1. Given Information and Assumptions:
- In triangle $ABC$, the difference between angles $B$ and $C$ is $90^\circ$.
- $AL$ is the angle bisector of $\angle BAC$.
- The bisector of the exterior angle at $A$ intersects $BC$ at point $F$.

2. Determine Angles:
- Let $\angle CBA = \beta$.
- Since $\angle B - \angle C = 90^\circ$, we have $\angle C = \beta - 90^\circ$.
- The sum of angles in triangle $ABC$ is $180^\circ$, so:
$$\angle BAC = 180^\circ - \angle CBA - \angle ACB = 180^\circ - \beta - (\beta - 90^\circ) = 90^\circ - 2\beta$$

3. **Angle Bisector $AL$:**
- The angle bisector $AL$ divides $\angle BAC$ into two equal parts:
$$\angle BAL = \angle CAL = \frac{1}{2} \angle BAC = \frac{1}{2} (90^\circ - 2\beta) = 45^\circ - \beta$$

4. **Angles in Triangle $CLA$:**
- In triangle $CLA$, we need to find $\angle CLA$:
$$\angle CLA = 180^\circ - \angle CAL - \angle ACB = 180^\circ - (45^\circ - \beta) - (\beta - 90^\circ) = 180^\circ - 45^\circ + \beta - \beta + 90^\circ = 45^\circ$$

5. **Perpendicularity of $AL$ and $AF$:**
- It is known that the interior angle bisector is perpendicular to the exterior angle bisector of the same angle in a triangle. Therefore, $AL \perp AF$.

6. **Isosceles Right Triangle $LAF$:**
- Since $\angle LAF = 90^\circ$ and $\angle CLA = 45^\circ$, triangle $LAF$ is an isosceles right triangle with $\angle ALF = 45^\circ$.
- In an isosceles right triangle, the legs are equal, so $AL = AF$.

$\blacksquare$