<s>[INST] Inscribed circle in a right-angled triangle 
In right-angled $\triangle ABC$ with catheti $a = 11\,\text{cm}, b=7\,\text{cm}$ a circle has been inscribed. Find the radius and altitude from $C$ to the hypotenuse.

I found that the hypotenuse is $c = \sqrt{170}$ and the radius is $r = \frac{a+b-c}{2}=\frac {18 - \sqrt{170}}{2}$. I think that the altitude $CH=$ the sum of radii of circles inscribed in $\triangle ABC, \triangle AHC, \triangle HBC$ but I don't understand how I should calculate them. Thank you in advance!
 [/INST] I assume you mean that $CH$ is the altitude from $C$ to the hypotenuse.
Since $\triangle ABC$ is a right triangle, and since 
$\triangle AHC$ and $\triangle HBC$ each share one angle with $\triangle ABC$,
the three triangles are similar. We have
$$
\frac{CH}{AC} = \frac{BC}{AB}
$$
and (equivalently)
$$
\frac{CH}{BC} = \frac{AC}{AB}.
$$
The radii of the three incircles are proportional to $AB,$ $AC,$ and $BC,$
so their sum is
\begin{align}
\frac{a+b-c}{2} + \frac{a(a+b-c)}{2c} + \frac{b(a+b-c)}{2c}
&= \frac{(a+b+c)(a+b-c)}{2c}
\\
&= \frac{(a+b)^2 - c^2}{2c} \\
&= \frac{ab}{c},
\end{align}
which is indeed equal to the altitude.
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