MathNet 2007-part-01 — 187 problems (EN–VI) Question 176

Let $ABCD$ be a convex quadrilateral with $AB = BC = CD$, $AC \neq BD$ and let $E$ be the intersection point of its diagonals. Prove that $AE = DE$ if and only if $\angle BAD + \angle ADC = 120^{\circ}$.