Proof: By Euclid

For (if) not then, if possible, let it fall like $FGH$ (in the figure), and let $AF$ and $AG$ have been joined.
Therefore, since $AG$ and $GF$ is greater than $FA$, that is to say $FH$ [Prop. 1.20], let $FG$ have been taken from both.
Thus, the remainder $AG$ is greater than the remainder $GH$.
And $AG$ (is) equal to $GD$.
Thus, $GD$ is also greater than $GH$, the lesser than the greater.
The very thing is impossible.
Thus, the straight line joining $F$ to $G$ will not fall outside (one circle but inside the other).
Thus, it will fall upon the point of union (of the circles) at point $A$.
Thus, if two circles touch one another internally, [and their centers are found], then the straight line joining their centers, [being produced], will fall upon the point of union of the circles.
(Which is) the very thing it was required to show.
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