5. In $\square A B C D$, $M$ is a point on the diagonal $B D$, and $M D=3 B M, A M$ intersects $B C$ at point $N$. Find the value of $\frac{S_{\triangle M N D}}{S_{\square A B C D}}$.
Notice,
$$
\begin{aligned}
& S_{\triangle A N D}=S_{\triangle B D A}, \\
& S_{\square A B C D}=2 S_{\triangle A B D}. \\
& \text{Since } A D \parallel B N \\
\Rightarrow & \frac{A M}{M N}=\frac{D M}{M B}=\frac{3}{1} \Rightarrow \frac{M N}{A N}=\frac{1}{4} \\
\Rightarrow & \frac{S_{\triangle M N D}}{S_{\triangle N A D}}=\frac{\frac{1}{2} M N h_{D}}{\frac{1}{2} A N h_{D}}=\frac{1}{4}.
\end{aligned}
$$

Therefore, $\frac{S_{\triangle M N D}}{S_{\triangle A B C D}}=\frac{S_{\triangle M N D}}{2 S_{\triangle N A D}}=\frac{1}{8}$.
5. As shown in Figure 1, connect $N D$, the distance from point $D$ to side $A N$ is $h_{D}$.