In triangle $A B C$, the angle bisectors $A M$ and $C N$ are drawn. It is known that $A C=6, A N=2, C M=3$. Find $M N$.

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Let $B N=x, B M=y$. By the property of the angle bisector of a triangle $B N: A N=C B: C A$ and $B M: C M=A B$ :

$$
A C \text {, or } x / 2=3+y / 6 \text { and } y / 3=x+2 / 6 \text {, from which }
$$

$x=8 / 5$ and $y=9 / 5$. Therefore, $A B=2+8 / 5=18 / 5, B C=3+9 / 5=24 / 5, A B: B C: A C=3: 4: 5$, hence, $\angle A B C$ $=90^{\circ}$. By the Pythagorean theorem

$$
M N^{2}=B M^{2}+B N^{2}=145 / 25
$$

## Answer