Let $ABCD$ be a cyclic quadrilateral whose center of the circumscribed circle is inside this quadrilateral, and its diagonals intersect in point $S{}$. Let $P{}$ and $Q{}$ be the centers of the curcimuscribed circles of triangles $ABS$ and $BCS$. The lines through the points $P{}$ and $Q{}$, which are parallel to the sides $AD$ and $CD$, respectively, intersect at the point $R$. Prove that the point $R$ lies on the line $BD$.