Let $ABC$ be an isosceles right triangle at $A$ with $AB=AC$. Let $M$ and $N$ be on side $BC$, with $M$ between $B$ and $N,$ such that $$BM^2+ NC^2= MN^2.$$ Determine the measure of the angle $\angle MAN.$
1. **Identify the properties of the isosceles right triangle:**
   Given that $\triangle ABC$ is an isosceles right triangle with $\angle A = 90^\circ$ and $AB = AC$, we know that $\angle B = \angle C = 45^\circ$.

2. **Reflect point $B$ across line $AM$:**
   Let $B'$ be the reflection of $B$ across $AM$. Since $AM$ is the angle bisector of $\angle BAC$, reflecting $B$ across $AM$ will place $B'$ such that $AB' = AB = AC$ and $\angle B'AM = \angle BAM$.

3. **Identify the congruent triangles:**
   Since $AB' = AB = AC$ and $\angle B'AM = \angle BAM$, triangles $AB'M$ and $ACM$ are congruent by the SAS (Side-Angle-Side) criterion. This implies that $B'M = CM$.

4. **Determine the position of $N$:**
   Let $N'$ be the point where the angle bisector of $\angle B'AC$ meets $BC$. Since $B'AN'$ and $CAN'$ are congruent (by the Angle Bisector Theorem), we have $AN' = AN'$ and $\angle B'AN' = \angle CAN'$. Therefore, triangles $B'AN'$ and $CAN'$ are congruent by the ASA (Angle-Side-Angle) criterion.

5. **Verify the given condition:**
   We need to verify that $BM^2 + NC^2 = MN^2$. Since $B'M = CM$ and $N'$ is on $BC$, we have:
   \[
   BM^2 + CN'^2 = MN'^2
   \]
   This holds true because $B'M = CM$ and $N'$ is the point where the angle bisector meets $BC$.

6. **Determine the measure of $\angle MAN$:**
   Since $N$ is the same as $N'$, we have:
   \[
   \angle MAN = \angle B'AM + \angle B'AN
   \]
   Given that $\angle B'AM = \angle BAM = 45^\circ$ and $\angle B'AN = \angle CAN = 45^\circ$, we get:
   \[
   \angle MAN = 45^\circ + 45^\circ = 90^\circ
   \]

The final answer is $\boxed{90^\circ}$