jmc

First of all, for our benefit, we should draw in the desired angle: We can see that $AB=BC=EB,$ thus $△ABE$ is an isosceles triangle, where $\angle ABE=90^{\circ }+60^{\circ }=150^{\circ }.$ Since the other two angles are equal and add up to $30^{\circ }$, we have that $\angle BAE=15^{\circ }.$ Likewise, $\angle CAD=15^{\circ }.$

Then, $\angle DAE=\angle CAB-\angle CAD-\angle BAE=60^{\circ }-15^{\circ }-15^{\circ }=\boxed{30^{\circ }.}$