Points $N$ and $M$ are on the sides $CD$ and $BC$ of square $ABCD$, respectively. The perimeter of triangle $MCN$ is equal to the double of the length of the square's side. Find $\angle MAN$.
1. Let the side length of the square \(ABCD\) be \(x\). We denote \(NC = a\) and \(MC = b\). Given that the perimeter of triangle \(MCN\) is equal to twice the side length of the square, we have:
   \[
   MC + CN + NM = 2x
   \]
   Substituting the given values, we get:
   \[
   b + a + NM = 2x
   \]
   Solving for \(NM\), we find:
   \[
   NM = 2x - a - b
   \]

2. Consider a point \(P\) on \(NM\) such that \(NP = x - a\) and \(MP = x - b\). This implies that the circles centered at \(N\) and \(M\) with radii \(x - a\) and \(x - b\) respectively are tangent at \(P\).

3. Using the Pythagorean theorem, we calculate the distances \(AN\) and \(AM\):
   \[
   AN^2 = x^2 + (x - a)^2
   \]
   \[
   AM^2 = x^2 + (x - b)^2
   \]

4. Since the circles are tangent at \(P\), the power of point \(A\) with respect to both circles is equal. Therefore:
   \[
   x^2 = AN^2 - (x - a)^2 = AM^2 - (x - b)^2
   \]
   This implies that point \(A\) lies on the radical axis of the two circles, and thus \(AP\) is perpendicular to \(NM\).

5. Since \(PM = MB\), \(AM = AM\), and \(\angle APM = \angle ABM = 90^\circ\), triangles \(PAM\) and \(BAM\) are congruent. Therefore, \(\angle PAM = \angle MAB\).

6. Similarly, \(\angle NAP = \angle NAD\).

7. Therefore, the angle \(\angle MAN\) can be expressed as:
   \[
   \angle MAN = \angle NAP + \angle PAM = \frac{\angle BAD}{2} + \frac{\angle BAC}{2} = \frac{90^\circ}{2} = 45^\circ
   \]

The final answer is \(\boxed{45^\circ}\)