10.252. The sides of a triangle are in the ratio $5: 4: 3$. Find the ratio of the segments of the sides into which they are divided by the point of tangency of the inscribed circle.
## Solution.

Since in $\triangle A B C$ (Fig. 10.58) $A B: A C: B C=5: 4: 3$, then $\angle C=90^{\circ}$. Let $B C=3 x$. Then $A C=4 x, A B=5 x$. The radius of the inscribed circle in a right triangle $r=\frac{B C+A C-A B}{2}=x$. $O$ is the center of the inscribed circle, $M, N, K$ are the points of tangency with the sides $B C, A B, A C$ of the triangle, respectively. Then $C K=C M=r=x, A N=A K=3 x$, $B N=B M=2 x$. Therefore, $C K: A K=1: 3 ; C M: B M=1: 2$; $B N: A N=2: 3$.

Answer: $1: 3 ; 1: 2 ; 2: 3$.