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?