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In triangle $A B C$, it is known that $A B=c, B C=a, A C=b$; $O$ is the center of the circle that touches side $A B$ and the extensions of sides $A C$ and $B C$, $D$ is the point of intersection of ray $C O$ with side $A B$. Find the ratio $\frac{C O}{O D}$

#
Note that at point $O$, the bisectors of the external angles of triangle $A B C$ at vertices $A$ and $B$ and the bisector of the internal angle at vertex $C$ intersect. By the property of the angle bisector of a triangle, $\frac{A D}{B D}=\frac{A C}{B C}=\frac{\hbar}{a}$, and since $A B=c$, then

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

The ray $A O$ is the bisector of the external angle at vertex $A$ of triangle $A B C$, therefore

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

## Answer

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