§6## ς΄.
## ς΄.
Εἴ κα ἐν ἁμικυκλίῳ σαμεῖόν τι ἐπὶ τᾶς διαμέτρου ᾖ, καὶ γραφέωντι ἀπὸ τῶν τμαμάτων τᾶς διαμέτρου δύο ἁμικύκλια ἐντός, τραφῇ δὲ ἐν τῷ ἀρβήλῳ κύκλος ἐπιψαύων τῶν τριῶν ἁμικυκλίων, τὸν λόγον τᾶς διαμέτρου τοῦ δοθέντος ἁμικυκλίου ποτὶ τὰν διάμετρον τοῦ ἐγγραφέντος κύκλου εὑρεῖν.
If in a semicircle there be any point on the diameter, and two semicircles are described internally from the segments of the diameter, and a circle is described in the arbelos touching the three semicircles, to find the ratio of the diameter of the given semicircle to the diameter of the inscribed circle.
ΑΒΓ, σαμεῖον δὲ τι ἐπὶ τᾶς διαμέτρου τὸ △ καὶ πεποιήσθω οὕτως, ὥστε τὸ μεῖζον τμᾶμα τὸ Α△ ἐλάσσονος τοῦ △Γ ἁμιόλιον εἶμεν, καὶ ἀπὸ τμαμάτων τῶν Α△, △Β ἀναγεγράφθων ἁμικύκλια, γεγράφθω δὲ ἐν τῷ ἀρβήλω κύκλος ὁ ΕΖ ἐπιψαύων τῶν τριῶν ἁμικυκλίων, καὶ ἄχθω διάμετρος αὐτοῦ παρὰ τὰν ΑΓ ἁ ΕΖ. Εὑρεῖν τὸν λόγον διαμέτρου τᾶς ΑΓ ποτὶ διάμετρον τὰν ΕΖ.
Ἐπεζεύχθωσαν αἱ ΑΕ, ΕΒ εὐθεῖαι καὶ αἱ ΓΖ, ΖΒ· εὐθεῖαι δή ἐντι αἱ ΑΒ, ΓΒ, ὡς ἐν τοῖς πρότερον ἐδείχθη.
Let there be [a semicircle] ABΓ, and any point Δ on the diameter, and let it be constructed such that the greater segment AΔ is one and a half times the lesser segment ΔΓ, and let semicircles be described from the segments AΔ, ΔB, and let the circle ΕΖ be described in the arbelos touching the three semicircles, and let its diameter ΕΖ be drawn parallel to AΓ. To find the ratio of the diameter AΓ to the diameter ΕΖ. Let the straight lines AE, EB and ΓΖ, ΖB be joined; the straight lines AB, ΓB are indeed straight lines, as was shown before.
Ἐπεζεύχθωσαν ἔτι αἱ ΖΗΑ, ΕΘΓ δείκνυνται δὴ αὗται εὐθεῖαι οὖσαι ἔτι δὲ ἐπεζεύχθωσαν αἱ △Ε, △Ζ, καὶ αἱ △Ι, △Λ, καὶ ἐπιζευχθεῖσαι αἱ ΕΜ, ΖΝ ἐκβεβλήσθωσαν ἐπὶ τὰ Ο, Ρ σαμεῖα.
Let also ΖΗA, EΘΓ be joined; these are shown to be straight lines. Let further ΔE, ΔΖ, and ΔI, ΔΛ be joined, and let the joined EM, ΖN be produced to the points O, P.
Ἐπεὶ οὖν ἐν τριγώνῳ τῷ ΑΕ△ ἁ ΑΗ τᾷ Ε△ ποτʼ ὀρθάς ἐστιν, καὶ ἁ △Ι τᾷ ΑΕ, τέμνοντι δὲ ἀλλάλας κατὰ τὸ Μ σαμεῖον, ἁ ΕΜΟ τᾷ ΑΓ ἐσσεῖται ποτʼ ὀρθάς, ὡς παῤ ἡμῶν ἐν τοῖς Περὶ τριγώνων ἐδείχθη καὶ τῷ πρότερον ὑπέκειτο· διὰ τὰ αὐτὰ δὴ καὶ ἁ ΖΝΡ τᾷ ΓΑ ἐσσεῖται ποτʼ ὀρθάς·
Since, then, in the triangle AEΔ, AH is perpendicular to EΔ, and ΔI to AE, and they cut each other at the point M, EMO will be perpendicular to AΓ, as was shown by us in our work "On Triangles" and assumed before; for the same reasons, indeed, ΖNP will also be perpendicular to ΓA.
ἔστι δὲ εὐθεῖα ἁ △Λ παρὰ τὰν ΑΒ καὶ ἁ △Ι παρὰ τὰν ΓΒ· ὥστε τὸν αὐτὸν λόγον ἔχει ἁ Α△ ποτὶ τὰν △Γ, ὃν ἔχει ἁ ΑΜ ποτὶ τὰν ΜΖ, τουτέστιν ἁ ΑΟ ποτὶ τὰν καὶ ἁ Γ△ ποτὶ τὰν △Α τὸν αὐτὸν λόγον ἔχει, ὃν ἔχει ἁ ΓΝ ποτὶ τὰν ΝΕ, τουτέστιν ἁ ΓΡ ποτὶ τὰν ΡΟ·
And the straight line ΔΛ is parallel to AB, and ΔI is parallel to ΓB; so that AΔ has the same ratio to ΔΓ as AM has to MΖ, that is, AO to [OP], and ΓΔ to ΔA has the same ratio as ΓN has to NE, that is, ΓP to PO.
ἦν δὲ ἁ Α△ ἁμιόλιος τᾶς △Γ· καὶ ἁ ΑΟ ἄρα τᾶς ΟΡ ἐστὶν ἁμιόλιος, καὶ ἁ ΟΡ τᾶς ΓΡ·
And AΔ was one and a half times ΔΓ; therefore AO is also one and a half times OP, and OP of ΓP.
εὐθεῖαι ἄρα αἱ ΑΟ, ΟΡ, ΡΓ ἑξῆς ἀνάλόγον ἐντι, ἇν ἁ μὲν ΡΓ ἴσα γίνεται τέσσαρα, ἁ δὲ ΟΡ ἕξ, ἁ δὲ ΑΟ ἐννέα, ἁ δὲ ΓΑ ἐννεακαίδεκα.
Therefore, the straight lines AO, OP, PΓ are in continued proportion, of which ΓP becomes equal to four, OP to six, AO to nine, and ΓA to nineteen.
Ἔστι δὲ ἁ ΡΟ τᾷ ΕΖ ἴσα ὥστε τὸν αὐτὸν λόγον ἔχει ἁ ΑΓ ποτὶ τὰν ΕΖ, ὃν ἔχει τὰ ἐννεακαίδεκα ποτὶ τὰ ἕξ·
And PO is equal to EΖ; so that AΓ has the same ratio to EΖ as nineteen has to six.
καί ἐστιν ἁ ΑΓ διάμετρος ἁμικυκλίου τοῦ ΑΒΓ, ἁ δὲ ΕΖ κύκλου τοῦ ΕΒΖ· εὑρέθη ἄρα ὁ αἰτούμενος λόγος.
And AΓ is the diameter of the semicircle ABΓ, and EΖ of the circle EBΖ; therefore the required ratio has been found.
Ὁμοίως δὴ δειχθήσεται εἴ κα ὁ λόγος τᾶς διαμέτρου τοῦ δοθέντος ἁμικυκλίου ποτὶ τὰν διάμετρον τοῦ ἐγγραφέντος κύκλου ἐπιμόριος ᾖ.
In like manner indeed it will be shown if the ratio of the diameter of the given semicircle to the diameter of the inscribed circle be superparticular.
§7## ζ΄.
## ζ΄.
Ὁ τετραγώνῳ περιγεγραμμένος κύκλος διπλασίων τοῦ ἐγγεγραμμένου ἐστίν.
The circle circumscribed about a square is double of the inscribed circle.
Ἔστω γὰρ κύκλος ὁ ΑΒ περὶ τετράγωνον τὸ ΑΒ καὶ ἐν αὐτῷ ἐγγεγραμμένος κύκλος ὁ Γ△, διάμετρος δὲ τοῦ περιγεγραμμένου κύκλου καὶ τοῦ τετραγώνου ἄχθω δὲ διάμετρος τοῦ ἐγγεγραμμένου κύκλου ἁ Γ△ παρὰ τὰν ΑΕ·
For let there be a circle AB about a square AB, and a circle ΓΔ inscribed in it, and let the diameter of the circumscribed circle and of the square [be AB], and let the diameter ΓΔ of the inscribed circle be drawn parallel to AE.
φαμὶ Et δή, ὁ περιγεγραμμένος κύκλος τοῦ ἐγγεγραμμένου ἐστὶ διπλασίων.
I say indeed that the circumscribed circle is double of the inscribed circle.
Ἐπεὶ οὖν τὸ ἀπὸ τᾶς ΑΒ διπλάσιον τοῦ ἀπὸ τᾶς ΑΕ, τουτέστι τοῦ ἀπὸ τᾶς Γ△, οἱ κύκλοι δὲ ἐντι ὡς τὰ ἀπὸ τᾶν διαμέτρων αὐτῶν τετράγωνα, ἐσσεῖται ἄρα καὶ ὁ περιγεγραμμένος κύκλος τοῦ ἐγγεγραμμένου διπλασίων·
Since, then, the square on AB is double of the square on AE, that is, of the square on ΓΔ, and circles are as the squares on their diameters, therefore the circumscribed circle will also be double of the inscribed circle.
δέδεικται οὖν τὸ προτεθέν.
Therefore, what was proposed has been proven.