[ [tangents drawn from one point] [ Midline of a triangle $\quad$]

A circle of radius 1 is inscribed in triangle $ABC$, where $\cos \angle B=0.8$. This circle touches the midline of triangle $ABC$, parallel to side $AC$. Find the side $AC$.
The distance from the vertex of a triangle to the nearest point of tangency with the inscribed circle is equal to the difference between the semiperimeter and the opposite side.

## Solution

Let $O$ be the center of the inscribed circle, and $K$ be its point of tangency with side $AB$. Then

$$
\operatorname{tg} \frac{1}{2} \angle B = \frac{\sin \angle B}{1 + \cos \angle B} = \frac{0.6}{1 + 0.8} = \frac{1}{3}
$$

Therefore,

$$
B K = \frac{O K}{\operatorname{tg} \frac{1}{2} \angle B} = 3
$$

On the other hand, the segment $BK$ is equal to the semiperimeter of the cut-off triangle, which is similar to the given triangle with a similarity coefficient of $\frac{1}{2}$. Therefore, the semiperimeter of the given triangle is 6. Consequently, $AC = 6 - BK = 3$.

![](https://cdn.mathpix.com/cropped/2024_05_06_c317f3f1f50d1d30304bg-44.jpg?height=432&width=717&top_left_y=2151&top_left_x=676)

Answer

3.