Ἐὰν ἐν κύκλῳ εὐθεῖά τις διὰ τοῦ κέντρου εὐθεῖάν τινα μὴ διὰ τοῦ κέντρου δίχα τέμνῃ, καὶ πρὸς ὀρθὰς αὐτὴν τέμνει: καὶ ἐὰν πρὸς ὀρθὰς αὐτὴν τέμνῃ, καὶ δίχα αὐτὴν τέμνει. Ἔστω κύκλος ὁ ΑΒΓ, καὶ ἐν αὐτῷ εὐθεῖά τις διὰ τοῦ κέντρου ἡ ΓΔ εὐθεῖάν τινα μὴ διὰ τοῦ κέντρου τὴν ΑΒ δίχα τεμνέτω κατὰ τὸ Ζ σημεῖον: λέγω, ὅτι καὶ πρὸς ὀρθὰς αὐτὴν τέμνει. Εἰλήφθω γὰρ τὸ κέντρον τοῦ ΑΒΓ κύκλου, καὶ ἔστω τὸ Ε, καὶ ἐπεζεύχθωσαν αἱ ΕΑ, ΕΒ. Καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΖ τῇ ΖΒ, κοινὴ δὲ ἡ ΖΕ, δύο δυσὶν ἴσαι [ εἰσίν ]. καὶ βάσις ἡ ΕΑ βάσει τῇ ΕΒ ἴση: γωνία ἄρα ἡ ὑπὸ ΑΖΕ γωνίᾳ τῇ ὑπὸ ΒΖΕ ἴση ἐστίν. ὅταν δὲ εὐθεῖα ἐπ' εὐθεῖαν σταθεῖσα τὰς ἐφεξῆς γωνίας ἴσας ἀλλήλαις ποιῇ, ὀρθὴ ἑκατέρα τῶν ἴσων γωνιῶν ἐστιν: ἑκατέρα ἄρα τῶν ὑπὸ ΑΖΕ, ΒΖΕ ὀρθή ἐστιν. ἡ ΓΔ ἄρα διὰ τοῦ κέντρου οὖσα τὴν ΑΒ μὴ διὰ τοῦ κέντρου οὖσαν δίχα τέμνουσα καὶ πρὸς ὀρθὰς τέμνει. Ἀλλὰ δὴ ἡ ΓΔ τὴν ΑΒ πρὸς ὀρθὰς τεμνέτω: λέγω, ὅτι καὶ δίχα αὐτὴν τέμνει, τουτέστιν, ὅτι ἴση ἐστὶν ἡ ΑΖ τῇ ΖΒ. Τῶν γὰρ αὐτῶν κατασκευασθέντων, ἐπεὶ ἴση ἐστὶν ἡ ΕΑ τῇ ΕΒ, ἴση ἐστὶ καὶ γωνία ἡ ὑπὸ ΕΑΖ τῇ ὑπὸ ΕΒΖ. ἐστὶ δὲ καὶ ὀρθὴ ἡ ὑπὸ ΑΖΕ ὀρθῇ τῇ ὑπὸ ΒΖΕ ἴση: δύο ἄρα τρίγωνά ἐστι τὰ ΕΑΖ, ΕΖΒ τὰς δύο γωνίας δυσὶ γωνίαις ἴσας ἔχοντα καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην κοινὴν αὐτῶν τὴν ΕΖ ὑποτείνουσαν ὑπὸ μίαν τῶν ἴσων γωνιῶν: καὶ τὰς λοιπὰς ἄρα πλευρὰς ταῖς λοιπαῖς πλευραῖς ἴσας ἕξει: ἴση ἄρα ἡ ΑΖ τῇ ΖΒ. Ἐὰν ἄρα ἐν κύκλῳ εὐθεῖά τις διὰ τοῦ κέντρου εὐθεῖάν τινα μὴ διὰ τοῦ κέντρου δίχα τέμνῃ, καὶ πρὸς ὀρθὰς αὐτὴν τέμνει: καὶ ἐὰν πρὸς ὀρθὰς αὐτὴν τέμνῃ, καὶ δίχα αὐτὴν τέμνει: ὅπερ ἔδει δεῖξαι.

If in a circle a straight line through the centre bisect a straight line not through the centre, it also cuts it at right angles; and if it cut it at right angles, it also bisects it. Let ABC be a circle, and in it let a straight line CD through the centre bisect a straight line AB not through the centre at the point F; I say that it also cuts it at right angles. For let the centre of the circle ABC be taken, and let it be E; let EA, EB be joined. Then, since AF is equal to FB, and FE is common, two sides are equal to two sides; and the base EA is equal to the base EB; therefore the angle AFE is equal to the angle BFE. [I. 8] But, when a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right; [I. Def. 10] therefore each of the angles AFE, BFE is right. Therefore CD, which is through the centre, and bisects AB which is not through the centre, also cuts it at right angles. Again, let CD cut AB at right angles; I say that it also bisects it. that is, that AF is equal to FB. For, with the same construction, since EA is equal to EB, the angle EAF is also equal to the angle EBF. [I. 5] But the right angle AFE is equal to the right angle BFE, therefore EAF, EBF are two triangles having two angles equal to two angles and one side equal to one side, namely EF, which is common to them, and subtends one of the equal angles; therefore they will also have the remaining sides equal to the remaining sides; [I. 26] therefore AF is equal to FB