Dear Darij

[DG]

> >> > (1b) The Euler lines of triangles AB'C', BC'A', CA'B' concur

> >> > at one point on the nine-point circle of triangle ABC.

> >> >

> >> > Note that (1a) follows from the definition of the Schiffler

> >> > point (H is the incenter of triangle A'B'C')

[JPE]

> >> Note that if an angle of ABC is obtuse, the incenter of A'B'C'

is

> >> not H, but the corresponding vertex of ABC.

[DG]

> Yes, of course I know this, the excenters of a triangle

> also lie on the Neuberg cubic.

I was just meaning that the common point is the Schiffler point of

A'B'C' only when ABC is acutangle

> >> > and (1b) is a result of Victor Thebault.

> >> Notethat the common point is the center of Jerabek hyperbola.

> I have met this before. The point M where the Euler lines

> of triangles AB'C', BC'A', CA'B' meet has the property

> that one of the equations MA' = MB' + MC' or cyclically

> holds. This was stated by Thebault. Can anybody find a

> SYNTHETIC proof?

I'm unable to find a synthetic proof of this fact but it is quite

easy to check that

MA' : MB' : MC' = |(b^2-c^2)SA| : |(c^2-a^2)SB| : |(a^2-b^2)SC| and

(b^2-c^2)SA+(c^2-a^2)SB+(a^2-b^2)SC = 0

Friendly. Jean-Pierre