A circle does not cut a circle at more points than two.

Κύκλος κύκλον οὐ τέμνει κατὰ πλείονα σημεῖα ἢ δύο. Εἰ γὰρ δυνατόν, κύκλος ὁ ΑΒΓ κύκλον τὸν ΔΕΖ τεμνέτω κατὰ πλείονα σημεῖα ἢ δύο τὰ Β, Η, Ζ, Θ, καὶ ἐπιζευχθεῖσαι αἱ ΒΘ, ΒΗ δίχα τεμνέσθωσαν κατὰ τὰ Κ, Λ σημεῖα: καὶ ἀπὸ τῶν Κ, Λ ταῖς ΒΘ, ΒΗ πρὸς ὀρθὰς ἀχθεῖσαι αἱ ΚΓ, ΛΜ διήχθωσαν ἐπὶ τὰ Α, Ε σημεῖα. Ἐπεὶ οὖν ἐν κύκλῳ τῷ ΑΒΓ εὐθεῖά τις ἡ ΑΓ εὐθεῖάν τινα τὴν ΒΘ δίχα καὶ πρὸς ὀρθὰς τέμνει, ἐπὶ τῆς ΑΓ ἄρα ἐστὶ τὸ κέντρον τοῦ ΑΒΓ κύκλου. πάλιν, ἐπεὶ ἐν κύκλῳ τῷ αὐτῷ τῷ ΑΒΓ εὐθεῖά τις ἡ ΝΞ εὐθεῖάν τινα τὴν ΒΗ δίχα καὶ πρὸς ὀρθὰς τέμνει, ἐπὶ τῆς ΝΞ ἄρα ἐστὶ τὸ κέντρον τοῦ ΑΒΓ κύκλου. ἐδείχθη δὲ καὶ ἐπὶ τῆς ΑΓ, καὶ κατ' οὐδὲν συμβάλλουσιν αἱ ΑΓ, ΝΞ εὐθεῖαι ἢ κατὰ τὸ Ο: τὸ Ο ἄρα σημεῖον κέντρον ἐστὶ τοῦ ΑΒΓ κύκλου. ὁμοίως δὴ δείξομεν, ὅτι καὶ τοῦ ΔΕΖ κύκλου κέντρον ἐστὶ τὸ Ο: δύο ἄρα κύκλων τεμνόντων ἀλλήλους τῶν ΑΒΓ, ΔΕΖ τὸ αὐτό ἐστι κέντρον τὸ Ο: ὅπερ ἐστὶν ἀδύνατον. Οὐκ ἄρα κύκλος κύκλον τέμνει κατὰ πλείονα σημεῖα ἢ δύο: ὅπερ ἔδει δεῖξαι. | A circle does not cut a circle at more points than two. For, if possible, let the circle ABC cut the circle DEF at more points than two, namely B, C, F, H; let BH, BG be joined and bisected at the points K, L, and from K, L let KC, LM be drawn at right angles to BH, BG and carried through to the points A, E. Then, since in the circle ABC a straight line AC cuts a straight line BH into two equal parts and at right angles, the centre of the circle ABC is on AC. [III. 1, Por.] Again, since in the same circle ABC a straight line NO cuts a straight line BG into two equal parts and at right angles, the centre of the circle ABC is on NO. But it was also proved to be on AC, and the straight lines AC, NO meet at no point except at P; therefore the point P is the centre of the circle ABC. Similarly we can prove that P is also the centre of the circle DEF; therefore the two circles ABC, DEF which cut one another have the same centre P: which is impossible. [III. 5] |