MathNet 2021 — Part 08 Question 79

Let $ABCD$ be a convex quadrilateral such that $AB\gt BC$, $CD=DA$, and $\angle ABD=\angle DBC$. Let $E$ be the point on the line $AB$ such that $\angle DEB=90^\circ$. Prove that
$$
AE=\frac{AB-BC}{2}.
$$