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15397Re: [EMHL] Re:Cevian Trace Equal Area Points

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  • Francois Rideau
    Aug 2, 2007
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      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


      On 8/2/07, Francois Rideau <francois.rideau@...> wrote:
      >
      > Dear Nikos and Tuan
      >
      >
      >
      > > > Two segments XY, X'Y' are on given two lines. To
      > > > find the locus of Z such that Area(ZXY) =
      > > > Area(ZX'Y').
      > > > If areas are not signed then the locus is two
      > > > lines passing intersection of two given lines.
      > >
      > > I note this fact:
      > >
      > > These 2 lines of this locus are harmonic conjugate wrt lines XY and
      > > X'Y'.
      >
      >
      > So if abc and a'b'c' are 2 inscribed triangles in ABC and if we use
      > ordinary area ( i.e not signed), the Tuan-Nikos six lines cut themselves
      > in 4 points forming an harmonic quad wrt ABC, each of these four points
      > beeing an equal area center wrt abc and a'b'c'. I will have to think about
      > it in choo-choo theory!
      > friendly
      > Francois
      >
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