4. 4.1. In a right triangle $ABC$ (right angle at $C$), the bisector $BK$ is drawn. Point $L$ on side $BC$ is such that $\angle CKL = \angle ABC / 2$. Find $KB$, if $AB = 18, BL = 8$.
Answer: 12.

## Solution.

Note that $\angle L K B=\angle C K B-\angle C K L=\angle C A B+\angle A B K-\angle C K L$ (the last equality holds because $\angle C K B$ is an exterior angle of triangle $A B K$). Since $\angle C K L=\angle A B C / 2=\angle A B K$, we have that $\angle L K B=\angle C A B$. From the fact that $\angle A B K=\angle K B L$, we conclude that triangles $K L B$ and $A K B$ are similar by two angles. We obtain the equality of ratios $\frac{K B}{B L}=\frac{A B}{K B}$. Therefore, $K B=\sqrt{B L \cdot A B}=\sqrt{18 \cdot 8}=12$.