If two circles touch one another internally, and their centers are taken, then the straight line joining their centers, being produced, falls on the point of contact of the circles.

Let the two circles *ABC* and *ADE* touch one another internally at the point *A,* and let the centers *F* and *G* of the circles *ABC* and *ADE* be taken.

I say that the straight line joined from *G* to *F* and produced falls on *A.*

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

Then, since the sum of *AG* and *GF* is greater than *FA,* that is, than *FH,* subtract *FG* from each, therefore the remainder *AG* is greater than the remainder *GH.*

But *AG* equals *GD,* therefore *GD* is also greater than *GH,* the less greater than the greater, which is impossible.

Therefore the straight line joined from *F* to *G* does not fall outside. Therefore it falls on *A,* the point of contact.

Therefore *if two circles touch one another internally, and their centers are taken, then the straight line joining their centers, being produced, falls on the point of contact of the circles.*

Q.E.D.

Various conclusions in the proof are based on the figure rather than rigorous deductive reasoning. Camerer and others have suggested ways of filling the gaps.

This proposition is used in III.13.