In the triangle $ABC$, we have that $\angle BAC$ is acute. Let $\Gamma$ be the circle that passes through $A$ and is tangent to the side $BC$ at $C$. Let $M$ be the midpoint of $BC$ and let $D$ be the other point of intersection of $\Gamma$ with $AM$. If $BD$ cuts back to$ \Gamma$ at $E$, show that $AC$ is the bisector of $\angle BAE$.