Example 10 Let the side length of square $ABCD$ be $AB=a$, and let $M$ and $N$ be points on $AB$ and $CD$ respectively. Fold the trapezoid $BCNM$ along $MN$ so that point $B$ lands on $AD$. How should it be folded to minimize the area of the folded part?

Fold the trapezoid $BCNM$ along $MN$ so that point $B$ lands on $AD$. How should it be folded to minimize the area of the folded part?
Analysis one: Since the factors affecting the area change of trapezoid $B C N M$ are the positions of $M$ or $N$, we can choose the length of $C N$ as $x$, and then use the symmetry of $B C N M$ and $B^{\prime} C^{\prime} N M$ about $M N$ to establish the functional relationship as $S_{B C N M} = f(x)$, and then find the minimum value of this function.

Solution one: Let $C N = x, x \in (0, a)$. Draw $N K \perp A B$, with the foot of the perpendicular at $K$, and connect $B B^{\prime}$. By the given condition, $B B^{\prime}$ is bisected perpendicularly by $M N$.
In the right triangles $\triangle M N K$ and $\triangle B^{\prime} B A$, $N K = B A = a$,
Since $\angle M N K$ and $\angle B^{\prime} B A$ have sides that are perpendicular to each other,
Therefore, $\angle M N K = \angle B^{\prime} B A$, thus we can deduce
Right $\triangle M N K \cong$ Right $\triangle B^{\prime} B A, M K = B^{\prime} A$.
Let $M B = u$, then $M B^{\prime} = u, A M = a - u, A B^{\prime} = M K = u$.
In the right triangle $\triangle M B^{\prime} A$, $A M^{2} + A B^{\prime 2} = M B^{\prime 2}$,
That is, $(a - u)^{2} + (u - x)^{2} = u^{2}$,
Since $u < a$, then $u = a + x - \sqrt{2 a x}$, thus $S = \frac{1}{2} a (2 x - \sqrt{2 a x} + a)$.
Let $\sqrt{2 a x} = t, 2 x = \frac{t^{2}}{a}$, then
$$
\begin{aligned}
S & = \frac{1}{2} a \left( \frac{t^{2}}{a} - t + a \right) \\
& = \frac{1}{2} \left( t - \frac{a}{2} \right)^{2} + \frac{3}{8} a^{2}, t \in (0, a) .
\end{aligned}
$$

When $t = \frac{a}{2}, C N = x = \frac{1}{8} a, B M = u = \frac{5}{8} a$, then $S_{\min} = \frac{3}{8} a^{2}$.
Therefore, on $B A$, take $B M = \frac{5}{8} a$, on $C D$, take $C N = \frac{1}{8} a$, connect $M N$, and fold along $M N$ to make the area of the folded part the smallest.
Analysis two: Use $\angle M N K$ as the independent variable to establish the functional relationship.
Solution two: Let $\angle M N K = \theta$, from solution one we know
$B B^{\prime} = M N = \frac{a}{\cos \theta}, B M = \frac{B B^{\prime}}{2 \cos \theta} = \frac{a}{2 \cos^{2} \theta}$, $C N = B M - M K = \frac{1}{2} a (\operatorname{tg}^{2} \theta - 2 \operatorname{tg} \theta + 1)$,
Thus $S = \frac{a^{2}}{2} (\operatorname{tg}^{2} \theta - \operatorname{tg} \theta + 1)$
$$
= \frac{a^{2}}{2} \left( \operatorname{tg} \theta - \frac{1}{2} \right)^{2} + \frac{3}{8} a^{2} .
$$

The rest is the same as solution one.
Analysis three: Use the analytical method.
Solution three: Take $B$ as the origin, and the lines $B C$ and $B A$ as the $x$-axis and $y$-axis to establish the coordinate system (as shown in Figure 2-5).
Let $B^{\prime}(t, a), a \in (0, a)$, the equation of the perpendicular bisector $M N$ of $B B^{\prime}$ is
$$
y - \frac{a}{2} = -\frac{t}{a} \left( x - \frac{t}{2} \right) \text{. }
$$

Let $x = 0$, we get
$$
B M = \frac{a}{2} + \frac{t^{2}}{2 a} ;
$$

Let $x = a$, we get
$$
C N = \frac{a}{2} - t + \frac{t^{2}}{2 a} \text{. }
$$

Thus $S = \frac{a}{2} (B M + C N)$
$$
= \frac{1}{2} \left( t - \frac{a}{2} \right)^{2} + \frac{3}{8} a^{2} .
$$

The rest is the same as solution one.