Call $\angle DEC\ = \angle DCE\ = \alpha$ and $\angle AED\ = \angle EAD\ = \theta$ We have $\angle AEC\ = \theta\ - \alpha$ and $\angle ECA$ = $45^\circ\ - \alpha$ adding up all the angles of $\vartriangle EAC$ $\therefore$ $\angle AEC\ = \theta\ - \alpha$ = $45^\circ$ $\square$