§4.prop.11εἰς τὸν δοθέντα κύκλον πεντάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον ἐγγράψαι.
To inscribe an equilateral and equiangular pentagon in the given circle.
ἔστω ὁ δοθεὶς κύκλος ὁ ΑΒΓΔΕ· δεῖ δὴ εἰς τὸν ΑΒΓΔΕ κύκλον πεντάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον ἐγγράψαι.
Let the given circle be ABCDE; it is required then to inscribe an equilateral and equiangular pentagon in the circle ABCDE.
Ἐκκείσθω τρίγωνον ἰσοσκελὲς τὸ ΖΗΘ διπλασίονα ἔχον ἑκατέραν τῶν πρὸς τοῖς Η, Θ γωνιῶν τῆς πρὸς τῷ Ζ, καὶ ἐγγεγράφθω εἰς τὸν ΑΒΓΔΕ κύκλον τῷ ΖΗΘ τριγώνῳ ἰσογώνιον τρίγωνον τὸ ΑΓΔ, ὥστε τῇ μὲν πρὸς τῷ Ζ γωνίᾳ ἴσην εἶναι τὴν ὑπὸ ΓΑΔ, ἑκατέραν δὲ τῶν πρὸς τοῖς Η, Θ ἴσην ἑκατέρᾳ τῶν ὑπὸ ΑΓΔ, ΓΔΑ· καὶ ἑκατέρα ἄρα τῶν ὑπὸ ΑΓΔ, ΓΔΑ τῆς ὑπὸ ΓΑΔ ἐστι διπλῆ.
Let an isosceles triangle FGH be set out having each of the angles at G, H double of the angle at F, and let there be inscribed in the circle ABCDE the triangle ACD equiangular to the triangle FGH, so that the angle CAD is equal to the angle at F, and each of the angles at G, H is equal to each of the angles ACD, CDA; therefore each of the angles ACD, CDA is also double of the angle CAD.
τετμήσθω δὴ ἑκατέρα τῶν ὑπὸ ΑΓΔ, ΓΔΑ δίχα ὑπὸ ἑκατέρας τῶν ΓΕ, ΔΒ εὐθειῶν, καὶ ἐπεζεύχθωσαν αἱ ΑΒ, ΒΓ,, ΔΕ, ΕΑ.
ἐπεὶ οὖν ἑκατέρα τῶν ὑπὸ ΑΓΔ, ΓΔΑ γωνιῶν διπλασίων ἐστὶ τῆς ὑπὸ ΓΑΔ, καὶ τετμημέναι εἰσὶ δίχα ὑπὸ τῶν ΓΕ, ΔΒ εὐθειῶν, αἱ πέντε ἄρα γωνίαι αἱ ὑπὸ ΔΑΓ, ΑΓΕ, ΕΓΔ, ΓΔΒ, ΒΔΑ ἴσαι ἀλλήλαις εἰσίν.
Let then each of the angles ACD, CDA be bisected by each of the straight lines CE, DB, and let AB, BC,, DE, EA be joined. Since then each of the angles ACD, CDA is double of the angle CAD, and they have been bisected by the straight lines CE, DB, therefore the five angles DAC, ACE, ECD, CDB, BDA are equal to one another.
αἱ δὲ ἴσαι γωνίαι ἐπὶ ἴσων περιφερειῶν βεβήκασιν· αἱ πέντε ἄρα περιφέρειαι αἱ ΑΒ, ΒΓ, ΓΔ, ΔΕ, ΕΑ ἴσαι ἀλλήλαις εἰσίν.
But equal angles stand on equal circumferences; therefore the five circumferences AB, BC, CD, DE, EA are equal to one another.
ὑπὸ δὲ τὰς ἴσας περιφερείας ἴσαι εὐθεῖαι ὑποτείνουσιν· αἱ πέντε ἄρα εὐθεῖαι αἱ ΑΒ, ΒΓ, ΓΔ, ΔΕ, ΕΑ ἴσαι ἀλλήλαις εἰσίν· ἰσόπλευρον ἄρα ἐστὶ τὸ ΑΒΓΔΕ πεντάγωνον.
And under equal circumferences equal straight lines subtend; therefore the five straight lines AB, BC, CD, DE, EA are equal to one another; therefore the pentagon ABCDE is equilateral.
λέγω δή, ὅτι καὶ ἰσογώνιον.
I say then that it is also equiangular.
ἐπεὶ γὰρ ἡ ΑΒ περιφέρεια τῇ ΔΕ περιφερείᾳ ἐστὶν ἴση, κοινὴ προσκείσθω ἡ ΒΓΔ· ὅλη ἄρα ἡ ΑΒΓΔ περιφέρεια ὅλῃ τῇ ΕΔΓΒ περιφερείᾳ ἐστὶν ἴση.
For since the circumference AB is equal to the circumference DE, let the circumference BCD be added to each; therefore the whole circumference ABCD is equal to the whole circumference EDCB.
καὶ βέβηκεν ἐπὶ μὲν τῆς ΑΒΓΔ περιφερείας γωνία ἡ ὑπὸ ΑΕΔ, ἐπὶ δὲ τῆς ΕΔΓΒ περιφερείας γωνία ἡ ὑπὸ ΒΑΕ· καὶ ἡ ὑπὸ ΒΑΕ ἄρα γωνία τῇ ὑπὸ ΑΕΔ ἐστιν ἴση.
And on the circumference ABCD stands the angle AED, and on the circumference EDCB stands the angle BAE; therefore the angle BAE is also equal to the angle AED.
διὰ τὰ αὐτὰ δὴ καὶ ἑκάστη τῶν ὑπὸ ΑΒΓ, ΒΓΔ, ΓΔΕ γωνιῶν ἑκατέρᾳ τῶν ὑπὸ ΒΑΕ, ΑΕΔ ἐστιν ἴση· ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΓΔΕ πεντάγωνον.
For the same reasons each of the angles ABC, BCD, CDE is also equal to each of the angles BAE, AED; therefore the pentagon ABCDE is equiangular.
ἐδείχθη δὲ καὶ ἰσόπλευρον.
And it has also been proved equilateral.
εἰς ἄρα τὸν δοθέντα κύκλον πεντάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον ἐγγέγραπται· ὅπερ ἔδει ποιῆσαι.
Therefore an equilateral and equiangular pentagon has been inscribed in the given circle; which was required to do.