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7411Re: Thebault point

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  • jpehrmfr
    Aug 2 1:14 AM
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      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
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