$ABC$ is triangle. $l_1$- line passes through $A$ and parallel to $BC$, $l_2$ -  line passes through $C$ and parallel to $AB$. Bisector of $\angle B$ intersect $l_1$ and $l_2$ at $X,Y$. $XY=AC$. What  value can take $\angle A- \angle C$ ?
1. Let $\angle A \leq \angle C$. Let $Z = AX \cap BY$. We know that $\angle BCZ = \angle A + \angle C = 180^\circ - \angle B$ and $\angle YBC = \frac{\angle B}{2}$, so $\angle BYC = \frac{\angle B}{2}$ and $BC = CY$.

2. Similarly, we can prove that $AX = AB$ and $\angle YXZ = \frac{\angle B}{2}$. Therefore, $\triangle XYZ$ has angles $\frac{\angle B}{2}, \frac{\angle B}{2}, 180^\circ - \angle B$ and sides equal to $AC, BC - AB, BC - AB$.

3. Let $BY$ intersect the circumcircle of $\triangle ABC$ at $D$. Then $\triangle ADC \cong \triangle XZY$.

4. Let $E$ be a point on $BA$ such that $BE = BC$. Therefore, $AE = BC - AB = XZ = AD$.

5. Let $F$ be the point of intersection of the circumcircle of $\triangle ABC$ and $CE$. We have $\angle BEC = 90^\circ - \frac{\angle B}{2}$, $\angle AFC = 180^\circ - \angle B$, $\angle AFE = \angle B$, and $\angle EAF = 180^\circ - \angle B - (90^\circ - \frac{\angle B}{2}) = 90^\circ - \frac{\angle B}{2} = \angle AEF$. Therefore, $AF = EF$.

6. We have $\angle AFD = \angle AFC + \angle CFD = 180^\circ - \angle B + \frac{\angle B}{2} = 180^\circ - \frac{\angle B}{2}$.

7. We also have $\angle EFD = 360^\circ - \angle AFD - \angle AFE = 360^\circ - 180^\circ + \frac{\angle B}{2} - \angle B = 180^\circ - \frac{\angle B}{2} = \angle AFD$.

8. Therefore, $\triangle EFD \cong \triangle AFD \implies ED = AD = AE \implies \angle BAD = 120^\circ$.

9. We have $120^\circ = \angle BAD = \angle A + \frac{\angle B}{2} = \angle A + \frac{180^\circ - \angle A - \angle C}{2} = \frac{180^\circ + \angle A - \angle C}{2}$.

10. Solving for $\angle A - \angle C$, we get:
\[
120^\circ = \frac{180^\circ + \angle A - \angle C}{2}
\]
\[
240^\circ = 180^\circ + \angle A - \angle C
\]
\[
\angle A - \angle C = 60^\circ
\]

The final answer is $\boxed{60^\circ}$