MathNet 2025-part-01 — 187 problems (EN–VI) Question 111

Let $ABCD$ be a quadrilateral with $AD = BC$. Point $E$ lies such that $E$ and $A$ are on opposite sides of line $CD$ such that $BC = CE = DE$. If $\angle BAE + \angle CBE = 90^\circ$, prove that $\angle ABE + \angle DAE = 90^\circ$.