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

Solution: First, note
$\dang EDF = 180\dg - \dang BOC = 180\dg - 2A$,
so $\dang FDE = 2A$.
\begin{center}
  \begin{asy}
    pair A = Drawing("A", dir(110), dir(110));
    pair B = Drawing("B", dir(210), dir(210));
    pair C = Drawing("C", dir(330), dir(330));
    draw(A--B--C--cycle);
    pair O = Drawing("O", origin, dir(45));
    pair D = Drawing("D", 0.4*B+0.6*C, dir(-90));
    pair F = Drawing("F", extension(A,B,D,foot(D,O,B)), dir(-C));
    pair E = Drawing("E", extension(A,C,D,foot(D,O,C)), dir(-B));
    pair K = Drawing("K", circumcenter(A,E,F), dir(90));
    draw(E--D--F--cycle);
    draw(E--K--F, dashed);
    draw(B--O--C);
    draw(D--K);
    draw(circumcircle(D,E,F), dotted);
  \end{asy}
\end{center}
Observe that $\dang FKE = 2A$ as well; hence $KFDE$ is cyclic.
Hence
\begin{align*}
  \dang KDB &= \dang KDF + \dang FDB \\
  &= \dang KEF + \left( 90\dg - \dang DBO \right) \\
  &= (90\dg-A) + (90\dg-(90\dg-A)) \\
  &= 90\dg.
\end{align*}
and the proof ends here.