Berkeley Math Circle: Monthly Contest 6

Solution:

By AA, $\triangle A C D \sim \triangle B C E$, as $\angle C$ is shared and both are right triangles. Thus $A C / C D = B C / C E$. It follows by SAS that
$$\triangle A B C \sim \triangle D E C \Longrightarrow \angle C D E = \angle B A C.$$
Since $O$ is the circumcenter of $\triangle A B C$, $\angle B O C = 2 \angle B A C$, and since $\triangle B O C$ is isosceles,
$$\angle O C B = \angle O B C = 90^\circ - \angle B A C.$$
Thus, $\angle C D E + \angle O C B = 90^\circ$, which implies that $D E \perp O C$.