Question

Let ∆ABC is divided by the altitude CC' - it is easy to see that
∆ABC ~ ∆ACC' ~ ∆CBC' and Area(∆ABC) = Area(∆ACC' ) + Area(∆CBC' )
Let the inradiuses of the above triangles are r, r_{a}, r_{b} respectively.
The similarity of these triangles implies
Area(Incircle(∆ABC)) : Area(Incircle(∆ACC' )) : Area(Incircle(∆CBC' )) =
= Area(∆ABC) : Area(∆ACC' ) : Area(∆CBC' ), so
Area(Incircle(∆ABC)) = Area(Incircle(∆ACC' )) + Area(Incircle(∆CBC' )), or
πr² = πr²_{a} + πr²_{b}
Connect C' with centers of both incircles I_{a} and I_{b}; ∆C'I_{a}I_{b} is right with sides
|C' I_{a}| = r_{a}√2, |C' I_{b