$C D$ is the median of triangle $A B C$. The circles inscribed in triangles $A C D$ and $B C D$ touch the segment $C D$ at points $M$ and $N$. Find $M N$, if $A C - B C = 2$.
If the circle inscribed in triangle $P Q R$ touches side $P Q$ at point $S$, then $P S=1 / 2(P Q+P R-R Q)$.

## Solution

Since $A D=D B$, and

$$
C M=1 / 2(A C+C D-A D) \text{ and } C N=1 / 2(B C+C D-B D)
$$

then

$$
\begin{aligned}
M N=|C M-C N| & =|1 / 2(A C+C D-A D)-1 / 2(B C+C D-B D)|= \\
& =1 / 2|A C-B C|=1 / 2 \cdot 2=1
\end{aligned}
$$

## Answer

1.