Let $\triangle ABC$ be a right-angled triangle and $BC > AC$. $M$ is a point on $BC$ such that $BM = AC$ and $N$ is a point on $AC$ such that $AN = CM$. Find the angle between $BN$ and $AM$.

1. **Define the triangle and points:**
   Let $\triangle ABC$ be a right-angled triangle with $\angle A = 90^\circ$. Let $BC = b$ and $AC = a$, where $b > a$. Point $M$ is on $BC$ such that $BM = AC = a$, and point $N$ is on $AC$ such that $AN = CM$.

2. **Determine the lengths:**
   Since $BM = a$, we have $CM = b - a$. Given $AN = CM$, it follows that $AN = b - a$.

3. **Find coordinates of points:**
   - Assume $A$ is at the origin $(0, 0)$.
   - $C$ is at $(a, 0)$.
   - $B$ is at $(0, b)$.
   - $M$ is on $BC$ such that $BM = a$, so $M$ is at $(0, a)$.
   - $N$ is on $AC$ such that $AN = b - a$, so $N$ is at $(b - a, 0)$.

4. **Find slopes of lines $BN$ and $AM$:**
   - Slope of $BN$: 
     \[
     \text{slope of } BN = \frac{0 - b}{(b - a) - 0} = \frac{-b}{b - a}
     \]
   - Slope of $AM$: 
     \[
     \text{slope of } AM = \frac{a - 0}{0 - 0} = \infty \quad \text{(vertical line)}
     \]

5. **Find the angle between $BN$ and $AM$:**
   - Since $AM$ is a vertical line, the angle between $BN$ and $AM$ is the angle between the vertical line and the line with slope $\frac{-b}{b - a}$.
   - The angle $\theta$ between a vertical line and a line with slope $m$ is given by:
     \[
     \theta = \tan^{-1} \left| \frac{1}{m} \right|
     \]
   - Here, $m = \frac{-b}{b - a}$, so:
     \[
     \theta = \tan^{-1} \left| \frac{1}{\frac{-b}{b - a}} \right| = \tan^{-1} \left| \frac{b - a}{-b} \right| = \tan^{-1} \left| \frac{a - b}{b} \right|
     \]
   - Since $b > a$, $\frac{a - b}{b}$ is negative, and the absolute value is:
     \[
     \theta = \tan^{-1} \left( \frac{b - a}{b} \right)
     \]
   - Simplifying further:
     \[
     \theta = \tan^{-1} \left( 1 - \frac{a}{b} \right)
     \]

6. **Calculate the angle:**
   - Since $\tan^{-1}(x) + \tan^{-1}(y) = 45^\circ$ when $x = y = 1$, and here $x = 1 - \frac{a}{b}$, the angle between $BN$ and $AM$ is:
     \[
     \theta = 45^\circ
     \]

The final answer is $\boxed{45^\circ}$