Let $\ell$ be the perpendicular bisector of $P Q$, and denote by $\omega$ the circle $C F P Q$. By $D P \perp B C$ and $D Q \perp A C$, the circle $\omega$ passes through $D$; moreover, $C D$ is a diameter of $\omega$. The lines $Q E$ and $P E$ are symmetric about $\ell$, and $\ell$ is a symmetry axis of $\omega$ as well; it follows that the chords $C Q$ and $F P$ are symmetric about $\ell$, hence $C$ and $F$ are symmetric about $\ell$. Therefore, the perpendicular bisector of $C F$ coincides with $\ell$. Thus $\ell$ passes through the circumcenter $O$ of $A B C$. Let $M$ be the midpoint of $A B$. Since $C M \perp D M, M$ also lies on $\omega$. By $\angle A C M=\angle B C M$, the chords $M P$ and $M Q$ of $\omega$ are equal. Then, from $M P=M Q$ it follows that $\ell$ passes through $M$. ! Finally, both $O$ and $M$ lie on lines $\ell$ and $C M$, therefore $O=M$, and $\angle A C B=90^{\circ}$ follows.