Problem 5. In triangle $\mathrm{KLM}$ with angle $\mathrm{L}=120^{\circ}$, the angle bisectors LA and $\mathrm{KB}$ of angles KLM and LKM are drawn respectively. Find the measure of angle KBA.
# Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_f7a16de7ae2a30fbf517g-4.jpg?height=591&width=1106&top_left_y=184&top_left_x=475)

1). Let KS be the extension of KL beyond point L. Then LM is the bisector of angle MLS, since $\angle M L S = \angle M L A = \angle A L K = 60^{\circ}$. Point B is the intersection of the bisectors LM and KB of angles SLA and SKM, respectively $\Rightarrow$ the intersection of the bisectors of the external angles of triangle KLA $\Rightarrow \mathrm{AB}$ is the bisector of angle LAM.

2). Similarly, it can be proven that AC is the bisector of angle LAK, but angles LAM and LAK are adjacent $\Rightarrow \angle C A B = 90^{\circ}$.

3). In triangle KLA: E is the point of intersection of the bisectors $\Rightarrow$ $\angle K E A = \frac{1}{2} \angle K L A + 90^{\circ} = \frac{1}{2} 60^{\circ} + 90^{\circ} = 120^{\circ} \Rightarrow \angle B E A = 180^{\circ} - 120^{\circ} = 60^{\circ}$.

4). In triangle ВЕА: $\angle A = 90^{\circ} \Rightarrow \angle B = 180^{\circ} - 90^{\circ} - 60^{\circ} = 30^{\circ}$.

Answer: $\angle K B A = 30^{\circ}$.