RU-ALLRUSSIAN 2009 Problem 117

Let be given a parallelogram $ABCD$ and two points $A_1$, $C_1$ on its sides $AB$, $BC$, respectively. Lines $AC_1$ and $CA_1$ meet at $P$. Assume that the circumcircles of triangles $AA_1P$ and $CC_1P$ intersect at the second point $Q$ inside triangle $ACD$. Prove that $\angle PDA = \angle QBA$.