Problem 1502

1. Given that point $L$ inside triangle $ABC$ is such that $CL = AB$ and $\angle BAC + \angle BLC = 180^\circ$. This implies that $\angle BAC$ and $\angle BLC$ are supplementary angles. Therefore, we have:
$$\sin \angle BAC = \sin \angle BLC$$

2. Using the Law of Sines in $\triangle BLC$ and $\triangle ABC$, we get:
$$\frac{LC}{\sin \angle LBC} = \frac{BC}{\sin \angle BLC} \quad \text{and} \quad \frac{AB}{\sin \angle BCA} = \frac{BC}{\sin \angle BAC}$$

3. Since $\sin \angle BAC = \sin \angle BLC$, we can equate the two expressions:
$$\frac{LC}{\sin \angle LBC} = \frac{AB}{\sin \angle BCA}$$

4. Given $LC = AB$, we substitute $LC$ with $AB$:
$$\frac{AB}{\sin \angle LBC} = \frac{AB}{\sin \angle BCA}$$

5. Simplifying, we find:
$$\sin \angle LBC = \sin \angle BCA$$

6. Since $L$ is inside $\triangle ABC$, the angles $\angle LBC$ and $\angle BCA$ cannot sum to $180^\circ$. Therefore, we must have:
$$\angle LBC = \angle BCA$$

7. Since $KL \parallel BC$, $\angle KLC = \angle BCA$ by the Alternate Interior Angles Theorem. Thus, $\angle KLC = \angle LBC$.

8. This implies that $\triangle KLC$ is isosceles with $KL = LC$. Given $LC = AB$, we have:
$$KL = AB$$

9. Since $KL \parallel BC$ and $K$ lies on $AC$, $CKLB$ forms an isosceles trapezoid with $CK = LB$ and $KL = AB$.

10. The diagonals of an isosceles trapezoid are equal, so:
$$BK = LC = AB$$

Therefore, we have proven that $AB = BK$.

$\blacksquare$

The final answer is $\boxed{ AB = BK }$