jmc

First, we draw line $CG$. Notice that $\angle CBG$ consists of a $60^{\circ }$ angle and a $90^{\circ }$ angle, hence, $\angle CBG=150^{\circ }$.

Since $CB=BG$, $△CBG$ is isosceles, with $\angle BCG=(1/2)(180^{\circ }-150^{\circ })=15^{\circ }$.

Finally, to find $\angle GCE$, we subtract $\angle BCG$ from $\angle BCE$ (which measures $60^{\circ }$ as it is an angle of the triangle.)

Hence our desired answer is $\angle GCE=60^{\circ }-15^{\circ }=\boxed{45^{\circ }}$.