Given an acute-angled triangle $ABC$ and a point $P$ inside it such that $\angle PBA=\angle PCA$. The lines $PB$ and $PC$ intersect the circumcircles of triangles $PCA$ and $PAB$ secondly at points $M$ and $N$, respectively. Let the rays $MC$ and $NB$ intersect at a point $S$, $K$ is the center of the circumscribed circle of the triangle $SMN$. Prove that the lines $AK$ and $BC$ are perpendicular.