Let $ABCD$ be an inscribed quadrilateral and $P$ be an inner point for it so that $\angle PAB=\angle PBC=\angle PCD=\angle PDA$. The lines $AD$ and $BC$ intersect in point $Q$ and lines $AB$ and $CD$ – in point $R$. Prove that $\angle (PQ,PR)=\angle (AC,BD)$.