jmc

First we draw $BD$: First, we see that $△ABC$ is isosceles; therefore $\angle ACB=\angle ABC=\frac{1}{2}\cdot (180^{\circ }-20^{\circ })=80^{\circ }.$ Likewise, $\angle ACD=\angle ADC=80^{\circ }.$ Now, we see that $\angle BCD=\angle ACB+\angle ACD=160^{\circ }.$

Then, we see that $△BCD$ is isosceles as well. That means $\angle CBD=\angle BDC=\frac{1}{2}\cdot (180^{\circ }-160^{\circ })=\boxed{10^{\circ }},$ as desired.