Triangle and circle geometry question

Question

I let $AE=x$ and $EC=y$, , then drew line $AD$ to get right $\triangle BAD$, but I don't know where to go from here. I was thinking maybe power of a point and similar triangles, but I'm not really sure.

Answer 1

Let $\measuredangle B=\alpha$.

Thus, since $$\tan\measuredangle C=\frac{OA}{AC},$$ we obtain $$\frac{5}{10}=\cot2\alpha$$ or $$\tan2\alpha=2$$ or $$\frac{2\tan\alpha}{1-\tan^2\alpha}=2$$ or $$\tan^2\alpha+\tan\alpha-1=0,$$ which gives $$\tan\alpha=\frac{\sqrt5-1}{2}$$ or $$\frac{\sqrt{a}-\sqrt{b}}{10}=\frac{\sqrt5-1}{2}$$ or...