Four, As shown in the figure, points $M, N$ are on the sides $B C$, $C D$ of square $A B C D$, respectively. It is known that the perimeter of $\triangle M C N$ is equal to half the perimeter of square $A B C D$. Find $\angle M A N$
Four, Solution: Rotate $\triangle A B M$ counterclockwise by $90^{\circ}$ around point $A$, so that $A B$ falls on $A D$, and point $M$ moves to point $M^{\prime}$.
Since $\angle A D M^{\prime}=\angle A B M=90^{\circ}$,
thus $\angle N D M^{\prime}=\angle N D A+\angle A D M^{\prime}$ $=180^{\circ}$,

which means $N, D, M^{\prime}$
are on a straight line.
In $\triangle A N M^{\prime}$

and $\triangle A N M$,
$A M^{\prime}=A M$,
$A N=A N$,
also, the perimeter of $\triangle N M C$ $=B C+C D$ (half the perimeter of square $A B C D$),
$$
M N=B M+N D=M^{\prime} N \text {, }
$$

therefore, $\triangle A N M^{\prime} \cong \triangle A N M$.
Hence $\angle M A N=\angle N A M^{\prime}=\frac{1}{2} \angle B A D$ $=45^{\circ}$