If two circles touch one another externally, then the straight line joining their centers passes through the point of contact.

Let the two circles *ABC* and *ADE* touch one another externally at the point *A.* Take the center *F* of *ABC,* and the center *G* of *ADE.*

I say that the straight line joined from *F* to *G* passes through the point of contact at *A.*

For suppose it does not, but, if possible, let it pass as *FCDG.* Join *AF* and *AG.*

Then, since the point *F* is the center of the circle *ABC, FA* equals *FC.*

Again, since the point *G* is the center of the circle *ADE, GA* equals GD.

But *FA* was also proved equal to *FC,* therefore *FA* and *AG* equal *FC* and *GD,* so that the whole *FG* is greater than *FA* and *AG,* but it is also less, which is impossible.

Therefore the straight line joined from *F* to *G* does not fail to pass through the point of contact at *A,* therefore it passes through it.

Therefore *if two circles touch one another externally, then the straight line joining their centers passes through the point of contact.*

Q.E.D.

This proposition is not used in the rest of the *Elements.*