difficult angle problem with a square on the edge of a circle
Solution 1:
Notice that $\angle ACG=\angle AGC=\dfrac{1}{2}\left(\pi-\dfrac{\alpha}{2}\right)$.
Therefore $\angle CGE=\pi-\angle AGC=\dfrac{\pi}{2}+\dfrac{\alpha}{4}$.
$\angle GEC=\dfrac{\pi}{4}$ since the diagonal of a square bisects the angle.
Your angle $\theta=\angle ECG=\pi-\left(\dfrac{\pi}{4}+\dfrac{\pi}{2}+\dfrac{\alpha}{4}\right)=\dfrac{\pi-\alpha}{4}$
