Proof: By Euclid

For, if possible, let circle $ABDC$¹ touch circle $EBFD$ - first of all, internally - at more than one point, $D$ and $B$.
And let the center $G$ of circle $ABDC$ have been found [Prop. 3.1], and (the center) $H$ of $EBFD$ [Prop. 3.1].
Thus, the (straight line) joining $G$ and $H$ will fall on $B$ and $D$ [Prop. 3.11].
Let it fall like $BGHD$ (in the figure).

fig13e

And since point $G$ is the center of circle $ABDC$, $BG$ is equal to $GD$.
Thus, $BG$ (is) greater than $HD$.
Thus, $BH$ (is) much greater than $HD$.
Again, since point $H$ is the center of circle $EBFD$, $BH$ is equal to $HD$.
But it was also shown (to be) much greater than it.
The very thing (is) impossible.
Thus, a circle does not touch a(nother) circle internally at more than one point.
So, I say that neither (does it touch) externally (at more than one point).
For, if possible, let circle $ACK$ touch circle $ABDC$ externally at more than one point, $A$ and $C$.
And let $AC$ have been joined.
Therefore, since two points, $A$ and $C$, have been taken at random on the circumference of each of the circles $ABDC$ and $ACK$, the straight line joining the points will fall inside each (circle) [Prop. 3.2].
But, it fell inside $ABDC$, and outside $ACK$ [Def. 3.3] .
The very thing (is) absurd.
Thus, a circle does not touch a(nother) circle externally at more than one point.
And it was shown that neither (does it) internally.
Thus, a circle does not touch a(nother) circle at more than one point, whether they touch internally or externally.
(Which is) the very thing it was required to show.

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The Greek text has "$ABCD$", which is obviously a mistake (translator's note). ↩