As shown in Figure 2, given a square $ABCD$, extend $BC$ and $DC$ to $M$ and $N$ respectively, such that $S_{\triangle QMN} = S_{\text{square } ABCD}$. Determine the degree measure of $\angle MAN$.
Solution: As shown in Figure 3, since $AD = AB$, we can rotate $\triangle ABM$ $90^{\circ}$ around point $A$ to the position of $\triangle ADE$. Connect $EM$. Then points $C$, $D$, and $E$ are collinear, $\triangle AME$ is an isosceles right triangle, and $\angle AME = 45^{\circ}$.

Let the side length of the square $ABCD$ be $a$, $CM = m$, and $CN = n$. Then
$$
\begin{array}{l}
DN = a + n, \quad CE = 2a + m. \\
\because S_{\triangle ONV} = S_{\text{square } ABCD}, \\
\therefore \frac{1}{2} mn = a^2,
\end{array}
$$

which implies
$$
\begin{array}{l}
\frac{m}{a} = \frac{2a}{n} \Rightarrow \frac{m}{a} = \frac{2a + m}{a + n}. \\
\text{Therefore, } \frac{CM}{AD} = \frac{CE}{DN}.
\end{array}
$$

Thus, Rt $\triangle MCE \sim \text{Rt} \triangle ADN$,
so $\angle CEM = \angle DNA$,
and $ME \parallel AN$.
Hence, $\angle MAN = \angle AME = 45^{\circ}$.
(Wan Xiren, BaiSha Township Government, Yuanjiang City, Hunan Province, 413105)