In triangle $A B C$, the bisector $C D$ of the right angle $A C B$ is drawn; $D M$ and $D N$ are the altitudes of triangles $A D C$ and $B D C$, respectively.

Find $A C$, given that $A M=4, B N=9$.

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Note that $M C=M D=D N$. From the similarity of triangles $A M D$ and $D N B$, it follows that $A M: M D=D N: N B$, that is, $M C^{2}=A M \cdot N B=36$. Therefore,

$A C=A M+M C=4+6=10$.

## Answer

10.