§3#2Ἔστω κύκλος καὶ διάμετρος ἡ ΑΓ, ἡ δὲ ὑπὸ ΒΑΓ τρίτου ὀρθῆς· ἡ ΑΒ ἄρα πρὸς ΒΓ ἐλάσσονα λόγον ἔχει ἢ ὃν αταν πρὸς ψπ.
Let there be a circle, and the diameter ΑΓ, and the angle ΒΑΓ one-third of a right angle; therefore ΑΒ has to ΒΓ a less ratio than 1351 has to 780.
Δίχα ἡ ὑπὸ ΒΑΓ τῇ ΑΗ. Ἐπεὶ οὖν ἴση ἐστὶν ἡ ὑπὸ ΒΑΗ τῇ ὑπὸ ΗΓΒ, ἀλλὰ καὶ τῇ ὑπὸ ΗΑΓ, καὶ ἡ ὑπὸ ΗΓΒ τῇ ὑπὸ ΗΑΓ ἐστὶν ἴση.
Let the angle ΒΑΓ be bisected by ΑΗ. Since, then, the angle ΒΑΗ is equal to the angle ΗΓΒ, but is also equal to the angle ΗΑΓ, the angle ΗΓΒ is also equal to the angle ΗΑΓ.
Καὶ κοινὴ ἡ ὑπὸ ΑΗΓ ὀρθή· καὶ τρίτη ἄρα ἡ ὑπὸ ΗΖΓ τρίτῃ τῇ ὑπὸ ΑΓΗ ἴση.
And the common angle ΑΗΓ is a right angle; therefore the third angle ΗΖΓ is also equal to the third angle ΑΓΗ.
Ἰσογώνιον ἄρα τὸ ΑΗΓ τῷ ΓΗΖ τριγώνῳ· ἔστιν ἄρα, ὡς ἡ ΑΗ πρὸς ΗΓ, ἡ ΓΗ πρὸς ΗΖ καὶ ἡ ΑΓ πρὸς ΓΖ. Ἀλλʼ ὡς ἡ ΑΓ πρὸς ΓΖ, συναμφότερος ἡ ΓΑΒ πρὸς ΒΓ· καὶ ὡς συναμφότερος ἄρα ἡ ΒΑΓ πρὸς ΒΓ, ἡ ΑΗ πρὸς ΗΓ. Διὰ τοῦτο οὖν ἡ ΑΗ πρὸς ΗΓ ἐλάσσονα λόγον ἔχει ἤπερ βϡια πρὸς ψπ, ἡ δὲ ΑΓ πρὸς τὴν ΓΗ ἐλάσσονα ἢ ὃν γιγ U+2220΄ δ΄ πρὸς ψπ.
Therefore the triangle ΑΗΓ is equiangular with the triangle ΓΗΖ; therefore, as ΑΗ is to ΗΓ, so is ΓΗ to ΗΖ and ΑΓ to ΓΖ. But as ΑΓ is to ΓΖ, so is the sum of ΓΑ and ΑΒ to ΒΓ; and as the sum of ΒΑ and ΑΓ is to ΒΓ, so is ΑΗ to ΗΓ. Therefore on this account ΑΗ has to ΗΓ a less ratio than 2911 has to 780, while ΑΓ has to ΓΗ a less ratio than 3013 and 1/2 and 1/4 has to 780.
Δίχα ἡ ὑπὸ ΓΑΗ τῇ ΑΘ· ἡ ΑΘ ἄρα διὰ τὰ αὐτὰ πρὸς τὴν ΘΓ ἐλάσσονα λόγον ἔχει ἢ ὃν εϡκδ U+2220΄ δ΄ πρὸς ψπ ἢ ὃν αωκγ πρὸς σμ· ἑκατέρα γὰρ ἑκατέρας δ ιγ΄· ὥστε ἡ ΑΓ πρὸς τὴν ΓΘ ἢ ὃν αωλη θ ια΄ πρὸς σμ.
Let the angle ΓΑΗ be bisected by ΑΘ; therefore, for the same reasons, ΑΘ has to ΘΓ a less ratio than 5924 and 1/2 and 1/4 has to 780, or than 1823 has to 240 (for each of the latter is 4/13 of each of the former); so that ΑΓ has to ΓΘ a less ratio than 1838 and 9/11 has to 240.
Ἔτι δίχα ἡ ὑπὸ ΘΑΓ τῇ ΚΑ· καὶ ὁ ΑΚ πρὸς τὴν ΚΓ ἐλάσσονα λόγον ἔχει ἢ ὃν αζ πρὸς ξς· ἑκατέρα γὰρ ἑκατέρας ια μ΄.
Further, let the angle ΘΑΓ be bisected by ΚΑ; and ΑΚ has to ΚΓ a less ratio than 1007 has to 66 (for each of the latter is 11/40 of each of the former).
Ἡ ΑΓ ἄρα πρὸς ΚΓ ἢ ὃν αθ ϛ΄ πρὸς ξς.
Therefore ΑΓ has to ΚΓ a less ratio than 1009 and 1/6 has to 66.
Ἔτι δίχα ἡ ὑπὸ ΚΑΓ τῇ ΛΑ· ἡ ΑΛ ἄρα πρὸς ΛΓ ἐλάσσονα λόγον ἔχει ἢ ὃν τὰ βις ϛ΄ πρὸς ξς, ἡ δὲ ΑΓ πρὸς ΓΛ ἐλάσσονα ἢ τὰ βιζ δ΄ πρὸς ξς.
Further, let the angle ΚΑΓ be bisected by ΛΑ; therefore ΑΛ has to ΛΓ a less ratio than 2016 and 1/6 has to 66, while ΑΓ has to ΓΛ a less ratio than 2017 and 1/4 has to 66.
Ἀνάπαλιν ἄρα ἡ περίμετρος τοῦ πολυγώνου πρὸς τὴν διάμετρον μείζονα λόγον ἔχει ἤπερ ςτλς πρὸς βιζ δ΄. ἅπερ τῶν βιζ δ΄ μείζονά ἐστιν ἢ τριπλασίονα καὶ δέκα οα΄· καὶ ἡ περίμετρος ἄρα τοῦ (??)ςγώνου τοῦ ἐν τῷ κύκλῳ τῆς διαμέτρου τριπλασίων ἐστὶ καὶ μείζων ἢ ι οα΄· ὥστε καὶ ὁ κύκλος ἔτι μᾶλλον τριπλασίων ἐστὶ καὶ μείζων ἢ ι οα΄.
Conversely, therefore, the perimeter of the polygon has to the diameter a greater ratio than 6336 has to 2017 and 1/4, which is greater than three times and ten seventy-firsts of 2017 and 1/4; and therefore the perimeter of the 96-sided polygon inscribed in the circle is three times the diameter and greater than ten seventy-firsts; so that the circle also is much more three times and greater than ten seventy-firsts.
Ἡ ἄρα τοῦ κύκλου περίμετρος τῆς διαμέτρου τριπλασίων ἐστὶ καὶ ἐλάσσονι μὲν ἢ ἑβδόμῳ μέρει, μείζονι δὲ ἢ ι οα΄ μείζων.
Therefore the perimeter of the circle is three times the diameter, and exceeds it by less than a seventh part, but by more than ten seventy-firsts.