§3.prop.34ἀπὸ τοῦ δοθέντος κύκλου τμῆμα ἀφελεῖν δεχόμενον γωνίαν ἴσην τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ.
From a given circle to cut off a segment admitting an angle equal to a given rectilinear angle.
ἔστω ὁ δοθεὶς κύκλος ὁ ΑΒΓ, ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος ἡ πρὸς τῷ Δ· δεῖ δὴ ἀπὸ τοῦ ΑΒΓ κύκλου τμῆμα ἀφελεῖν δεχόμενον γωνίαν ἴσην τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ τῇ πρὸς τῷ Δ.
ἤχθω τοῦ ΑΒΓ ἐφαπτομένη ἡ ΕΖ κατὰ τὸ Β σημεῖον, καὶ συνεστάτω πρὸς τῇ ΖΒ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Β τῇ πρὸς τῷ Δ γωνίᾳ ἴση ἡ ὑπὸ ΖΒΓ.
ἐπεὶ οὖν κύκλου τοῦ ΑΒΓ ἐφάπτεταί τις εὐθεῖα ἡ ΕΖ, καὶ ἀπὸ τῆς κατὰ τὸ Β ἐπαφῆς διῆκται ἡ ΒΓ, ἡ ὑπὸ ΖΒΓ ἄρα γωνία ἴση ἐστὶ τῇ ἐν τῷ ΒΑΓ ἐναλλὰξ τμήματι συνισταμένῃ γωνίᾳ.
Let the given circle be ΑΒΓ, and the given rectilinear angle that at Δ; thus it is required from the circle ΑΒΓ to cut off a segment admitting an angle equal to the given rectilinear angle at Δ. Let ΕΖ be drawn touching ΑΒΓ at the point Β, and let the angle ΖΒΓ be constructed on the straight line ΖΒ and at the point Β on it equal to the angle at Δ. Since then a straight line ΕΖ touches the circle ΑΒΓ, and from the contact at Β, ΒΓ has been drawn across, therefore the angle ΖΒΓ is equal to the angle constructed in the alternate segment ΒΑΓ of the circle.
ἀλλʼ ἡ ὑπὸ ΖΒΓ τῇ πρὸς τῷ Δ ἐστιν ἴση· καὶ ἡ ἐν τῷ ΒΑΓ ἄρα τμήματι ἴση ἐστὶ τῇ πρὸς τῷ Δ.
But the angle ΖΒΓ is equal to the angle at Δ; therefore the angle in the segment ΒΑΓ is also equal to the angle at Δ.
ἀπὸ τοῦ δοθέντος ἄρα κύκλου τοῦ ΑΒΓ τμῆμα ἀφῄρηται τὸ ΒΑΓ δεχόμενον γωνίαν ἴσην τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ τῇ πρὸς τῷ Δ· ὅπερ ἔδει ποιῆσαι.
Therefore, from the given circle ΑΒΓ, a segment ΒΑΓ has been cut off admitting an angle equal to the given rectilinear angle at Δ; which was to be done.
§3.prop.35ἐὰν ἐν κύκλῳ δύο εὐθεῖαι τέμνωσιν ἀλλήλας, τὸ ὑπὸ τῶν τῆς μιᾶς τμημάτων περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ὑπὸ τῶν τῆς ἑτέρας τμημάτων περιεχομένῳ ὀρθογωνίῳ.
If in a circle two straight lines cut one another, the rectangle contained by the segments of the one is equal to the rectangle contained by the segments of the other.
ἐν γὰρ κύκλῳ τῷ ΑΒΓΔ δύο εὐθεῖαι αἱ ΑΓ, ΒΔ τεμνέτωσαν ἀλλήλας κατὰ τὸ Ε σημεῖον· λέγω, ὅτι τὸ ὑπὸ τῶν ΑΕ, ΕΓ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ὑπὸ τῶν ΔΕ, ΕΒ περιεχομένῳ ὀρθογωνίῳ.
For in the circle ΑΒΓΔ let two straight lines ΑΓ, ΒΔ cut one another at the point Ε; I say that the rectangle contained by ΑΕ, ΕΓ is equal to the rectangle contained by ΔΕ, ΕΒ.
εἰ μὲν οὖν αἱ ΑΓ, ΒΔ διὰ τοῦ κέντρου εἰσὶν ὥστε τὸ Ε κέντρον εἶναι τοῦ ΑΒΓΔ κύκλου, φανερόν, ὅτι ἴσων οὐσῶν τῶν ΑΕ, ΕΓ, ΔΕ, ΕΒ καὶ τὸ ὑπὸ τῶν ΑΕ, ΕΓ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ὑπὸ τῶν ΔΕ, ΕΒ περιεχομένῳ ὀρθογωνίῳ.
If then ΑΓ, ΒΔ are through the center, so that Ε is the center of the circle ΑΒΓΔ, it is manifest that, since ΑΕ, ΕΓ, ΔΕ, ΕΒ are equal, the rectangle contained by ΑΕ, ΕΓ is also equal to the rectangle contained by ΔΕ, ΕΒ.
μὴ ἔστωσαν δὴ αἱ ΑΓ, ΔΒ διὰ τοῦ κέντρου, καὶ εἰλήφθω τὸ κέντρον τοῦ ΑΒΓΔ, καὶ ἔστω τὸ Ζ, καὶ ἀπὸ τοῦ Ζ ἐπὶ τὰς ΑΓ, ΔΒ εὐθείας κάθετοι ἤχθωσαν αἱ ΖΗ, ΖΘ, καὶ ἐπεζεύχθωσαν αἱ ΖΒ, ΖΓ, ΖΕ.
καὶ ἐπεὶ εὐθεῖά τις διὰ τοῦ κέντρου ἡ ΗΖ εὐθεῖάν τινα μὴ διὰ τοῦ κέντρου τὴν ΑΓ πρὸς ὀρθὰς τέμνει, καὶ δίχα αὐτὴν τέμνει· ἴση ἄρα ἡ ΑΗ τῇ ΗΓ. ἐπεὶ οὖν εὐθεῖα ἡ ΑΓ τέτμηται εἰς μὲν ἴσα κατὰ τὸ Η, εἰς δὲ ἄνισα κατὰ τὸ Ε, τὸ ἄρα ὑπὸ τῶν ΑΕ, ΕΓ περιεχόμενον ὀρθογώνιον μετὰ τοῦ ἀπὸ τῆς ΕΗ τετραγώνου ἴσον ἐστὶ τῷ ἀπὸ τῆς ΗΓ· προσκείσθω τὸ ἀπὸ τῆς ΗΖ· τὸ ἄρα ὑπὸ τῶν ΑΕ, ΕΓ μετὰ τῶν ἀπὸ τῶν ΗΕ, ΗΖ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΓΗ, ΗΖ. ἀλλὰ τοῖς μὲν ἀπὸ τῶν ΕΗ, ΗΖ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΖΕ, τοῖς δὲ ἀπὸ τῶν ΓΗ, ΗΖ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΖΓ· τὸ ἄρα ὑπὸ τῶν ΑΕ, ΕΓ μετὰ τοῦ ἀπὸ τῆς ΖΕ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΖΓ. ἴση δὲ ἡ ΖΓ τῇ ΖΒ· τὸ ἄρα ὑπὸ τῶν ΑΕ, ΕΓ μετὰ τοῦ ἀπὸ τῆς ΕΖ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΖΒ. διὰ τὰ αὐτὰ δὴ καὶ τὸ ὑπὸ τῶν ΔΕ, ΕΒ μετὰ τοῦ ἀπὸ τῆς ΖΕ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΖΒ. ἐδείχθη δὲ καὶ τὸ ὑπὸ τῶν ΑΕ, ΕΓ μετὰ τοῦ ἀπὸ τῆς ΖΕ ἴσον τῷ ἀπὸ τῆς ΖΒ· τὸ ἄρα ὑπὸ τῶν ΑΕ, ΕΓ μετὰ τοῦ ἀπὸ τῆς ΖΕ ἴσον ἐστὶ τῷ ὑπὸ τῶν ΔΕ, ΕΒ μετὰ τοῦ ἀπὸ τῆς ΖΕ. κοινὸν ἀφῃρήσθω τὸ ἀπὸ τῆς ΖΕ· λοιπὸν ἄρα τὸ ὑπὸ τῶν ΑΕ, ΕΓ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ὑπὸ τῶν ΔΕ, ΕΒ περιεχομένῳ ὀρθογωνίῳ.
Next, let ΑΓ, ΔΒ not be through the center, and let the center of ΑΒΓΔ be found, and let it be Ζ, and from Ζ let ΖΗ, ΖΘ be drawn perpendicular to the straight lines ΑΓ, ΔΒ, and let ΖΒ, ΖΓ, ΖΕ be joined. And since a straight line ΗΖ through the center cuts a straight line ΑΓ not through the center at right angles, it also bisects it; therefore ΑΗ is equal to ΗΓ. Since then the straight line ΑΓ has been cut into equal parts at Η, and into unequal parts at Ε, therefore the rectangle contained by ΑΕ, ΕΓ together with the square on ΕΗ is equal to the square on ΗΓ; let the square on ΗΖ be added; therefore the rectangle contained by ΑΕ, ΕΓ together with the squares on ΗΕ, ΗΖ is equal to the squares on ΓΗ, ΗΖ. But the square on ΖΕ is equal to the squares on ΕΗ, ΗΖ, and the square on ΖΓ is equal to the squares on ΓΗ, ΗΖ; therefore the rectangle contained by ΑΕ, ΕΓ together with the square on ΖΕ is equal to the square on ΖΓ. And ΖΓ is equal to ΖΒ; therefore the rectangle contained by ΑΕ, ΕΓ together with the square on ΕΖ is equal to the square on ΖΒ. For the same reasons, the rectangle contained by ΔΕ, ΕΒ together with the square on ΖΕ is also equal to the square on ΖΒ. And the rectangle contained by ΑΕ, ΕΓ together with the square on ΖΕ was also proved equal to the square on ΖΒ; therefore the rectangle contained by ΑΕ, ΕΓ together with the square on ΖΕ is equal to the rectangle contained by ΔΕ, ΕΒ together with the square on ΖΕ. Let the square on ΖΕ be subtracted from each; therefore the remaining rectangle contained by ΑΕ, ΕΓ is equal to the rectangle contained by ΔΕ, ΕΒ.
ἐὰν ἄρα ἐν κύκλῳ εὐθεῖαι δύο τέμνωσιν ἀλλήλας, τὸ ὑπὸ τῶν τῆς μιᾶς τμημάτων περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ὑπὸ τῶν τῆς ἑτέρας τμημάτων περιεχομένῳ ὀρθογωνίῳ· ὅπερ ἔδει δεῖξαι.
Therefore, if in a circle two straight lines cut one another, the rectangle contained by the segments of the one is equal to the rectangle contained by the segments of the other; which was to be proved.