10.364 In an isosceles triangle, a circle is inscribed. The points of tangency divide each lateral side into segments of length $\boldsymbol{m}$ and $\boldsymbol{n}$, measured from the vertex. Three tangents are drawn to the circle, parallel to each of the triangle's sides. Find the lengths of the segments of the tangents, enclosed between the sides of the triangle.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-226.jpg?height=308&width=458&top_left_y=1187&top_left_x=85)

Let the tangents $M N, K L$ to the circle inscribed in the given $\triangle A B C$ with base $A C$ pass through points $E$ and $F$ of the circle; $P, Q, R$ are the points of tangency of the circle with sides $A B, A C, B C$; $M N \| A C, K L \| B C$.

By Theorem 3, $M E = M P, N E = N R \Rightarrow$ the perimeter $P_{1}$ of $\triangle B M N$ is equal to $B P + B R = 2 m$.

Similarly, $K F = K P, L F = L Q \Rightarrow$ the perimeter $P_{2}$ of $\triangle A K L$ is equal to $A P + A Q = 2 n$. The perimeter $P_{3}$ of $\triangle A B C$ is equal to $2 \cdot A B + 2 \cdot A Q = 2 m + 4 n$.

The perimeters of similar triangles are proportional to the corresponding sides $\Rightarrow \frac{P_{1}}{P_{3}} = \frac{M N}{A C} (\triangle B M N \sim \triangle A B C)$,
$\frac{P_{2}}{P_{3}} = \frac{K L}{B C} (\triangle A K L \sim \Delta A B C) \Rightarrow M N = \frac{2 m n}{m + 2 n}, K L = \frac{n(m + n)}{m + 2 n}$.

Answer: $\frac{2 m n}{m + 2 n} ; \frac{n(m + n)}{m + 2 n}$.