SHARYGIN 2019 Problem 18

A quadrilateral $ABCD$ without parallel sidelines is circumscribed around a circle centered at $I$ . Let $K, L, M$ and $N$ be the midpoints of $AB, BC, CD$ and $DA$ respectively. It is known that $AB \cdot CD = 4IK \cdot IM$ . Prove that $BC \cdot AD = 4IL \cdot IN$ .