Problem

Source: IMO 2019 SL G5

Tags: geometry, IMO Shortlist, IMO Shortlist 2019



Let $ABCDE$ be a convex pentagon with $CD= DE$ and $\angle EDC \ne 2 \cdot \angle ADB$. Suppose that a point $P$ is located in the interior of the pentagon such that $AP =AE$ and $BP= BC$. Prove that $P$ lies on the diagonal $CE$ if and only if area $(BCD)$ + area $(ADE)$ = area $(ABD)$ + area $(ABP)$. (Hungary)