10.201. Given a triangle $A B C$ such that $A B=15 \text{~cm}, B C=12$ cm and $A C=18 \text{~cm}$. Calculate in what ratio the incenter of the triangle divides the angle bisector of angle $C$.
## Solution.

Let $C K$ be the bisector of $\angle C, O$ be the center of the circle inscribed in $\triangle A B C$, the intersection point of its bisectors (Fig. 10.11). Let $B K=x$ cm, $x>0$. Then $A K=(15-x)$ cm. By the property of the bisector $\frac{B K}{B C}=\frac{A K}{A C}$; $\frac{x}{12}=\frac{15-x}{18} ; x=6$. $B O$ is the bisector of $\triangle B C K$. Then $\frac{C O}{O K}=\frac{B C}{B K}=\frac{12}{6}=2: 1$.

Answer: $2: 1$.