In an isosceles triangle with a lateral side equal to $b$, the angle bisectors of the angles at the base are drawn. The segment of the line between the points of intersection of the bisectors with the lateral sides is equal to $t$. Find the base of the triangle.
The bisector of a triangle divides its side into segments proportional to the other two sides.

## Solution

Let $B M$ and $C K$ be the bisectors of triangle $A B C$, $A B = A C = b$, and $K M = m$. Since $\angle K M B = \angle M B C = \angle K B M$, triangle $K B M$ is isosceles. Therefore, $B K = K M = m$. Similarly, $M C = B K = m$. By the property of the bisector of a triangle, $A B : B C = A M : M C$, or $b : B C = (b - m) : m$. From this, we find that $B C = b m / (b - m)$.

![](https://cdn.mathpix.com/cropped/2024_05_06_96d74c2697cd905bd948g-40.jpg?height=457&width=504&top_left_y=1921&top_left_x=777)

Answer

$b m / (b - m)$