1. Given: Let $AD$ be the altitude of a triangle $\triangle ABC$ and $E$, $F$ be the incenters of the triangles $\triangle ABD$ and $\triangle ACD$, respectively. Line $EF$ meets $AB$ and $AC$ at $K$ and $L$. We need to prove that $AK = AL$ if and only if $AB = AC$ or $\angle A = 90^\circ$.

2. Incenter and Angle Bisector: Let $I$ be the incenter of the triangle $\triangle ABC$. We need to show that the bisector $AI$ of the angle $\angle A$ is perpendicular to $EF$ if and only if either $AB = AC$ or $\angle CAB = 90^\circ$.

3. Concurrent Cevians: The bisectors $AI, BI \equiv EI, CI \equiv FI$ are concurrent cevians of the triangle $\triangle AEF$. Let $X \in AF$ and $Y \in AE$ be the feet of $EI$ and $FI$ in this triangle.

4. Angle Calculation:
$$\angle EXA \equiv \angle BXA = 180^\circ - \frac{\angle B}{2} - (90^\circ - \angle B) - \frac{90^\circ - \angle C}{2} = \frac{90^\circ + \angle B + \angle C}{2} = 90^\circ + \frac{90^\circ - \angle A}{2}$$
$$\angle FYA \equiv \angle CYA = 180^\circ - \frac{\angle C}{2} - (90^\circ - \angle C) - \frac{90^\circ - \angle B}{2} = \frac{90^\circ + \angle C + \angle B}{2} = 90^\circ + \frac{90^\circ - \angle A}{2}$$

5. Equality of Angles: Hence, $\angle EXA = \angle FYA$.

6. **Case 1: $\angle A = 90^\circ$**:
- If $\angle A = 90^\circ$, then $EX$ and $FY$ are two altitudes of the triangle $\triangle AEF$, $I$ is its orthocenter, and $AI \perp EF$ is its remaining altitude.

7. **Case 2: $\angle A \neq 90^\circ$**:
- Assume now that $\angle A \neq 90^\circ$ and that we still have $AI \perp EF$, i.e., $AI$ is the A-altitude of the triangle $\triangle AEF$, but $I$ is no longer its orthocenter.
- Since $\angle EXA = \angle FYA \neq 90^\circ$, then also $\angle EXF = \angle FYE \neq 90^\circ$, i.e., the quadrilateral $EFXY$ is cyclic.

8. Cyclic Quadrilateral:
- Let $(P)$ be its circumcircle, which is centered on the perpendicular bisector of the segment $EF$.
- Using parallel projection, we can project the triangle $\triangle AEF$ into a triangle $\triangle A'EF$, so that the point $I$ is projected into its orthocenter $I'$.
- Then the cevians $EX$, $FY$ are projected into its altitudes $EX'$, $FY'$, i.e., the quadrilateral $EFX'Y'$ is also cyclic, with the circumcircle $(O')$ centered at the midpoint $O'$ of the segment $EF$.

9. Symmetry and Congruence:
- Thus the points $X$, $Y$ lie both on the circle $(P)$ and on an ellipse $o$ with the main axis $EF$ (which is projected into the circle $(O')$ in our parallel projection), not identical with the circle $(P)$.
- Since both the ellipse $o$ and the circle $(P)$ are symmetrical with respect to the perpendicular bisector of the segment $EF$, so are their intersections $X$, $Y$, i.e., $EX = FY$ and $EY = FX$.
- Hence, the triangles $\triangle EFX \cong \triangle FEY$ are congruent and it immediately follows that the triangle $\triangle AEF$ is isosceles with $AE = AF$.

10. Conclusion:
- This and $AD \perp BC$ implies that the incircles $(E) \cong (F)$ are congruent, which means that the triangle $\triangle ABC$ itself is isosceles with $AB = AC$.

$\blacksquare$