10.276. A right triangle $ABC$ is divided by the altitude $CD$, drawn to the hypotenuse, into two triangles $BCD$ and $ACD$. The radii of the circles inscribed in triangles $BCD$ and $ACD$ are 4 and $3 \text{ cm}$, respectively. Find the distance between their centers.
## Solution.

Let $O_{2}$ be the center of the circle inscribed in $\triangle B C D$, and $O_{1}$ be the center of the circle inscribed in $\triangle A C D$ (Fig. 10.71), $N$ and $M$ be the points of tangency of these circles with the line $A B$. Then $O_{2} N \perp A B, O_{1} M \perp A B, O_{2} N=r_{2}=4$ cm, $O_{1} N=r_{1}=3$ cm. Drop the perpendicular $O_{1} K$ to $O_{2} N$. Then $O_{2} K=r_{2}-r_{1}=1$ cm, $O_{1} K=r_{2}+r_{1}=7$ cm. From $\Delta O_{1} K O_{2}\left(\angle O_{1} K O_{2}=90^{\circ}\right):$ $O_{1} O_{2}=\sqrt{O_{1} K^{2}+O_{2} K^{2}}=\sqrt{50}=5 \sqrt{2}$ cm.

Answer: $5 \sqrt{2}$ cm.