Consider an isosceles triangle $ABC$ ($AB=AC$), with points $D, E$ lying on sides $AB, AC$, respectively. Suppose that $\angle BAC=\alpha, \angle EBC=\beta, \angle DCB=\gamma, \angle DEB=\omega$.
Suppose that the value (in degrees) of each of the angles $\alpha, \beta, \gamma, \omega$ is an integer.
a) Prove that $\alpha$ must be divisble by four!
b) Prove it is impossible that exactly three of the angles $\alpha, \beta, \gamma, \omega$ is divisible by three!
Using lots of trigonometry I proved that $$\cot \omega=\dfrac{2\cos \frac{\alpha}{2}\cos (\alpha-\gamma)}{\sin \gamma(\cos \alpha+\cos2\beta)}-\tan (\alpha+\beta)$$
Can someone help to prove a) and b)? Thans in advance. Maybe with trigonometry (perhaps using my fact)? I would be really thankful if someone would post a solution without trigonometry!
The statements are not true, here is a figure.