15 (12 points) As shown in the figure, plane $m \parallel$ plane $n$, line segment $AD$ intersects $m$ and $n$ at points $B$ and $C$, respectively. Another line through point $A$ intersects $m$ and $n$ at points $M$ and $P$, respectively, and another line through point $D$ intersects $m$ and $n$ at points $N$ and $Q$, respectively. Given $\frac{S_{\triangle BMN}}{S_{\triangle CPQ}}=\frac{3}{2}$, find the minimum value of $\frac{AD}{CD}$.
(15) From the fact that plane $m //$ plane $n$, we know $B M // C P, B N // C Q$, so we have
$$
\sin \angle M B N=\sin \angle P C Q \text {, and } \frac{B M}{C P}=\frac{A B}{A C}, \frac{B N}{C Q}=\frac{B D}{C D} \text {. }
$$

Also, because
$$
\begin{aligned}
S_{\triangle B M N} & =\frac{1}{2} B M \cdot B N \sin \angle M B N, \\
S_{\triangle C P Q} & =\frac{1}{2} C P \cdot C Q \sin \angle P C Q,
\end{aligned}
$$
$$
\begin{array}{l} 
\text { From } \frac{S_{\triangle B M N}}{S_{\triangle C P Q}}= \frac{3}{2}, \text { we get } \frac{A B}{A C} \cdot \frac{B D}{C D}=\frac{3}{2} . \\
\text { Let } \frac{A C}{A B}=\alpha, \frac{B D}{C D}=\beta, \text { then } 2 \beta=3 \alpha, \text { and } \beta>\alpha>1 \cdot \frac{B C}{A B}=\frac{A C-A B}{A B}= \\
\alpha-1, A B=\frac{1}{\alpha-1} B C .
\end{array}
$$
$$
\begin{aligned}
\frac{A D}{C D} & =\frac{A B+B C+C D}{C D}=1+\frac{\alpha}{\alpha-1} \cdot \frac{B C}{C D} \\
& =1+\frac{\alpha}{\alpha-1} \cdot \frac{B D-C D}{C D} \\
& =1+\frac{\alpha}{\alpha-1}(\beta-1) \\
& =1+\frac{\alpha}{\alpha-1}\left(\frac{3}{2} \alpha-1\right)=\frac{3 \alpha^{2}-2}{2(\alpha-1)} \\
& =\frac{3}{2}(\alpha+1)+\frac{1}{2(\alpha-1)} \\
& =\frac{3}{2}\left[(\alpha-1)+\frac{1}{3(\alpha-1)}\right]+3 \\
& \geqslant \frac{3}{2} \cdot 2 \sqrt{\frac{1}{3}(\alpha-1) \cdot \frac{1}{\alpha-1}}+3 \\
& =3+\sqrt{3} .
\end{aligned}
$$

Clearly, when $\alpha-1=\frac{\sqrt{3}}{3}$, i.e., $\alpha=1+\frac{\sqrt{3}}{3}$, we have $\frac{A D}{C D}=3+\sqrt{3}$, so the minimum value of $\frac{A D}{C D}$ is $3+\sqrt{3}$.