Let $ABCD$ be a parallelogram and let $P{}$ be a point inside it such that $\angle PDA= \angle PBA$. Let $\omega_1$ be the excircle of $PAB$ opposite to the vertex $A{}$. Let $\omega_2$ be the incircle of the triangle $PCD$. Prove that one of the common tangents of $\omega_1$ and $\omega_2$ is parallel to $AD$. Ivan Frolov