High school proofs in plane geometry

Question

I've already established that $\triangle EOD$ is similar to $\triangle EBC$ and that $\frac{EO}{EB} = \frac{1}{2}$, then I know I need to show that $\frac{CD}{BC} = \frac{BC}{CE}$, but I must have gotten something wrong because it seems that I am comparing different sides of the two triangles.

Any help will be appreciated, thanks!

Answer 1

1. Since $BE$ is a diameter of $C_1$, we obtain $\measuredangle EDB=90^{\circ}$, which gives $\measuredangle BDC=90^{\circ}$.

2. Since $FB||DC$, we get $\measuredangle FDB=\measuredangle FBA=\measuredangle C$

and since $BE\perp FD$ and $BD\perp FC$, we obtain $\measuredangle E=\measuredangle FDB.$

Thus, $\measuredangle E=\measuredangle C$, which gives $BE=BC$ and $\frac{OE}{BC}=\frac{OE}{BE}=\frac{1}{2}$.

3. Since $BD$ is an altitude to hypotenuse of $\Delta EBC$, we obtain $BC^2=CD\cdot CE$.

Done!