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15399Θέμα: Re: [EMHL] Re:Cevian Trace Equal Area Points

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  • Nikolaos Dergiades
    Aug 2, 2007
      Dear Francois,
      I've sent you a message
      and now I saw your message that you sent it
      before mine.
      As you see we use almost the same symbolism.
      You conclude the harmonic relation using vectors
      and I conclude it because my lines La+, La-, AB, AC
      are parallel to the sides and diagonals of a
      parallelogram.
      Now I have an interesting result.
      I investigated pedal triangles of
      isogonal conjugate points.
      If abc is the pedal triangle of P and
      a'b'c' is the pedal triangle of P' isogonal
      conjugate of P then the areal point Q of these
      triangles lies on the circumcircle of ABC.
      This is from a proof.
      The following is from observation:
      The point Q is the fourth intersection
      of circumcircle with the rectangular circumhyperbola
      of the circumcevian triangle of P, that passes
      through P.
      Similarly Q lies on the rectangular circumhyperbola
      of the circumcevian triangle of P', that passes
      through P'.

      Best regards
      Nikos Dergiades

      > Dear Nikos
      > I give you my own construction of the equal area
      > axis based on vectors.
      > So you could compare it with yours.
      > Given 4 points A, B, A', B' such that line AB and
      > A'B' are through point O,
      > let L+ be the line locus of points M such that:
      > Area(MAB) = Area(MA'B')
      > and L- be the line locus of points M such that:
      > Area(MAB) = -Area(MA'B')
      > where Area means signed area.
      > Then L+ is the line through O directed by the
      > vector V(AB) - V(A'B')
      > and L- is the line through O directed by the vector
      > V(AB) + V(A'B').
      > That's why the four lines AB, A'B', L+, L- are in an
      > harmonic range.
      > Friendly
      > Francois




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