My solution: Suppose the tangent to $\Gamma$ and $(ABD)$ intersect $AC$ at $L.$ We know $AB^2=AC^2=AD\cdot AE\to \angle ABD=\angle BED.$ $(1)$ We know $\angle BEC=\angle BED-\angle CED=^{(1)} \angle ABD-\angle LDC=180-\angle BAD-\angle BDA-\angle LDC$ $=180-\angle BDL-\angle LDC-\angle BDA=180-\angle CDA=\angle CDE.$ As desired.

This was my solution during the test, I solved it in the first 10 minutes: By Power of a Point, $AC^2=AD\cdot AE$. Invert about $A$ with radius $r=AB=AC$. $B$, $C$ and $\Gamma$ are invariant under the inversion, while $D$ maps to $E$ and $E$ maps to $D$. The image of circle $ABD$ will be a line passing through the images of $B$ and $D$ and is tangent to $\Gamma$. The image of $B$, $D$ is $B$, $E$. Thus the circle maps to line $BE$, and thus $BE$ is tangent to $\Gamma$

Let the tangent at $E$ intersect circle $ABD$ at $B',K$ with $B'$ on the opposite side of $AC$ to $D$. We will prove $AB'=AC \implies B=B'$ which completes the problem. By a well-known lemma (similar triangles) we have $AB'^2=AE \cdot AD=AC^2 \implies AB'=AC$ as desired.

Ferid.---. wrote: My solution: Suppose the tangent to $\Gamma$ and $(ABD)$ intersect $AC$ at $L.$ We know $AB^2=AC^2=AD\cdot AE\to \angle ABD=\angle BED.$ $(1)$ We know $\angle BEC=\angle BED-\angle CED=^{(1)} \angle ABD-\angle LDC=180-\angle BAD-\angle BDA-\angle LDC$ $=180-\angle BDL-\angle LDC-\angle BDA=180-\angle CDA=\angle CDE.$ As desired. Is there some mistakes?

My students solution: $BD$ meet $\Gamma$ at another point L It is easy to show that $EL$$//$$AB$,then $\angle ABE = \angle BEL$ $AB^2=AC^2=AD\cdot AE\to \angle ABE=\angle ADB$, then $\angle BEL=\angle EDL$,DONE.

How can we draw the diagram?

Let $G=AC\cap (ABD) , E=AD\cap \Gamma , F=DG\cap \Gamma ,\quad \text{and} \quad H=DC\cap (ABD)$ Claim: $EF\parallel AG$ Proof. by Star Lemma , $HD$ bisects $\angle{ADG}$ so $CF=CE$ and the result follows. $\blacksquare$ Now $AE.AD=\text{Pow}_\Gamma(A)=AC^2=AB^2$ so $\Delta ABE\sim \Delta ABD$ $\implies \angle{AEB}=\angle{ABD}=\angle{AGD}=\angle{EFD}.$ so we're clear @above you could draw the two circles firstly

Attachments:

As $\Gamma$ is tangent to $\overline{AC}$, then $AB^2=AC^2=AD\cdot AE$, which means that $\overline{AB}$ is tangent to $(BED)$. Now we do angle chase, \begin{align*} \measuredangle DEB=\measuredangle DBA=\measuredangle DCE, \end{align*}where the last equality comes from the fact that $(ABD)$ and $(CED)$ are tangent to each other. Done.

YanYau wrote: Let $ABC$ be a triangle with $AB=AC$. A circle $\Gamma$ lies outside triangle $ABC$ and is tangent to line $AC$ at $C$. Point $D$ lies on $\Gamma$ such that the circumcircle of triangle $ABD$ is internally tangent to $\Gamma$. Segment $AD$ meets $\Gamma$ secondly at $E$. Prove that $BE$ is tangent to $\Gamma$ Let $BD \cap \Gamma=F$ by PoP $AE \cdot AD=AC^2=AB^2$ so $AB$ is tangent to $BED$ now becuase $(ABD)$ and $\Gamma$ are tangent at $D$ we have $AB \parallel EF$ so by angle chase $$\angle EDF=\angle ABE=\angle BEF \implies BD \; \text{is tangent to} \; \Gamma$$Thus we are done