Let $ABCD$ be a square, point $E$ be the midpoint of the side $BC$. On the side $AB$ mark a point $F$ such that $FE \perp DE$. Prove that $AF + BE = DF$. (Ercole Suppa, Italy)
Problem
Source: V.A. Yasinsky Geometry Olympiad 2020 X-XI p2 , Ukraine
Tags: perpendicular, square, geometry
