IGO 2014 J4

In a triangle ABC we have $\angle C = \angle A + 90^o$. The point $D$ on the continuation of $BC$ is given such that $AC = AD$. A point $E$ in the side of $BC$ in which $A$ doesn't lie is chosen such that $\angle EBC = \angle A, \angle EDC = \frac{1}{2} \angle A$. Prove that $\angle CED = \angle ABC$.

by Morteza Saghafian