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[EMHL] Hagge Theorem (0, 1, 2, 3)

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  • José Carlos Chávez Sandoval
    Dear Ricardo, Could you upload any proof of the Hagge s theorems you show us? Thanks José Carlos ... LLama Gratis a cualquier PC del Mundo. Llamadas a fijos y
    Message 1 of 4 , Feb 28, 2006
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      Dear Ricardo, Could you upload any proof of the Hagge's theorems you show us?

      Thanks

      José Carlos


      ---------------------------------

      LLama Gratis a cualquier PC del Mundo.
      Llamadas a fijos y móviles desde 1 céntimo por minuto.
      http://es.voice.yahoo.com

      [Non-text portions of this message have been removed]
    • Ricardo Barroso
      Dear José Carlos, and all: Jonhson proposes them like exercises. Only them I have seen with Cabri. Ricardo ...
      Message 2 of 4 , Feb 28, 2006
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        Dear José Carlos, and all:

        Jonhson proposes them like exercises.

        Only them "I have seen" with Cabri.

        Ricardo


        --- José Carlos Chávez Sandoval <hucht_sc@...>
        escribió:

        > Dear Ricardo, Could you upload any proof of the Hagge's
        > theorems you show us?
        >
        > Thanks
        >
        > José Carlos
        >
        >
        > ---------------------------------
        >
        > LLama Gratis a cualquier PC del Mundo.
        > Llamadas a fijos y móviles desde 1 céntimo por minuto.
        > http://es.voice.yahoo.com
        >
        > [Non-text portions of this message have been removed]
        >
        >




        ______________________________________________
        LLama Gratis a cualquier PC del Mundo.
        Llamadas a fijos y móviles desde 1 céntimo por minuto.
        http://es.voice.yahoo.com
      • Nikolaos Dergiades
        Dear José Carlos, Let ABC be a triangle H its orthocenter H1H2H3 its cevian triangle. Let P be an arbitrary point with cevian triangle A B C . We know that
        Message 3 of 4 , Feb 28, 2006
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          Dear José Carlos,

          Let ABC be a triangle H its orthocenter H1H2H3 its cevian triangle.
          Let P be an arbitrary point with cevian triangle A'B'C'.
          We know that with signed segments we have
          HA.HH1 = HB.HH2 = HC.HH3 = -2cosA.cosB.cosC

          1) Let the perpendicular from H meets the circle with diameter AA'
          at the points A1,A2 and similarly define on the circles (BB'), (CC')
          the points B1, B2, C1, C2.
          Since the points A, H1, A1, A2 are concyclic then
          HA1.HA2 = HA.HH1 similarly
          HB1.HB2 = HB.HH2
          HC1.HC2 = HC.HH3.
          Hence HA1.HA2 = HB1.HB2
          the points A1, A2, B1, B2 are concyclic and the
          center of the circle lies on the lines AA', BB' and
          this is the point P.
          Similarly the points A1, A2, C1, C2 are lying on a
          circle with center P that is same as above.

          2) The circle (BC) passes through H2, H3
          meets the line HA1A2 at the points A3, A4.
          Similarly define the points B3, B4, C3, C4.
          Hence HA3.HA4 = HB.HH2
          Thinking as in (1) we conclude that
          the points A3, A4, B3, B4 are concyclic and the
          center of the circle lies on the line from the mid point
          of BC parallel to AA' and on the line parallel from the
          mid point of CA to BB'. This point Q is the complement of P.
          (perhaps there is an error in Jonson I think.
          The phrase
          whose center, with reference to triangle A1 A2 A3 is
          homologous to the position of P in the triangle A1A2A3
          must be
          whose center, with reference to medial triangle of
          A1A2A3 is homologous to the position of P in the triangle A1A2A3)
          (A1A2A3 in Jonson is the triangle ABC)


          3) The circle (BC) passes through H2, H3
          meets the circle (AA') at A5, A6 and
          since HA.HH1 = HB.HH2 the point H lies
          on the radical axis of the circles (BC), (AA').
          Hence H lies on the line A5A6 and we have
          HA5.HA6 = HA.HH1 = HB5.HB6 = HC5.HC6
          Similarly thinking as in (1) and (2)
          the points are concyclic and the center of the
          circle is the point that lies on the lines
          AoA", BoB", CoC" where AoBoCo is the medial
          triangle of ABC and A" is the intersection of BoCo
          with the line AP. Similarly define B", C".
          This center is the isotomic conjugate of Q wrt
          the medial triangle AoBoCo.

          Best regards
          Nikos Dergiades


          > Dear Ricardo, Could you upload any proof of the Hagge's theorems
          > you show us?
          >
          > Thanks
          >
          > José Carlos
          >
        • Nikolaos Dergiades
          A correction please: I wrote ... the correct is HA.HH1 = HB.HH2 = HC.HH3 = -4R^2cosA.cosB.cosC Best regards Nikos Dergiades
          Message 4 of 4 , Feb 28, 2006
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            A correction please:
            I wrote

            > We know that with signed segments we have
            > HA.HH1 = HB.HH2 = HC.HH3 = -2cosA.cosB.cosC

            the correct is
            HA.HH1 = HB.HH2 = HC.HH3 = -4R^2cosA.cosB.cosC

            Best regards
            Nikos Dergiades
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