Proof: By Euclid

For (if) not then, if possible, let it go like $FCDG$ (in the figure), and let $AF$ and $AG$ have been joined.
Therefore, since point $F$ is the center of circle $ABC$, $FA$ is equal to $FC$.
Again, since point $G$ is the center of circle $ADE$, $GA$ is equal to $GD$.
And $FA$ was also shown (to be) equal to $FC$.
Thus, the (straight lines) $FA$ and $AG$ are equal to the (straight lines) $FC$ and $GD$.
So the whole of $FG$ is greater than $FA$ and $AG$.
But, (it is) also less [Prop. 1.20].
The very thing is impossible.
Thus, the straight line joining $F$ to $G$ cannot not go through the point of union at $A$.
Thus, (it will go) through it.
Thus, if two circles touch one another externally then the [straight line] joining their centers will go through the point of union.
(Which is) the very thing it was required to show.
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