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BC x DE = CD x PQ wanted, if BP xQE = PQ^ 2 inside a cyclic pentagon

Source: 2014 Saudi Arabia GMO TST day II p3

July 31, 2020
geometryCyclicpentagon

Problem Statement

Let ABCDEABCDE be a cyclic pentagon such that the diagonals ACAC and ADAD intersect BEBE at PP and QQ, respectively, with BP⋅QE=PQ2BP \cdot QE = PQ^2. Prove that BC⋅DE=CD⋅PQBC \cdot DE = CD \cdot PQ.
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