In triangle $A B C$, it is known that $A B=10, B C=24$, and the median $B D$ is 13. The circles inscribed in triangles $A B D$ and $B D C$ touch the median $B D$ at points $M$ and $N$ respectively. Find $M N$.
Let's first prove the following statement. If a circle is inscribed in a triangle $X Y Z$, and $x$ is the distance from vertex $X$ to the point of tangency of the circle with side $X Y$, and $Y Z=a$, then $x=p-a$, where $p$ is the semiperimeter of the triangle.

Let the points of tangency of the inscribed circle with sides $X Y, Y Z$, and $X Z$ be denoted by $Z_{1}, X_{1}$, and $Y_{1}$, respectively. Let $X Z=b$ and $X Y=c$. Then,

\[
\begin{gathered}
Y X_{1}=Y Z_{1}=X Y-X Z_{1}=c-x, \quad Z X_{1}=Z Y_{1}=X Z-X Y_{1}=X Z-X Z_{1}=b-x, \\
Y Z=Y X_{1}+Z X_{1}=c-x+b-x=b+c-2 x .
\end{gathered}
\]

Therefore,

\[
x=\frac{b+c-a}{2}=\frac{b+c+a}{2}-a=p-a
\]

This completes the proof.

Now, let's return to our problem. The circle inscribed in triangle $A B D$ touches its side $B D$ at point $M$. By the proven statement,

\[
B M=\frac{A B+B D+A D}{2}-A D=\frac{A B+B D-A D}{2}=\frac{23-A D}{2}
\]

Similarly,

\[
B N=\frac{B C+B D-C D}{2}=\frac{37-C D}{2}
\]

Since $A D=C D$, we have

\[
M N=|B N-B M|=\left|\frac{37-C D}{2}-\frac{23-A D}{2}\right|=\frac{37-23}{2}=7
\]

## Answer

7.