9. A circle of radius 1 is inscribed in triangle $A B C$, where $\cos \widehat{A B C}=0.8$. This circle touches the midline of triangle $A B C$, parallel to side $A C$. Find the length of side $A C$.
9. It follows from the condition that the height to side $AC$ is equal to two diameters of the inscribed circle, i.e., 4. If $M, N$ and $K$ are the points of tangency with $AB, BC$ and $CA$, then

$$
|BM|=|BN|=r \operatorname{ctg} \frac{\widehat{ABC}}{2}=1 \cdot \sqrt{\frac{1+0.8}{1-0.8}}=\sqrt{9}=3
$$

If $|MA|=|AK|=x,|KC|=|NC|=y$, then, expressing the area through the semiperimeter and the radius of the inscribed circle and through the base and height, we get

$$
(3+x+y)=(x+y) 2, \quad x+y=3
$$

Answer: $|AC|=3$.