Let $ABC$ be a triangle and $Q$ a point on the internal angle bisector of $\angle BAC $. Circle $\omega_1$ is circumscribed to triangle $BAQ$ and intersects the segment $AC$ in point $P \neq C$. Circle $\omega_2$ is circumscribed to the triangle $CQP$. Radius of the cirlce $\omega_1$ is larger than the radius of $\omega_2$. Circle centered at $Q$ with radius $QA$ intersects the circle $\omega_1$ in points $A$ and $A_1$. Circle centered at $Q$ with radius $QC$ intersects $\omega_1$ in points $C_1$ and $C_2$. Prove $\angle A_1BC_1 = \angle C_2PA $. Proposed by Matija Bucić.