In the figure, ABCD is a square circumscribing a circle ($\pi_1$) whose center is E, the point of intersection of the diagonals AC and BD. With A as center, AB as radius, sector ABD is drawn cutting $\pi_1$ at G (and also at G').
I have proved CG = CE in (problem #994185) by analytic method. The question is :- can this fact be proven by Geometrical method?
Use the cosine theorem (law of cosines) in $\triangle AEG$ and calculate $\cos \theta$ ($\theta$ is the measure of angle $AEG$).
Use the cosine theorem again in $\triangle GEC$, and calculate $GC$.
Recall that $-\cos \theta = \cos (\pi- \theta)$.