In a right $\Delta ABC$ ($\angle C = 90^{\circ} $), $CD$ is the height. Let $r_1$ and $r_2$ be the radii of inscribed circles of $\Delta ACD$ and $\Delta DCB$. Find the radius of inscribed circle of $\Delta ABC$
1. Given a right triangle $\Delta ABC$ with $\angle C = 90^\circ$, let $CD$ be the height from $C$ to the hypotenuse $AB$. Let $r_1$ and $r_2$ be the radii of the inscribed circles of $\Delta ACD$ and $\Delta DCB$, respectively.

2. The key observation is that in a right triangle, the hypotenuse is the diameter of the circumcircle. However, for this problem, we are interested in the inscribed circles of the smaller triangles $\Delta ACD$ and $\Delta DCB$.

3. By the properties of the inscribed circle in a right triangle, the radius $r$ of the inscribed circle of a right triangle with legs $a$ and $b$, and hypotenuse $c$ is given by:
   \[
   r = \frac{a + b - c}{2}
   \]

4. In $\Delta ACD$, let $AC = 2r_1$ and in $\Delta DCB$, let $BC = 2r_2$. By the Pythagorean theorem, the hypotenuse $AB$ of $\Delta ABC$ is:
   \[
   AB = \sqrt{(2r_1)^2 + (2r_2)^2} = 2\sqrt{r_1^2 + r_2^2}
   \]

5. Now, we apply the formula for the radius of the inscribed circle of $\Delta ABC$. The legs of $\Delta ABC$ are $AC = 2r_1$ and $BC = 2r_2$, and the hypotenuse is $AB = 2\sqrt{r_1^2 + r_2^2}$.

6. Therefore, the radius $r$ of the inscribed circle of $\Delta ABC$ is:
   \[
   r = \frac{AC + BC - AB}{2} = \frac{2r_1 + 2r_2 - 2\sqrt{r_1^2 + r_2^2}}{2} = r_1 + r_2 - \sqrt{r_1^2 + r_2^2}
   \]

Conclusion:
\[
\boxed{r_1 + r_2 - \sqrt{r_1^2 + r_2^2}}
\]