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 $\frac{5}{4} a^{2}$.