TUYMAADA 2018 Problem 2

A circle touches the side $AB$ of the triangle $ABC$ at $A$ , touches the side $BC$ at $P$ and intersects the side $AC$ at $Q$ . The line symmetrical to $PQ$ with respect to $AC$ meets the line $AP$ at $X$ . Prove that $PC=CX$ .

Proposed by S. Berlov