, 2014

Let $\Gamma$ be the circumcircle of quadrilateral $ABCD$. Let $\alpha = \angle PAB = \angle PBC = \angle PCD = \angle PDA$ and let $T_1, T_2, T_3$ and $T_4$ denote the circumcircles of triangles $APD$, $BPC$, $APB$ and $CPD$, respectively. Let $M$ be the intersection of $T_1$ with line $RP$ and let $N$ be the intersection of $T_3$ with line $SP$. Also let $X$ denote the intersection of diagonals $AC$ and $BD$.

By power of a point for circles $T_1$ and $\Gamma$, it follows that $RM \cdot RP = RA \cdot RD = RB \cdot RC$ which implies that the quadrilateral $BMPC$ is cyclic and $M$ lies on $T_2$. Therefore $\angle PMB = \angle PCB = \alpha = \angle PAB = \angle DMP$ where all angles are directed. This implies that $M$ lies on the diagonal $BD$ and also that $\angle XMP = \angle DMP = \alpha$. By a symmetric argument applied to $S$, $T_3$ and $T_4$, it follows that $N$ lies on $T_4$ and that $N$ lies on diagonal $AC$ with $\angle XNP = \alpha$. Therefore $\angle XMP = \angle XNP$ and $X, M, P$ and $N$ are concyclic. This implies that the angle formed by lines $MP$ and $NP$ is equal to one of the angles formed by lines $MX$ and $NX$. The fact that $M$ lies on $BD$ and $RP$ and $N$ lies on $AC$ and $SP$ now implies the desired result.