In triangle $A B C$ with sides $A B=3, B C=4$ and $A C=5$, the bisector $B D$ is drawn. Circles are inscribed in triangles $A B D$ and $B C D$, touching $B D$ at points $M$ and $N$ respectively. Find $M N$.

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By the property of the bisector of triangle $A D=3 / 7 A C=15 / 7, D C=4 / 7 A C=20 / 7$. Therefore, $M N=|M D-N D|=|1 / 2(B D+A D-A B)-1 / 2(B D+C D-B C)|=|1 / 2(B D+15 / 7-3)-1 / 2(B D+20 / 7-4)|=1 / 7$.

## Answer

$1 / 7$