Question 5
Point $P$ lies inside parallelogram $ABCD$ and satisfies $\angle PDA=\angle PBA$. Let $\omega_1$ be the excircle opposite $A$ of triangle $PAB$, and $\omega_2$ the incircle of triangle $PCD$. Prove that these circles have a common tangent parallel to $AD$. (9 points)

Neutral editorial illustration.

Neutral editorial illustration.