Given an acute triangle $ABC$. Points $B'$ and $C'$ are symmetrical, respectively, to vertices $B$ and $ C$ wrt straight lines $AC$ and $AB$. Let $P$ be the intersection point of the circumcircles of triangles $ABB'$ and $ACC'$, different from $A$. Prove that the center of the circumcircle of triangle $ABC$ lies on line $PA$.