10.197. A circle is inscribed in a right-angled triangle with sides 6, 8, and $10 \mathrm{~cm}$. Lines are drawn through the center of the circle parallel to the sides of the triangle. Calculate the lengths of the segments of the sides of the triangle cut off by the constructed lines.
Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0714.jpg?height=446&width=588&top_left_y=84&top_left_x=671)

Fig. 10.7

Let's find the radius of the inscribed circle. We have $r=\frac{S}{p}=\frac{0.5 \cdot 6 \cdot 8}{0.5(6+8+10)}=\frac{24}{12}=2$ cm. Next, since $\triangle A B C \sim \triangle M P O$ (Fig. 10.7), we have $\frac{A B}{M P}=\frac{A C}{M O}$, from which $M P=\frac{A B \cdot M O}{A C}=\frac{6 \cdot 2}{8}=\frac{3}{2}$ cm. Similarly, $\triangle E O Q \sim \triangle M P O: \frac{E Q}{M O}=\frac{O E}{P M}$, from which $E Q=\frac{O E \cdot M O}{P M}=\frac{2 \cdot 2}{3 / 2}=\frac{8}{3}$ cm. Draw $O L \perp B C$; then $D N=D L+L N$. $\triangle E O Q=\triangle M P O$ (by a leg and an acute angle); hence, $L N=E Q=\frac{8}{3}$ cm. Similarly, from the equality of triangles $M P O$ and $D L O$, we find $D L=M P=\frac{3}{2}$ cm, and therefore, $D N=\frac{3}{2}+\frac{8}{3}=\frac{25}{6}$ cm.

Answer: $\frac{3}{2}$, $\frac{8}{3}$, and $\frac{25}{6}$ cm.