Dear Jean-Pierre Ehrmann,

Thanks for the mail. You wrote:

>> > (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')

>>

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

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

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

also lie on the Neuberg cubic.

>> > 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?

Thanks,

Sincerely,

Darij Grinberg