Problem 1073

7. Draw a line through point $K$ parallel to $B C$, intersecting $A B$ and $A C$ at points $R$ and $S$ respectively. Then $I K \perp R S$.
Since $I F \perp A B$, points $I, K, F, R$ are concyclic.
Similarly, points $I, K, S, E$ are concyclic.
Thus, $\angle I R K = \angle I F K = \angle I E K = \angle I S K$
$\Rightarrow K R = K S$.
Since $R S \parallel B C$, $G$ is the midpoint of $B C$.
Given $\angle C N E = \angle A E F - \angle E C N$
$= 90^{\circ} - \frac{\angle B A C}{2} - \frac{\angle A C B}{2}$
$$= \frac{\angle A B C}{2} = \angle F B I,$$

we know that points $B, I, N, F$ are concyclic.
Since points $B, D, I, F$ are concyclic, points $B, D, I, N, F$ are concyclic. Thus, $I N \perp B N$.
Similarly, $I M \perp C M$.
Therefore, $I$ is the orthocenter of $\triangle P B C$, and hence, $P I \perp B C$.

Thus, points $P, I, D$ are collinear.
Let $B I$ intersect $P Q$ at point $T$.
Since points $D, C, M, I$ are concyclic,
$$\begin{array}{l} \angle P I M = \angle P C B. \\ \text{Also, } \angle T P I = 90^{\circ} - \angle B P D = \angle P B C \\ \Rightarrow \triangle T P I \sim \triangle P B C . \end{array}$$

Take the midpoint $L$ of segment $P I$ and connect $L T$. Then
$$\begin{array}{l} \triangle T P L \sim \triangle P B G \Rightarrow \angle P T L = \angle B P G \\ \Rightarrow \angle P T L + \angle G P T = \angle B P G + \angle G P T \\ = \angle B P Q = 90^{\circ} \\ \Rightarrow T L \parallel Q I . \end{array}$$

Therefore, $P T = T Q$, i.e., $B I$ bisects $P Q$.