1. Define the orthocenter and projections: Let $H$ be the orthocenter of $\triangle ABC$, and let $D$ be the projection of $A$ onto $BC$.

2. Consider the inversion: Consider the inversion through pole $H$ with power $k = \overline{HA} \cdot \overline{HD} = \overline{HE} \cdot \overline{HB} = \overline{HC} \cdot \overline{HF}$. This inversion maps $P$ to $P'$ and $Q$ to $Q'$.

3. **Mapping of line $BC$**: The inversion $\mathcal{I}(H, k)$ maps the line $BC$ to the circumcircle of $\triangle AEF$. Therefore, $P'$ and $Q'$ both lie on the circumcircle of $\triangle AEF$.

4. Orthocenter properties: Since $H$ is the orthocenter of $\triangle ABC$, it will also be the orthocenter of $\triangle AQP$. Thus, $Q'$ and $P'$ are the projections of $P$ and $Q$ onto $AQ$ and $AP$, respectively.

5. Radical axis: The line $AF$ is the radical axis of the two circles passing through $A$ and $F$ and tangent to $BC$ at $P$ and $Q$, respectively. Therefore, $AF$ will go to the midpoint $B$ of $PQ$.

6. Equal power of point: We have $BP^2 = BQ^2 = \overline{BF} \cdot \overline{BA} = \overline{BD} \cdot \overline{BC}$. If $C'$ is the intersection of $P'Q'$ with $PQ$, then $(QPDC')$ is harmonic.

7. MacLaurin identity: By the MacLaurin identity, $BQ^2 = BP^2 = \overline{BD} \cdot \overline{BC'}$, which implies $C$ must coincide with $C'$. Therefore, $P'$, $Q'$, and $C$ are collinear.

8. Harmonic division: Since $(CDPQ)$ is harmonic, we have $\overline{CA} \cdot \overline{CE} = \overline{CP} \cdot \overline{CQ} = \overline{CD} \cdot \overline{CB}$. This implies $E \in (AQP)$. Similarly, $F \in (HPQ)$.

9. Angle chasing: Finally, we have $\angle (FQ, EP) = \angle (FQ, QP) + \angle (QP, EP) = \angle (HF, HP') + \angle (AQ, AE) = \angle (AF, AP') + \angle (AP', AE) = \angle (AF, AE)$.

10. Intersection point: Thus, if $W$ is the intersection of $EP$ with $FQ$, then $W \in (AEF)$.

$\blacksquare$