, 2021

Let $\ell$ be the perpendicular bisector of $PQ$, and denote by $\omega$ the circle $CFPQ$. By $DP \perp BC$ and $DQ \perp AC$, the circle $\omega$ passes through $D$; moreover, $CD$ is a diameter of $\omega$.

The lines $QE$ and $PE$ are symmetric about $\ell$, and $\ell$ is a symmetry axis of $\omega$ as well; it follows that the chords $CQ$ and $FP$ are symmetric about $\ell$, hence $C$ and $F$ are symmetric about $\ell$. Therefore, the perpendicular bisector of $CF$ coincides with $\ell$. Thus $\ell$ passes through the circumcenter $O$ of $ABC$.

Let $M$ be the midpoint of $AB$. Since $CM \perp DM$, $M$ also lies on $\omega$. By $\angle ACM = \angle BCM$, the chords $MP$ and $MQ$ of $\omega$ are equal. Then, from $MP = MQ$ it follows that $\ell$ passes through $M$.

Finally, both $O$ and $M$ lie on lines $\ell$ and $CM$, therefore $O = M$, and $\angle ACB = 90^\circ$ follows.

Like in the first solution, we conclude that points $C, P, Q, D, F$ and the midpoint $M$ of $AB$ lie on one circle $\omega$ with diameter $CD$, and $M$ lies on $\ell$, the perpendicular bisector of $PQ$.

Let $BF$ and $CM$ meet at $G$ and let $\alpha = \angle ABF$. Then, since $E$ lies on $\ell$, and the quadrilaterals $FCBA$ and $FCPQ$ are cyclic, we have
$$\angle CQP = \angle FPQ = \angle FCQ = \angle FCA = \angle FBA = \alpha.$$
Since points $P, E, F$ are collinear, we have
$$\angle FEM = \angle FEQ + \angle QEM = 2\alpha + (90^\circ - \alpha) = 90^\circ + \alpha.$$
But $\angle FGM = 90^\circ + \alpha$, so $FEGM$ is cyclic. Hence
$$\angle EGC = \angle EFM = \angle PFM = \angle PCM.$$
Thus $GE \parallel BC$. It follows that $\angle FAC = \angle CBF = \angle EGF$, so $FEGA$ is cyclic, too. Hence $\angle ACB = \angle AFB = \angle AFG = 180^\circ - \angle AMG = 90^\circ$, that completes the proof.