XXIX Rioplatense Mathematical Olympiad

First, by the tangency condition and $AMBC$ being a cyclic quadrilateral we have that $$\angle AEM=180^{\circ }-\angle AMB=\angle ACB$$ and hence $ME$ is parallel to $BC$. Next, we claim that $M$ and $E$ are symmetric with respect to $AD$. This is because $AH$ is a diameter of the circumcircle of $△AFE$ and it is also perpendicular to $ME$.

To finish the proof observe that $\angle ADM=\angle ADE=90^{\circ }-\angle BAC=\angle ADF$, where the second and third equalities hold in any triangle, by angle chasing.