Hello Vladimir,

> I've been asked the following question:

>

> do the three cevians each of which divides a given triangle into

two

> triangles with equal inradii concur?

>

> A quick GSP experiment shows that they are very close to

concurrence, but

> there's a lot of room for doubt.

>

> Does anybody know the answer?

I have done a similar experiment and find that although close, I do

not think they are perspective.

Call A' the meet of the A cevian with BC. Then

A' = {0, c (b + Sqrt[s^2 - a s]), b (c + Sqrt[s^2 - a s])}

B' = {c (a + Sqrt[s^2 - b s]), 0, a (c + Sqrt[s^2 - b s])}

C' = {b (a + Sqrt[s^2 - c s]), a (b + Sqrt[s^2 - c s]), 0}

from which A'B'C' is not perspective.

The line I of BAA' to I of BCC'

The line I of CBB' to I of CAA'

The line I of ACC' to I of ABB'

are of course concurrent in I of ABC.

The radii of the incircles are rA' = r (s - Sqrt[s^2 - a s]) / a

Best regards,

Peter.