[ The ratio in which the bisector divides the side.]

In triangle $A B C$, it is known that $A B=c, B C=a, A C=b$. In what ratio does the center of the inscribed circle of the triangle divide the bisector $C D$?

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By the property of the bisector of a triangle $\frac{A D}{B D}=\frac{A C}{B C}=\frac{\bar{b}}{a}$, and since $A B=c$, then

$$
A D=b \cdot \frac{c}{a+b}=\frac{b c}{a+b}
$$

Let $O$ be the center of the inscribed circle of triangle $A B C$. Then $O$ is the point of intersection of the bisectors of the triangle, so $A O$ is the bisector of triangle $A C D$. Therefore,

$$
\frac{C O}{O D}=\frac{A C}{A D}=\frac{\frac{b}{b c}}{a+b}=\frac{a+b}{c}
$$

Answer

$\frac{a+b}{c}$.