10.403 In a right triangle $A B C$ ( $\angle C=90^{\circ}$ ), the altitude $CD$ is drawn. The radii of the circles inscribed in triangles $ACD$ and $BCD$ are 0.6 and 0.8 cm, respectively. Find the radius of the circle inscribed in triangle $\boldsymbol{A} \boldsymbol{\text { B }}$.
Solution. Let $x, r_{1}, r_{2}$ be the radii of the circles inscribed in $\triangle A B C$, $\triangle A C D, \triangle B C D$ respectively. $\triangle A C D \sim \triangle B C D \Rightarrow \frac{A C}{B C}=\frac{r_{1}}{r_{2}}=\frac{0.6}{0.8}$ (the radii of the circles inscribed in similar triangles are proportional to the corresponding sides) $\Rightarrow B C=\frac{4}{3} A C, \quad A B=\sqrt{A C^{2}+B C^{2}}=\frac{5}{3} A C$.

$$
\triangle A C D \sim \triangle A B C \Rightarrow \frac{r_{1}}{r}=\frac{A C}{A B} \Rightarrow r=1
$$

Answer: 1 cm.