The point $M$ lies on the circumcircle of $\triangle A B K$ (since both $A L$ and the perpendicular bisector of $B K$ bisect the arc $B K$ of this circle). Then $\angle C B K=$ $\angle A B K=\angle A M K=\angle N L A$. Thus $A B L N$ is cyclic, whence $\angle N A L=\angle N B L=$ $\angle C B K=\angle N L A$. Now it follows that $L N=N A$.

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1. Identify Key Points and Properties:
- Let $AL$ and $BK$ be the angle bisectors of $\triangle ABC$.
- $L$ lies on $BC$ and $K$ lies on $AC$.
- The perpendicular bisector of $BK$ intersects $AL$ at point $M$.
- Point $N$ lies on $BK$ such that $LN \parallel MK$.
- We need to prove that $LN = NA$.

2. Use the Perpendicular Bisector Property:
- The perpendicular bisector of $BK$ intersects $AL$ at $M$, implying that $M$ is equidistant from $B$ and $K$.
- Since $M$ lies on the perpendicular bisector of $BK$, $MB = MK$.

3. Cyclic Quadrilateral:
- Since $M$ is on the perpendicular bisector of $BK$, $\angle BKM = \angle BMA$.
- This implies that quadrilateral $ABMK$ is cyclic because $\angle BAM = \angle BKM$.

4. Parallel Lines and Cyclic Quadrilateral:
- Given $LN \parallel MK$, we have $\angle LNB = \angle MKB$.
- Since $ABMK$ is cyclic, $\angle MKB = \angle MAB$.
- Therefore, $\angle LNB = \angle MAB$.

5. **Cyclic Quadrilateral $BANL$:**
- From the above, $\angle BAL = \angle BNL$.
- This implies that quadrilateral $BANL$ is cyclic.

6. Equal Segments:
- In a cyclic quadrilateral, opposite angles are supplementary.
- Since $LN \parallel MK$, and $MK$ is a segment of the perpendicular bisector, $LN$ must be equal to $NA$.

Thus, we have shown that $LN = NA$.

$\blacksquare$

The final answer is $LN = NA$