Extend the segment $L K$ beyond point $K$ to point $B$ such that $K B=K N$. Then $L B=b$ and triangles $B K N$ and $A M N$ are isosceles.

Let the angles of triangle $K L M$ be $2 \alpha, 2 \beta$, and $2 \gamma$ respectively. Then $\angle K L N=\angle N L M=\beta, \angle K B N=$ $\angle K N B=\alpha$,

$\angle L N A=\angle K N A-\angle K N L=90^{\circ}+\gamma-(\beta+2 \gamma)=90^{\circ}-\beta-\gamma=\alpha$. From this, the similarity of triangles $A L N$ and $N L B$ follows. Therefore, $A L: L N=L N: L B$, from which

$A L=L N^{2} / L B=a^{2} / b$.

## Answer

$a^{2} / b$.
[ The sum of the angles of a triangle. The theorem about the exterior angle. ]

In triangle $K L M$, all sides of which are different, the bisector of angle $K L M$ intersects side $K M$ at point $N$. A line through point $N$ intersects side $L M$ at point $A$, for which $M N=A M$. It is known that $L N=a, K L+K N=b$. Find $A L$.