TUYMAADA 2012 S3

Point $P$ is taken in the interior of the triangle $ABC$ , so that $$\angle PAB = \angle PCB = \dfrac {1} {4} (\angle A + \angle C).$$ Let $L$ be the foot of the angle bisector of $\angle B$ . The line $PL$ meets the circumcircle of $\triangle APC$ at point $Q$ . Prove that $QB$ is the angle bisector of $\angle AQC$ .

Proposed by S. Berlov