14. As shown in Figure 5, in $\square A B C D$, $P_{1}$, $P_{2}, \cdots$, $P_{n-1}$ are the $n$ equal division points on $B D$. Connect $A P_{2}$ and extend it to intersect $B C$ at point $E$, and connect $A P_{n-2}$ and extend it to intersect $C D$ at point $F$.
(1) Prove that $E F \parallel B D$;
(2) Let the area of $\square A B C D$ be $S$. If $S_{\triangle A E F} = \frac{3}{8} S$, find the value of $n$.
14. (1) Since $A D / / B C, A B / / D C$. Therefore, $\left.\triangle P_{n-2} F I\right) \triangle \triangle P_{n-2} A B, \triangle P_{2} B E \subset \triangle P_{2} D . A$.
Thus, we know
$$
\frac{A P_{n-2}}{P_{n+2} F}=\frac{B P_{n}}{P_{n-2} D}=\frac{n-2}{2}, \frac{A P_{2}}{P_{2} E}=\frac{D P_{2}}{P_{2} B}=\frac{n-2}{2},
$$

which means $\frac{A P_{n-2}}{P_{n} \frac{{ }_{2}}{} F}=\frac{A P_{2}}{P_{2} E}$; hence $E F / / B D$.
(2) From (1), we know $\frac{A P_{2}}{A E}=\frac{n-2}{n}, \frac{S_{\triangle A P_{2} P_{n-2}}}{S_{\text {ZAFF }}}=$ $\left(\frac{A P_{2}}{A E}\right)^{2}$. Therefore,
$$
\begin{aligned}
& S_{\triangle_{2} P_{n-2}}=\left(\frac{n-2}{n}\right)^{2} \cdot S_{\text {IAE }} \\
= & \left(\frac{n-2}{n}\right)^{2} \cdot \frac{3}{8} S .
\end{aligned}
$$

Also, since $S_{\triangle A A S}=\frac{1}{2} S_{\triangle A / \times D}=\frac{1}{2} S$, and $P_{1}, P_{2}, \cdots$ 。 $P_{n}$ 。are the $n$ equal division points of $B D$. Therefore,
$$
\begin{array}{l}
S_{-N_{2} p_{n}}=\left(\frac{n-4}{n}\right) \cdot S_{\text {U(S) }} \\
=\left(\frac{n-4}{n}\right) \cdot \frac{1}{2} S .
\end{array}
$$

From (1) and (2), we get $\left(\frac{n-2}{n}\right)^{2} \cdot \frac{3}{8} S=\left(\frac{n-4}{n}\right) \cdot \frac{1}{2} S$,
which simplifies to $n^{2}-4 n-12=0$.
Solving, we get $n=6, n=-2$ (discard). Therefore, $n=6$.