Line Joining Centers of Two Circles Touching Internally

Theorem
Let two circles touch internally.

Then the straight line joining their centers passes through the point where they touch.

Proof
Let the circles $ABC$ and $ADE$ touch internally at $A$.

Let $F$ be the center of $ABC$ and let $G$ be the center of $ADE$.

We are to show that the straight line joining $F$ to $G$ passes through $A$.


 * Euclid-III-11.png

Suppose, as in the diagram above, that it does not.

Let $FG$ fall on $H$ instead.

It will also pass through $D$ on its way, which lies on circle $ADE$.

Join $AF$ and $AG$.

From Sum of Two Sides of Triangle Greater than Third Side $AG + GF$ is greater than $AF$.

Therefore $AG + GF$ is greater than $FH$ as $F$ is the center and both $AF$ and $FH$ are radii.

So, subtract $FG$ from both $AG + GF$ and $FH$.

It follows that $AG$ is greater than $GH$, and hence also greater than $GD$.

But as $G$ is the center of $ADE$, it follows that $AG$ is supposed to be equal to $GD$.

So the $AG + GF$ can not be greater than $AF$ and so must be coincident with it.

Hence the result.