Example 9. Fold a regular hexagonal paper sheet $A B C D$ with side length $a$, so that $B$ falls on $A D$. At what position on $A D$ should $B$ fall to minimize the area of the folded part? Find this minimum area.
From the given, we know that $M N$ is the fold line, and the trapezoid $B^{\prime} C^{\prime} N M$ is the folded part. Therefore, $B^{\prime}, B$ are symmetric about $M N$, so $B B^{\prime} \perp M N$. From $\mathrm{Rt} \triangle M P B$:
$$
\begin{array}{l}
\text { In } \triangle B^{\prime} A B \text {. Let } A B^{\prime} \\
= x, X A B=a, \text { then } \\
\frac{\sqrt{x^{2}+a^{2}}}{2} \\
= \frac{v^{2}+a^{2}}{a} .
\end{array}
$$

Thus, $B M=\frac{x^{2}+a^{2}}{2 a}$.
Also, from Rt $\triangle M N = R t \therefore B^{\prime} B A$.
$$
\begin{array}{l}
\text { Then } Q M=A B^{\prime}=x . \\
C N=B M-Q M \\
= \frac{x^{2}+a^{2}}{2 a}-x=\frac{x^{2}+a^{2}-2 a x}{2 a} .
\end{array}
$$

Therefore, $S_{B C Y M}=\frac{1}{2}(B M+C N) a$
$$
\begin{array}{l}
=\frac{1}{2}(x+a-a x) \\
=\frac{1}{2}\left(x-\frac{a}{2}\right)^{2}+\frac{3}{8} a^{2}
\end{array}
$$

That is, when $x=\frac{a}{2}$, $S_{\text {trapezoid }}=\frac{3 a^{2}}{8}$.