθᾶς γωνία ἄρα ἁ ὑπὸ Α△Ε γωνίᾳ τᾷ ὑπὸ Η△Θ ἐστὶν ἴσα.
Again, since the angle A△E is an exterior angle of the cyclic quadrilateral A△ΓB, therefore the angle A△E will be equal to the angle ABΓ; and the angle ABΓ is two-fifths of a right angle, therefore the angle A△E is equal to the angle H△Θ.
Καὶ ἐπεὶ ἐν δυσὶ τριγώνοις τοῖς Ε△Α, Θ△Η δύο γωνίαι αἱ ὑπὸ Ε△Α, △ΑΕ δυσὶ ταῖς ὑπὸ Θ△Η. △ΗΘ ἑκατέρα ἑκατέρᾳ ἴσαι ἐντί, βάσις δὲ ἁ △Α βάσει τᾷ △Η ἴσα, πλευρὰ ἄρα ἁ ΕΑ πλευρᾷ τᾷ ΘΗ ἴσα ἐστίν.
And since in the two triangles E△A, Θ△H the two angles E△A, △AE are equal respectively to the two angles Θ△H, △HΘ, and the base △A is equal to the base △H, therefore the side EA is equal to the side ΘH.
Κοινὰ ποτικείσθω ἁ ΑΗ εὐθεῖα ἄρα ἁ ΕΗ εὐθείᾳ τᾷ ΑΘ, τουτέστι τᾷ ἐκ τοῦ κέντρου τοῦ κύκλου, ἴσα ἐστίν· δέδεικται οὖν τὸ προτεθέν.
Let the common part AH be added; therefore, the straight line EH is equal to the straight line AΘ, that is, to the radius of the circle; therefore, what was proposed has been shown.
ΠΟΡΙΣΜΑ
Ἐκ τούτου δὴ φανερὸν ὅτι εὐθεῖα ἁ △Ε τᾷ ἐκ τοῦ κέντρου τοῦ κύκλου ἐστὶν ἴσα.
COROLLARY From this indeed it is manifest that the straight line △E is equal to the radius of the circle.
Ἐπεὶ γὰρ γωνία ἁ ὑπὸ △ΑΕ γωνίᾳ τᾷ ὑπὸ △ΗΘ ἴσα ἐστίν, ἐσσεῖται καὶ πλευρὰ ἁ △Θ πλευρᾷ τᾷ △Ε, τουτέστι τᾷ ΑΘ, ἴσα.
For since the angle △AE is equal to the angle △HΘ, the side △Θ will also be equal to the side △E, that is, to AΘ.
ΠΟΡΙΣΜΑ
Καὶ ἔτι δῆλον ὅτι εὐθεῖα ἁ ΑΓ ἄκρον καὶ μέσον τέτμαται κατὰ τὸ △ σαμεῖον τμᾶμα δὲ τὸ △Ε τὸ μεῖζόν ἐστιν, ἐπεὶ ἁ Ε△ πλευρὰ τοῦ ἑξαγώνου, ἁ δὲ △Γ πλευρὰ τοῦ δεκαγώνου τῶν ἐν τῷ κύκλῳ ἐγγραφομένων.
COROLLARY And it is further manifest that the straight line AΓ is cut in extreme and mean ratio at the point △, and the segment △E is the greater part, since E△ is a side of the hexagon, and △Γ is a side of the decagon inscribed in the circle.