What is the measure of angle alpha? If AB=CD then $\alpha=$?

I tried to draw auxiliary lines but none provides an equation that allows the solution

 A: Denote the angles at point $D$ by $D_1$ and $D_2$ so that
$$\angle D = D_1+D_2$$
We see that (which is shown in your second picture):
$$D_2 = 180-3a$$
Which means that:
$$D_1 = 3a$$
HINT: Now use the Law of sines. Can you take it from here?
(Just comment if you can't do it, I'll fill in the rest of the anwser then)
A: Here is a path with Euclidean geometry.


*

*Draw line $BE$ so that $\measuredangle CBE = 2\alpha$, and let $F$ on $BE$ such that $DE \cong DF$.

*Since $BE \cong EC$, we also have $BF \cong DC \cong AB$. Hence $\triangle BAF$ is isoscles. And since $\measuredangle ABE = 180^\circ -  12\alpha$, then $\measuredangle BAF = 6\alpha$ and $\measuredangle FAD = 2\alpha$.

*Since $\triangle FED$ is also isosceles, it is $\measuredangle FDE = 2\alpha$.

*By 3. and External Angle Theorem on $\triangle BDC$, $\measuredangle FDB = \alpha$.

*By 2., 3., and 4. conclude that $\triangle FAD$ and $\triangle BFD$ are isosceles triangles, and therefore $\triangle BAF$ is equilateral, which yields $\alpha = 10^\circ$.

