$\begin{array}{ll}{\left[\begin{array}{l}\text { Inscribed and circumscribed circles } \\ \text { [Two tangents drawn from one point] }\end{array}\right]}\end{array}$

In an isosceles triangle $A B C$ on the base $A C$, a point $M$ is taken such that $A M=a, M C=b$. Circles are inscribed in triangles $A B M$ and $C B M$. Find the distance between the points of tangency of these circles with the segment $B M$.
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 $P$ and $Q$ be the points of tangency of the circles inscribed in triangles $A B M$ and $C B M$ with side $B M$. Then the desired distance is $|B P - B Q|$.

Let $p_{1}$ and $p_{2}$ be the semiperimeters of these triangles. Then $B P = p_{1} - a, B Q = p_{2} - b$. Therefore,

$$
\begin{aligned}
& |B P - B Q| = \left| p_{1} - a - p_{2} + b \right| = \\
= & \left| b - a - \frac{b - a}{2} \right| = \frac{|a - b|}{2}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_cb0f0d605674d837f9abg-16.jpg?height=369&width=423&top_left_y=407&top_left_x=818)

Answer

$\frac{|a - b|}{2}$. Send a comment

Problem