Ἐὰν δύο κύκλοι ἐφάπτωνται ἀλλήλων ἐκτός, ἡ ἐπὶ τὰ κέντρα αὐτῶν ἐπιζευγνυμένη διὰ τῆς ἐπαφῆς ἐλεύσεται. Δύο γὰρ κύκλοι οἱ ΑΒΓ, ΑΔΕ ἐφαπτέσθωσαν ἀλλήλων ἐκτὸς κατὰ τὸ Α σημεῖον, καὶ εἰλήφθω τοῦ μὲν ΑΒΓ κέντρον τὸ Ζ, τοῦ δὲ ΑΔΕ τὸ Η: λέγω, ὅτι ἡ ἀπὸ τοῦ Ζ ἐπὶ τὸ Η ἐπιζευγνυμένη εὐθεῖα διὰ τῆς κατὰ τὸ Α ἐπαφῆς ἐλεύσεται. Μὴ γάρ, ἀλλ' εἰ δυνατόν, ἐρχέσθω ὡς ἡ ΖΓΔΗ, καὶ ἐπεζεύχθωσαν αἱ ΑΖ, ΑΗ. Ἐπεὶ οὖν τὸ Ζ σημεῖον κέντρον ἐστὶ τοῦ ΑΒΓ κύκλου, ἴση ἐστὶν ἡ ΖΑ τῇ ΖΓ. πάλιν, ἐπεὶ τὸ Η σημεῖον κέντρον ἐστὶ τοῦ ΑΔΕ κύκλου, ἴση ἐστὶν ἡ ΗΑ τῇ ΗΔ. ἐδείχθη δὲ καὶ ἡ ΖΑ τῇ ΖΓ ἴση: αἱ ἄρα ΖΑ, ΑΗ ταῖς ΖΓ, ΗΔ ἴσαι εἰσίν: ὥστε ὅλη ἡ ΖΗ τῶν ΖΑ, ΑΗ μείζων ἐστίν: ἀλλὰ καὶ ἐλάττων: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἡ ἀπὸ τοῦ Ζ ἐπὶ τὸ Η ἐπιζευγνυμένη εὐθεῖα διὰ τῆς κατὰ τὸ Α ἐπαφῆς οὐκ ἐλεύσεται: δι' αὐτῆς ἄρα. Ἐὰν ἄρα δύο κύκλοι ἐφάπτωνται ἀλλήλων ἐκτός, ἡ ἐπὶ τὰ κέντρα αὐτῶν ἐπιζευγνυμένη [εὐθεῖα] διὰ τῆς ἐπαφῆς ἐλεύσεται: ὅπερ ἔδει δεῖξαι.
If two circles touch one another externally, the straight line joining their centres will pass through the point of contact. For let the two circles ABC, ADE touch one another externally at the point A, and let the centre F of ABC, and the centre G of ADE, be taken; I say that the straight line joined from F to G will pass through the point of contact at A. For suppose it does not, but, if possible, let it pass as FCDG, and let AF, AG be joined. Then, since the point F is the centre of the circle ABC, FA is equal to FC. Again, since the point G is the centre of the circle ADE, GA is equal to GD. But FA was also proved equal to FC; therefore FA, AG are equal to FC, GD, so that the whole FG is greater than FA, AG; but it is also less [I. 20]: which is impossible. Therefore the straight line joined from F to G will not fail to pass through the point of contact at A; therefore it will pass through it.