The problem:
Answer:
Extend $AC$ to the side of $C$, such that $BC=CE$. Then, construct the equilateral triangle $BEF$. Triangles $CAB$ and $EFC$ are congruent, thus $AC=BE$. Now, observe that $AC=DE$, thus $BE=DE$. It follows that $\triangle BED$ is isosceles, thus $\angle EDB=80$.
I was given this problem. The provided construction is smart, but I have a feeling that there should be a simpler non-trigonometric approach. Can anyone think of one?
Say $O$ is a circumcenter of circumcircle $(ABC)$ and let $BO$ cut $AC$ at $E$. Then $$\angle COB = 2 \angle CAB = 60$$ so triangle $BCO$ is equaliteral. So we have $$ BC =OC=OB = OA$$ Since $$\angle AOB = 2 \angle ACB = 80$$ and since $\angle OAE = 20$ (note that $ACO$ is isosceles) we have $\angle AEO = 80$ so $AO=AE$ which means $E =D$, and finaly we have $\angle CDB =80$.