Proof: By Euclid
(related to Proposition: 3.13: Circles Touch at One Point at Most)
 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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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"
Footnotes