As shown in Figure 5, connect $D M$ and $D N$. Draw $R K / / B C$ through point $A$.
First, prove: $\angle N D A = \angle M D A$.
From $\triangle A R N \backsim \triangle B D N$ and $\triangle A K M \backsim \triangle C D M$, we have
$$A R = \frac{A N}{B N} \cdot B D,$$
$$A K = \frac{A M}{M C} \cdot D C.$$
By Ceva's Theorem, we get $\frac{A N}{N B} \cdot \frac{B D}{D C} \cdot \frac{C M}{M A} = 1$, which means
$$\frac{A N}{N B} \cdot B D = \frac{A M}{M C} \cdot D C.$$

From equations (1), (2), and the above equation, we get $A R = A K$.
Since $A D \perp B C$, we have $A D \perp R K$, thus
$$\angle N D A = \angle M D A.$$
Therefore, we only need to prove
$$\frac{\sin \angle M D F}{\sin \angle A D F} = \frac{\sin \angle N D E}{\sin \angle A D E}.$$

Let $\angle A D M = \angle A D N = \theta$, $\angle A D F = \alpha$, and $\angle A D E = \beta$.
Equation (3) becomes $\frac{\sin (\theta - \alpha)}{\sin \alpha} = \frac{\sin (\theta - \beta)}{\sin \beta}$, which simplifies to $\sin \theta \cdot \cot \alpha = \sin \theta \cdot \cot \beta$.
Since $\sin \theta \neq 0$, we get $\cot \alpha = \cot \beta$, hence $\alpha = \beta$.
Now, we prove equation (3).
From $\frac{D M \sin \angle M D F}{A D \sin \angle A D F} = \frac{S_{\triangle M D F}}{S_{\triangle A D F}} = \frac{F M}{A F}$, we get
$$\frac{\sin \angle M D F}{\sin \angle A D F} = \frac{A D \cdot F M}{D M \cdot A F}.$$
Similarly, $\frac{\sin \angle N D E}{\sin \angle A D E} = \frac{P D \cdot N E}{P E \cdot N D}$.
Thus, we only need to prove
$$\frac{A D \cdot F M}{A F \cdot D M} = \frac{P D \cdot N E}{P E \cdot N D}$$
or
$$\frac{F M \cdot P E}{A F \cdot N E} = \frac{D M \cdot P D}{A D \cdot N D}.$$
Now, we prove the above equation holds.
$$\begin{array}{l} \frac{F M \cdot P E}{A F \cdot N E} = \frac{S_{\triangle Q F M}}{S_{\triangle Q Q F}} \cdot \frac{S_{\triangle Q P E}}{S_{\triangle Q N E}} \\ = \frac{Q M \sin \angle F Q M}{A Q \sin \angle A Q F} \cdot \frac{P Q \sin \angle E Q P}{N Q \sin \angle N Q E} \\ = \frac{Q M \cdot P Q}{A Q \cdot N Q}. \end{array}$$

Since $D Q$ bisects $\angle M D N$, we have $\frac{D M}{D N} = \frac{Q M}{Q N}$.
Thus, we only need to prove $\frac{P D}{A D} = \frac{P Q}{A Q}$.
By Menelaus' Theorem, we know
$$\begin{array}{l} \frac{P Q}{A Q} = \frac{P M}{B M} \cdot \frac{B N}{A N} \\ = \frac{S_{\triangle A P C}}{S_{\triangle A B C}} \cdot \frac{S_{\triangle B P C}}{S_{\triangle P C}} = \frac{S_{\triangle B P C}}{S_{\triangle A B C}} = \frac{P D}{A D}. \end{array}$$

Thus, the proposition is proved.