An inscribed circle is drawn in an isosceles triangle. The points of tangency divide each lateral side into segments of length $m$ and $n$, counting 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.

#
The ratio of the perimeters of similar triangles is equal to the similarity coefficient.

## Solution

Let $P, Q$ and $R$ be the points of tangency of the inscribed circle with the sides $B C, A B$ and $A C$ of triangle $A B C \ (A B = B C)$. Then $B P = B Q = m$,

$A Q = A R = C P = C R = n, A C = 2 n$, and the perimeter of triangle $A B C$ is $2(m+n) + 2n = 2(m+2n)$.

Let a line parallel to side $B C$ touch the circle at point $M$ and intersect sides $A B$ and $A C$ at points $K$ and $L$ respectively. Then $K M = K Q, L M = L R$. Therefore, the perimeter of triangle $A K L$ is $A K + K L + A L = A K + K M + M L + A L = A K + K Q + R L + A L = A Q + A R = 2 n$.

The similarity coefficient of triangles $A K L$ and $A B C$ is the ratio of their perimeters, that is, $n / (m + 2n)$.

Therefore, $K L = n \cdot B C / (m + 2n) = n(m + n) / (m + 2n)$.

Similarly, we find the other required segments.

![](https://cdn.mathpix.com/cropped/2024_05_06_0e468145bb05e34a17aag-46.jpg?height=361&width=872&top_left_y=1769&top_left_x=593)

## Answer

$2 m n / (m + 2n), n(m + n) / (m + 2n), n(m + n) / (m + 2n).$