Answer: $90^\circ - \beta/2$.
Let point $N$ be marked on the side $AC$ such that $AN = 0.5 AB$. Then, by condition, $NC = AC - AN = 0.5(AB + BC) - 0.5 AB = 0.5 BC$. Therefore, $AN : NC = AB : BC$. Since $BL$ is a bisector of the angle $ABC$ we have $AL : LC = AB : BC$, so $L$ and $N$ coincide. Therefore the triangles $KAL$ and $MCL$ are isosceles and
$$\angle AKL = \angle ALK = 0.5(180^\circ - \angle BAC), \quad \angle CML = \angle CLM = 0.5(180^\circ - \angle BCA).$$
Thus,
$$\begin{aligned} \angle KLM &= 180^\circ - \angle ALK - \angle CLM \\ &= 180^\circ - 0.5(180^\circ - \angle BAC) - 0.5(180^\circ - \angle BCA) \\ &= 0.5(\angle BAC + \angle BCA) \\ &= 0.5(180^\circ - \angle ABC) \\ &= 0.5(180^\circ - \beta) \\ &= 90^\circ - \beta/2. \end{aligned}$$