Problem

Source: Mexico 2023/3

Tags: geometry, 2023, P3, Mexico, Hi



Let $ABCD$ be a convex quadrilateral. If $M, N, K$ are the midpoints of the segments $AB, BC$, and $CD$, respectively, and there is also a point $P$ inside the quadrilateral $ABCD$ such that, $\angle BPN= \angle PAD$ and $\angle CPN=\angle PDA$. Show that $AB \cdot CD=4PM\cdot PK$.