7411Re: Thebault point
- Aug 2 1:14 AMDear Darij
> >> > (1b) The Euler lines of triangles AB'C', BC'A', CA'B' concur[JPE]
> >> > 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.[DG]
> Yes, of course I know this, the excenters of a triangleI was just meaning that the common point is the Schiffler point of
> also lie on the Neuberg cubic.
A'B'C' only when ABC is acutangle
> >> > and (1b) is a result of Victor Thebault.I'm unable to find a synthetic proof of this fact but it is quite
> >> 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?
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
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