Let $ABCD$ be a cyclic quadrilateral. Let $ P$ be the intersection of the lines $BC$ and $AD$. Line $AC$ cuts the circumscribed circle of the triangle $BDP$ in $S$ and $T$, with $S$ between $ A$ and $C$. The line $BD$ intersects the circumscribed circle of the triangle $ACP$ in $U$ and $V$, with $U$ between $ B$ and $D$. Prove that $PS = PT = PU = PV$.