jmc

First, observe that $AC=CD$. Therefore, triangle $ACD$ is isosceles and $\angle CAD$ is congruent to $\angle CDA$. Also, $m\angle ACD=m\angle ACB+m\angle BCD=60^{\circ }+90^{\circ }=150^{\circ }$. Since the three angles of triangle $ACD$ sum to 180 degrees, we have $$\begin{array}{rl}m\angle CAD+m\angle CDA+150^{\circ }& =180^{\circ }⟹\\ 2m\angle CAD& =30^{\circ }⟹\\ m\angle CAD& =\boxed{15}\text{ degrees}.\end{array}$$