Note that $C$ is the center of circumcircle $\triangle ABD$. From $CA=CB$ we get $\angle ABC=\angle BAC= 70^o$ and so $\angle BCA= 40^o$. Thus in circumcircle $\triangle ABD$ we have $\angle ADB=\frac{\overarc{AB}}{2}=\frac{\angle BCA}{2}= 20^o$.