MathNet 2016-part-07 — 182 problems (EN–VI) Question 33

Let $ABCDE$ be a convex pentagon with $CD = DE$ and $\angle BCD = \angle DEA = 90^{\circ}$. Point $F$ lies on $AB$ such that $\frac{AF}{AE} = \frac{BF}{BC}$. Prove that $\angle FCE = \angle ADE$ and $\angle FEC = \angle BDC$.