Geometry problem

Question

A friend emailed me this problem and I found out that it was taken from some math contest for high school students surprisingly. I was wondering if anyone could explain to me why $ \angle CDE = \angle BAC $, that was the issue she was referring to:

Let $ABC$ be a triangle with a circumcenter $O$. The points $D$, $E$ and $F$ lie in the interiors of the sides $BC$, $CA$ and $AB$ respectively such that $DE \perp CO$ and $DF \perp BO$. Let $K$ be the circumcenter of triangle $AFE$. Prove that $DK$ and $BC$ are $ \perp $

I also wouldn't mind a worked out solution since I am at a lost as to what kind of math this problem requires.

Answer 1

For your question of $\angle CDE = \angle BAC$, it is just simple angle chasing.

Let $\angle BAC = \alpha$. Then $\angle BOC = 2\alpha$ (angle at center), $\angle OCD = 90^\circ - \alpha$ (base angle of isosceles triangle) and hence $\angle EDC = \alpha$.

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To continue with the rest of the problem, since $\angle FKE = 2 \angle FAE = 2 \alpha$, hence $DFKE$ is concyclic (opposite angles sum to $180^\circ$). This is an important observation, that allows us to relate $K$ to the rest of the points.

Since $\angle FKE = 2 \alpha$ (angle at center), thus, $ 90^\circ - \alpha = \angle KEF = \angle KDF $ (cyclic quad).

Hence, $\angle KDB = \angle KDF + \angle FDB = (90^\circ - \alpha) + \alpha = 90^\circ$.

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In the diagram below, all red angles have measure $\alpha$, green angles have measure $2\alpha$, blue angles have measure $90^\circ - \alpha$.