Question 152
As shown below, let $ABCD$ be a cyclic quadrilateral such that the diagonals $AC$ and $BD$ are perpendicular with intersection point $E$. Point $F$ is on the side $AD$, the ray $FE$ meets the circumscribed circle of $ABCD$ at point $P$. Point $Q$ is on the segment $PE$ such that $PQ \cdot PF = PE^2$. The line through $Q$ perpendicular to $AD$ meets $BC$ at $R$. Show that $RP = RQ$.
