Based on the base $AB$ of isosceles triangle $ABC$, a point $D$ is taken such that $BD - AD = 4$. Find the distance between the points where the incircles of triangles $ACD$ and $BCD$ touch the segment $CD$.
If the circle inscribed in triangle $P Q R$ touches side $P Q$ at point $S$, then $P S=\frac{P Q+P R-R Q}{2}$.

## Solution

Let the circles inscribed in triangles $A C D$ and $B C D$ touch segment $C D$ at points $M$ and $N$ respectively. Since $A C=B C$, and

$$
C M=\frac{A C+C D-A D}{2}, C N=\frac{B C+C D-B D}{2},
$$

then

$$
\begin{aligned}
M N=|C M-C N|=\mid & \left.\frac{A C+C D-A D}{2}-\frac{B C+C D-B D}{2} \right\rvert\,= \\
& =\frac{|B D-A D|}{2}=\frac{4}{2}=2
\end{aligned}
$$

## Answer

2.