Observe inversion at $P$ with pover $r^2= PA\cdot PC$. It takes: - $A$ to $C$ and vice versa - smaler circle into it self - bigger circle to line wich touches smaler circle, since circles touch each other, and vice versa, so this line is tangent on smaler circle at $C$. Then it takes $E$ (which is on bigger circle and tangent) in $E'$ which is on tangent and and bigger circle, thus $E'=E$ so $PE^2=PA\cdot PC$ so $PE$ is tangent to circle $ACE$.

In fact, it is very classic. See http://www.cut-the-knot.org/ctk/Circle.shtml

Observe homothety at $A$ which takes smaller circle to biger, It takes: - $C$ to $P$ - tangent $t$ at $C$ on smaller circle to $t'$ which is tangent on bigger circle at $P$ thus $t'||t$. So $\angle ECP = \angle (CP, t') = \angle (AP, t') = \angle AEP$, thus $PE$ is tangent on circle $ACE$.

Same prove as above would work if we observe negative inversion at $C$ and power $\sqrt{CA\cdot CP}$, this is inversion composed with reflection over $C$. It takes bigger circle to it self and smaller to line which is paralel to $t$ and touches bigger circle at $P$ since it takes $A$ to $P$. I have to say, this is wonderful exercise in transformation geometry. I wil try to find, if any, solution with spiral similarity.

Dear Mathlinkers, for a synthetic proof, you can see http://perso.orange.fr/jl.ayme vol. 1 Le théorème de Feuerbach p. 5 Sincerely Jean-Louis

Let tangent at $E$ on circle $ACE = :\mathcal{C}$ (with center $S$) cuts bigger circle at $P'$. Let $S_1$ and $S_2$ be centers of small and big circle ($\mathcal{C}_1$ and $\mathcal{C}_2$). Since $\mathcal{C}_1$ and $\mathcal{C}$ intersects at $A$ and $C$ spiral similarity $\mathcal{S}_1$ at $A$ sends $S_1$ to $S$ and $C$ to $E$. Since $\mathcal{C}$ and $\mathcal{C}_2$ intersects at $A$ and $E$ spiral similarity $\mathcal{S}_2$ at $A$ sends $S$ to $S_2$ and $E$ to $P'$. Now composition $\mathcal{S}=\mathcal{S}_2\circ \mathcal{S}_1$ takes $S_1$ to $S_2$ so its rotation angle is $\pi$ thus $\mathcal{S}$ is homothety at $A$. So $A,C, P'$ are collinear since it takes $C$ to $P'$. So $P'=P$ and we are done. Note that every prove above works not only for internally tangent situation.

Let $AE$ cuts the smaller circle at $K$. Let $l$ be the common tangent line of 2 circles and $T$ be a point on it. We know $\angle EPA=\angle EAT=\angle KCA$ so $EP//CK$. Then $\angle CAE=\angle CAK=\angle ECK=\angle PEC$. Done.

Attachments:

Since $P$ is the midpoint of the arc $\overset{\huge\frown}{DPE}$, $\angle{DEP}=\angle{PAE}$ $ \implies$ $ PE^2=PC.PA$.