Problem with angles in an isosceles triangle Given an isosceles $ABC$ triangle ($AB=AC$), and points $D,E$ on sides $AC, AB$, respectively, such that $AD<AE$. Suppose that angles $\angle BAC, \angle DBC, \angle ECB, \angle EDB$ are integers in degrees. Also let $\angle BAC=\angle EDB$. 
What are the possible values $\angle BAC$ and $\angle DBC$? I’ve done some research and the possible $(\angle BAC, \angle DBC)$ pairs are $(12,42),(12,66),(20,70)$. My question is: how to find these with euclidean maths? Why these are the only possible values? If you just find some connection between these two angles, please post it, because it is really helpful.
Soon I post a figure I hope. Thanks in advance! 
(This would be my lemma in a problem)
 A: Set: 
$$
\alpha=\angle BAC=\angle BDE,\quad
\beta=\angle CBD,\quad \gamma=\angle BCE.
$$
From the sine rule applied to triangle $BCD$ you get:
$$
{BD\over BC}={\cos(\alpha/2)\over\cos(\beta-\alpha/2)},
$$
and from the sine rule applied to triangle $BDE$ you get:
$$
{BE\over BD}={\sin\alpha\over\cos(\beta-\alpha/2)},
\quad\text{that is:}\quad
{BE\over BC}={\sin\alpha\cos(\alpha/2)\over\cos^2(\beta-\alpha/2)}.
$$
On the other hand, from the sine rule applied to triangle $BCE$ you get:
$$
{BE\over BC}={\sin\gamma\over\cos(\gamma-\alpha/2)},
$$
and comparing with the preceding result one obtains an equation for $\gamma$, which can be easily solved to yield:
$$
\tan\gamma={2\sin\alpha\cos^2(\alpha/2)\over
2\cos^2(\beta-\alpha/2)-\sin^2\alpha}.
$$
The condition $AD<AE$ entails $\alpha<\beta$, which combined with $\beta<90°-\alpha/2$ gives
$$
\alpha<\beta<90°-{\alpha\over2},\quad 0°<\alpha<60°.
$$
Hence you need to check only a finite number of cases: it turns out that the only integer values of $\alpha$ and $\beta$ which yield an integer value for $\gamma$ (when angles are measured in degrees) are just those you mentioned in your question.
