Example $6 \triangle A B C$ and $\triangle A^{\prime} B^{\prime} C^{\prime}$ are two isosceles right triangles with legs both equal to $2 a$. As shown in Figure 6, they are stacked together, with $\triangle A B C$ fixed in position, and the midpoints of the legs $A C$ and $B C$ being $M$ and $N$, respectively. Keeping the hypotenuse $A^{\prime} B^{\prime}$ on the line $M N$, while moving $\triangle A^{\prime} B^{\prime} C^{\prime}$ (the shaded part), find the maximum and minimum values of the overlapping area.
Solution: In figure $\mathrm{o}$, it is easy to prove that quadrilateral $A E B^{\prime} M$ is a parallelogram.
Then $B^{\prime} E=A M=\frac{1}{2} A C=a$.
$$
\therefore C^{\prime} E=2 a-a=a \text {. }
$$

Therefore, $E$ is the midpoint of $B^{\prime} C^{\prime}$.
Similarly, $D$ is the midpoint of $A^{\prime} C$.
$$
\therefore D E=\frac{1}{2} A^{\prime} B^{\prime}=\frac{1}{2} \times 2 \sqrt{2} a=\sqrt{2} a \text {. }
$$

Let $A H=x$, then
$$
\begin{array}{l}
A D=\frac{x}{\cos 45^{\circ}}=\sqrt{2} x, \\
B E=2 \sqrt{2} a-\sqrt{2} a-\sqrt{2} x=\sqrt{2}(a-x) . \\
\begin{array}{l} 
\therefore B K=B E \cos 45^{\circ}=a-x . \\
\therefore S_{\text {pentagon ADDEEN }} \\
\quad=S_{\text{UAMAN }}-S_{\triangle A D H}-S_{\triangle E B K} \\
=\frac{3}{2} a^{2}-\frac{1}{2} x^{2}-\frac{1}{2}(a-x)^{2} \\
=-\left(x-\frac{a}{2}\right)^{2}+\frac{5}{4} a^{2}(0 \leqslant x \leqslant a) .
\end{array}
\end{array}
$$
$\therefore S_{\text {polygon HIDEEN }}$

Therefore, the maximum value of the overlapping area of the two triangles is $\frac{5}{4} a^{2}$.