Solution:
Since $B D I E$ is cyclic, and $B I$ is the bisector of $\angle D B E$, then $I D = I E$. Similarly, $I D = I F$, so $I$ is the circumcenter of the triangle $D E F$.

We also have
$$\angle I E A = \angle I D B = \angle I F C$$
which implies that $A E I F$ is cyclic. We can assume that $A, E, M$ and $A, N, F$ are collinear in that order. Then $\angle I E M = \angle I F N$. Since also $I M = I E = I N = I F$, the two isosceles triangles $I E M$ and $I N F$ are congruent, thus $E M = F N$ and therefore $E N$ is parallel to $F M$.

From that, we can also see that the two triangles $I E A$ and $I N A$ are congruent, which implies that $A I$ is the perpendicular bisector of $E N$ and $M F$.

Note that $\angle I D P = \angle I D E = \angle I B E = \angle I B D$, so the triangles $I P D$ and $I D B$ are similar, which implies that $\frac{I D}{I B} = \frac{I P}{I D}$ and $I P \cdot I B = I D^{2}$. Similarly, we have $I Q \cdot I C = I D^{2}$, thus $I P \cdot I B = I Q \cdot I C$. This implies that $B P Q C$ is cyclic, which leads to
$$\angle I P Q = \angle I C B = \frac{\hat{C}}{2}$$
But $\angle A I B = 90^{\circ} + \frac{\hat{C}}{2}$, so $A I$ is perpendicular to $P Q$. Hence, $P Q$ is parallel to $E N$ and $F M$.