## Book III, Proposition 4

If in a circle two straight lines cut one another which are not through the centre, they do not bisect one another.

 Ἐὰν ἐν κύκλῳ δύο εὐθεῖαι τέμνωσιν ἀλλήλας μὴ διὰ τοῦ κέντρου οὖσαι, οὐ τέμνουσιν ἀλλήλας δίχα. Ἔστω κύκλος ὁ ΑΒΓΔ, καὶ ἐν αὐτῷ δύο εὐθεῖαι αἱ ΑΓ, ΒΔ τεμνέτωσαν ἀλλήλας κατὰ τὸ Ε μὴ διὰ τοῦ κέντρου οὖσαι: λέγω, ὅτι οὐ τέμνουσιν ἀλλήλας δίχα. Εἰ γὰρ δυνατόν, τεμνέτωσαν ἀλλήλας δίχα ὥστε ἴσην εἶναι τὴν μὲν ΑΕ τῇ ΕΓ, τὴν δὲ ΒΕ τῇ ΕΔ: καὶ εἰλήφθω τὸ κέντρον τοῦ ΑΒΓΔ κύκλου, καὶ ἔστω τὸ Ζ, καὶ ἐπεζεύχθω ἡ ΖΕ. Ἐπεὶ οὖν εὐθεῖά τις διὰ τοῦ κέντρου ἡ ΖΕ εὐθεῖάν τινα μὴ διὰ τοῦ κέντρου τὴν ΑΓ δίχα τέμνει, καὶ πρὸς ὀρθὰς αὐτὴν τέμνει: ὀρθὴ ἄρα ἐστὶν ἡ ὑπὸ ΖΕΑ: πάλιν, ἐπεὶ εὐθεῖά τις ἡ ΖΕ εὐθεῖάν τινα τὴν ΒΔ δίχα τέμνει, καὶ πρὸς ὀρθὰς αὐτὴν τέμνει: ὀρθὴ ἄρα ἡ ὑπὸ ΖΕΒ. ἐδείχθη δὲ καὶ ἡ ὑπὸ ΖΕΑ ὀρθή: ἴση ἄρα ἡ ὑπὸ ΖΕΑ τῇ ὑπὸ ΖΕΒ ἡ ἐλάττων τῇ μείζονι: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα αἱ ΑΓ, ΒΔ τέμνουσιν ἀλλήλας δίχα. Ἐὰν ἄρα ἐν κύκλῳ δύο εὐθεῖαι τέμνωσιν ἀλλήλας μὴ διὰ τοῦ κέντρου οὖσαι, οὐ τέμνουσιν ἀλλήλας δίχα: ὅπερ ἔδει δεῖξαι. If in a circle two straight lines cut one another which are not through the centre, they do not bisect one another. Let ABCD be a circle, and in it let the two straight lines AC, BD, which are not through the centre, cut one another at E; I say that they do not bisect one another. For, if possible, let them bisect one another, so that AE is equal to EC, and BE to ED; let the centre of the circle ABCD be taken [III. 1], and let it be F; let FE be joined. Then, since a straight line FE through the centre bisects a straight line AC not through the centre, it also cuts it at right angles; [III. 3] therefore the angle FEA is right. Again, since a straight line FE bisects a straight line BD, it also cuts it at right angles; [III. 3] therefore the angle FEB is right. But the angle FEA was also proved right; therefore the angle FEA is equal to the angle FEB, the less to the greater: which is impossible. Therefore AC, BD do not bisect one another.