§8## ή.
## ή.
Εἴ κα ἐν κύκλῳ εὐθεῖά τις ΑΒ προσαρμοσμένα ᾖ, ἐκβληθῇ δὲ κατὰ τὸ Γ σαμεῖον, ὥστε τὰν ΒΓ εὐθεῖαν τᾷ ἐκ τοῦ κέντρου ἴσαν εἶμεν, διαχθῇ δὲ εὐθεῖά τις ἀπὸ τοῦ Γ διὰ τοῦ κέντρου τοῦ κύκλου ἐπὶ τὸ Ε σαμεῖον, ἐσσεῖται περιφέρεια ἁ ΑΕ περιφερείας τᾶς ΒΖ τριπλασίων.
If in a circle a straight line AB be fitted [as a chord], and be produced to the point Γ so that the straight line BΓ is equal to the radius, and a straight line is drawn from Γ through the center of the circle to the point E, the arc AE will be triple of the arc BΖ.
Ἄχθω γὰρ ἁ ΕΗ παρὰ τὰν ΑΒ καὶ ἐπεζεύχθων αἱ △Β, △Η. Ἐπεὶ οὖν γωνίαι αἱ ὑπὸ ΒΓ△, Β△Γ, △ΕΗ, △ΗΕ ἴσαι ἀλλάλαις ἐντί, γωνία δὲ ἁ ὑπὸ Γ△Η γωνίας τᾶς ὑπὸ △ΕΗ ἐστὶ διπλασίων, ἐσσεῖται ἄρα γωνία ἁ ὑπὸ Β△Η γωνίας τᾶς ὑπὸ Β△Γ τριπλασίων.
For let EH be drawn parallel to AB, and let ΔB, ΔH be joined. Since, then, the angles BΓΔ, BΔΓ, ΔEH, ΔHE are equal to one another, and the angle ΓΔH is double of the angle ΔEH, therefore the angle BΔH will be triple of the angle BΔΓ.
Ἐσσεῖται ἄρα περιφέρεια ἁ ΒΗ, τουτέστιν ἁ ΑΕ, περιφερείας τᾶς ΒΖ τριπλασίων·
Therefore, the arc BH, that is, AE, will be triple of the arc BΖ.
δέδεικται οὖν τὸ προτεθέν.
Therefore, what was proposed has been proven.
§9## θ΄.
## θ΄.
Εἴ κα ἐν κύκλῳ δύο εὐθεῖαι τέμνωντι ἀλλάλας ποτʼ ὀρθὰς μὴ διὰ τοῦ κέντρου οὖσαι, δύο αἱ ἀπεναντίον περιφέρειαι δυσὶ ταῖς ἀπεναντίον ἴσαι ἀλλάλαις ἐντί.
If in a circle two straight lines not passing through the center cut each other at right angles, the two opposite arcs [together] are equal to the [other] two opposite arcs [together].
Ἔστω κύκλος ὁ ΑΒΓ καὶ δύο εὐθεῖαι αἱ ΑΒ, Γ△ τέμνουσαι ἀλλάλας ποτʼ ὀρθάς, μὴ διὰ τοῦ κέντρου οὖσαι· φαμὶ δή, δύο αἱ ἀπεναντίον περιφέρειαι αἱ Α△, ΓΒ δυσὶ ταῖς ἀπεναντίον ταῖς ΑΓ, Β△ ἴσαι ἀλλάλαις ἐντί.
Let there be a circle ABΓ, and two straight lines AB, ΓΔ not passing through the center cutting each other at right angles; I say indeed that the two opposite arcs AΔ, ΓB [together] are equal to the [other] two opposite arcs AΓ, BΔ [together].
Τετμάσθω γὰρ δίχα ἁ Γ△ κατὰ τὸ H σαμεῖον καὶ διὰ τοῦ H διάχθω διάμετρος τοῦ κύκλου ἁ ΕΖ παρὰ τὰν ΑΒ.
Ἐπεὶ οὖν περιφέρεια ἁ ΕΓ περιφερείαις ταῖς ΕΑ, Α△ ἴσα ἐστίν, ἐσσοῦνται ἄρα περιφέρειαι αἱ ΓΖ, ΕΑ, Α△ ἁμικυκλίῳ ἴσαι·
For let ΓΔ be bisected at the point H, and through H let a diameter of the circle EΖ be drawn parallel to AB. Since, then, the arc EΓ is equal to the arcs EA, AΔ [together], therefore the arcs ΓΖ, EA, AΔ [together] will be equal to a semicircle.
ἔστι δὲ περιφέρεια ἁ ΕΑ περιφερείᾳ τᾷ ΒΖ ἴσα· συναμφότερος ἄρα περιφέρεια ἁ ΓΒ, Α△ ἁμικυκλίῳ ἐστὶν ἴσα·
And the arc EA is equal to the arc BΖ; therefore the combined arc ΓB, AΔ [together] is equal to a semicircle.
λοιπὴ ἄρα περιφέρεια ἁ ΑΓ, △Β ἁμικυκλίῳ ἐστὶν ἴσα δέδεικται οὖν τὸ προτεθέν.
Therefore, the remaining arc AΓ, ΔB [together] is equal to a semicircle. Therefore, what was proposed has been proven.
§10## ι΄.
## ι΄.
Εἴ κα ᾖ κύκλος καὶ δύο εὐθεῖαι αἱ △Α, △Γ ἐπιψαύουσαι αὐτοῦ κατὰ τὰ Α, Γ σαμεῖα, τέμνουσα δὲ εὐθεῖα ἁ △Β, ἀχθῇ δὲ ἁ ΕΓ παρὰ τὰν Β△, ἐπιζευχθῇ δὲ ἁ ΕΑ τέμνουσα τὰν △Β κατὰ τὸ Ζ, καὶ ἀπὸ τοῦ Ζ ποτʼ ὀρθὰς τᾷ ΕΓ ἀχθῇ ἁ ΖΗ, ἁ ἀγμένα τὰν ΕΓ δίχα τέμνει.
If there be a circle and two straight lines ΔA, ΔΓ tangent to it at the points A, Γ, and a secant line ΔB, and let EΓ be drawn parallel to BΔ, and let EA be joined cutting ΔB at the point Ζ, and from Ζ let ΖH be drawn perpendicular to EΓ, the drawn [line ΖH] bisects EΓ.
Ἐπεζεύχθω γὰρ ἁ ΑΓ. Ἐπεὶ οὖν εὐθεῖα ἁ △Α ἐπιψαύουσα τοῦ κύκλου ἐστίν, ἁ δὲ ΑΓ τέμνουσα αὐτόν, γωνία ἁ ὑπὸ △ΑΓ τᾷ ἐν τῷ ἐναλλὰξ τμάματι τοῦ κύκλου γωνίᾳ τᾷ ὑπὸ ΑΕΓ, τουτέστι τᾷ ὑπὸ ΑΖ△, ἐστὶν ἴσα.
For let AΓ be joined. Since, then, the straight line ΔA is tangent to the circle, and AΓ cuts it, the angle ΔAΓ is equal to the angle in the alternate segment of the circle, [namely] AEΓ, that is, AΖΔ.
Ἔστι γὰρ ἁ ΓΕ παρὰ τὰν Β△.
For ΓE is parallel to BΔ.
Καὶ ἐπεὶ ἐν δυσὶ τριγώνοις τοῖς △ΑΖ, ΑΘ△ δύο γωνίαι αἱ ὑπὸ ΑΖ△, ΘΑ△ ἴσαι ἀλλάλαις ἐντί, γωνία δὲ ἁ πρὸς τῷ △ κοινά, τὸ ἄρα ὑπὸ τῶν Ζ△, △Θ περιεχόμενον ὀρθογώνιον τῷ ἀπὸ τᾶς △Α, τουτέστι τῷ ἀπὸ τᾶς △Γ τετραγώνῳ, ἐστὶν ἴσον ἐπεὶ οὖν ὃν λόγον ἔχει ἁ Ζ△ ποτὶ τὰν △Γ, τοῦτον ἔχει καὶ ἁ △Γ ποτὶ τὰν △Θ, γωνία δὲ ἁ ποτὶ τὸ △ σαμεῖον κοινά ἐστιν, τρίγωνα ἄρα τὰ △ΖΓ, △ΓΘ ἐστὶν ὅμοια καὶ γωνίαι αἱ ὑπὸ △ΖΓ, △ΓΘ, △ΑΘ, ΑΖ△ ἴσαι ἀλλάλαις ἐντί·
And since in the two triangles ΔAΖ, AΘΔ two angles AΖΔ, ΘAΔ are equal to one another, and the angle at Δ is common, therefore the rectangle contained by ΖΔ, ΔΘ is equal to the square on ΔA, that is, to the square on ΔΓ. Since, then, as ΖΔ is to ΔΓ, so is ΔΓ to ΔΘ, and the angle at the point Δ is common, therefore the triangles ΔΖΓ, ΔΓΘ are similar, and the angles ΔΖΓ, ΔΓΘ, ΔAΘ, AΖΔ are equal to one another.
ἔστι δὲ καὶ γωνία ἁ ὑπὸ △ΖΓ τᾷ ὑπὸ ΖΓΕ ἴσα·
And the angle ΔΖΓ is also equal to ΖΓE.
ἦν δὲ καὶ ἁ ὑπὸ △ΖΑ τᾷ ὑπὸ ΑΕΓ ἴσα ἐν δυσὶ τριγώνοις ἄρα τοῖς ΕΗΖ, ΓΗΖ δύο γωνίαι αἱ ὑπὸ ΗΕΖ, ΗΓΖ ἴσαι ἀλλάλαις ἐντὶ καὶ αἱ ποτὶ τῷ Η σαμείῳ γωνίαι ὀρθαί·
And the angle ΔΖA was also equal to AEΓ. Therefore, in the two triangles EHΖ, ΓHΖ, the two angles HEΖ, HΓΖ are equal to one another, and the angles at the point H are right angles; and the side HΖ is common.
ἔστι ὲ πλευρὰ ἁ ΗΖ κοινά ἔστιν ἄρα ἁ ΕΗ τᾷ ΗΓ ἴσα·
Therefore, EH is equal to HΓ.
δέδεικται οὖν τὸ προτεθέν.
Therefore, what was proposed has been proven.