Example 4 As shown in Figure 8, given that the side length of square $ABCD$ is 1, points $M$ and $N$ are on $BC$ and $CD$ respectively, such that the perimeter of $\triangle CMN$ is 2. Find:
(1) the size of $\angle MAN$;
(2) the minimum value of the area of $\triangle MAN$.
(2003, (Yu Zhen Cup) Shanghai Junior High School Mathematics Competition)
Solution: (1) As shown in Figure 8, extend $CB$ to point $L$ such that $BL = DN$. Then, Rt $\triangle ABL \cong \text{Rt} \triangle ADN$.
Thus, $AL = AN$, $\angle LAB = \angle NAD$,
$\angle NAL = \angle DAB = 90^\circ$.
Also, $MN = 2 - CN - CM$
$= 1 - CN + 1 - CM = DN + BM$
$= BL + BM = ML$,
and $AM$ is a common side, so $\triangle AMN \cong \triangle AML$.
Therefore, $\angle MAN = \angle MAL = \frac{1}{2} \times 90^\circ = 45^\circ$.
(2) Let $CM = x$, $CN = y$, $MN = z$. Then $\left\{\begin{array}{l}x + y + z = 2, \\ x^2 + y^2 = z^2\end{array} \Leftrightarrow \left\{\begin{array}{l}x = 2 - y - z, \\ x^2 + y^2 = z^2 .\end{array}\right.\right.$
Thus, $(2 - y - z)^2 + y^2 = z^2$, which simplifies to $2y^2 + (2z - 4)y + (4 - 4z) = 0$.
Since $y > 0$, consider its discriminant
$\Delta = 4(z - 2)^2 - 32(1 - z) \geqslant 0$,
so, $(z + 2 + 2\sqrt{2})(z + 2 - 2\sqrt{2}) \geqslant 0$.
Since $z > 0$, then $z \geqslant 2\sqrt{2} - 2$.
Equality holds if and only if $x = y = 2 - \sqrt{2}$.
Note that
$S_{\triangle AMN} = S_{\triangle AML} = \frac{1}{2} ML \cdot AB = \frac{1}{2} z \geqslant \sqrt{2} - 1$.
Therefore, the minimum area of $\triangle AMN$ is $\sqrt{2} - 1$.
Note: In Figure 8, 
$\angle MAN = 45^\circ$
$\Leftrightarrow$ the perimeter of $\triangle CMN$ is half the perimeter of square $ABCD$.