3. In Rt $\triangle A B C$, $\angle B=90^{\circ}$, $P$ is a point on the angle bisector of $\angle A$ inside $\triangle A B C$, $M$ (different from $A, B$) is a point on side $A B$, and lines $A P, C P, M P$ intersect side $B C, A B, A C$ at points $D, E, N$ respectively. If $\angle M P B = \angle P C N, \angle N P C = \angle M B P$, find $\frac{S_{\triangle A P C}}{S_{A C D E}}$.
3. Let $\angle M P B=\alpha, \angle N P C=\beta, A B=c, B C=a, C A=b, r$ be the inradius of $\triangle A B C$.

Since $180^{\circ}=\angle M P N=\alpha+\left(\frac{\angle A}{2}+\beta\right)+\left(\frac{\angle A}{2}+\alpha\right)+\beta$
Therefore, $\alpha+\beta=\frac{\angle B}{2}+\frac{\angle C}{2}$
Therefore, $\angle B P C=180^{\circ}-\frac{\angle B}{2}-\frac{\angle C}{2}$
Since point $P$ lies on the angle bisector of $\angle A$
Therefore, point $P$ must be the incenter of $\triangle A B C$
Therefore, $\frac{S_{A C D E}}{S_{\triangle A P C}}=\frac{S_{\triangle A P C}+S_{\triangle D P C}+S_{\triangle A P E}+S_{\triangle E P D}}{S_{\triangle A P C}}=\left(1+\frac{P D}{P A}\right)\left(1+\frac{P E}{P C}\right)$
Also, $\frac{P D}{P A}=\frac{r}{c-r}, \frac{P E}{P C}=\frac{r}{a-r}$ and $r=\frac{a c}{a+b+c}$
Therefore, $\frac{S_{\triangle A P C}}{S_{A C D E}}=\frac{(c-r)(a-r)}{a c}=\frac{(a+b)(b+c)}{(a+b+c)^{2}}=\frac{b^{2}+b(a+c)+a c}{(a+b+c)^{2}}=\frac{1}{2}$