Let $ABCDE$ be a cyclic pentagon such that the diagonals $AC$ and $AD$ intersect $BE$ at $P$ and $Q$, respectively, with $BP \cdot QE = PQ^2$. Prove that $BC \cdot DE = CD \cdot PQ$.
Source: 2014 Saudi Arabia GMO TST day II p3
Tags: geometry, Cyclic, pentagon
Let $ABCDE$ be a cyclic pentagon such that the diagonals $AC$ and $AD$ intersect $BE$ at $P$ and $Q$, respectively, with $BP \cdot QE = PQ^2$. Prove that $BC \cdot DE = CD \cdot PQ$.