Question 81
Let $\triangle ABC$ be an acute triangle and $\omega$ be its circumcircle. Points $F$ on $AC$, $E$ on $AB$, and $P$ and $Q$ on $\omega$ are given such that
$$
\angle AFB = \angle AEC = \angle APE = \angle AQF = 90^\circ.
$$
Prove that the lines $BC$, $EF$, and $PQ$ are concurrent.
$$
\angle AFB = \angle AEC = \angle APE = \angle AQF = 90^\circ.
$$
Prove that the lines $BC$, $EF$, and $PQ$ are concurrent.