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Re: [EMHL] Perp. Bisectors + locus

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  • Antreas P. Hatzipolakis
    Let ABC be a triangle. The perp. bisector of BC intersects BA, CA at Ba,Ca, resp. The perp. bisector of CA intersects CB, AB at Cb,Ab, resp. The perp. bisector
    Message 1 of 11 , Jan 15 1:02 PM
      Let ABC be a triangle.

      The perp. bisector of BC intersects BA, CA at Ba,Ca, resp.

      The perp. bisector of CA intersects CB, AB at Cb,Ab, resp.

      The perp. bisector of AB intersects AC, BC at Ac,Bc, resp.

      Denote:

      A# = AbAc /\ BcCb [=BC]

      B# = BcBa /\ CaAc [=CA]

      C# = CaCb /\ AbBa [=AB]

      The triangles ABC, A#B#C# are orthologic.
      (Orthologic centers: O, ??)

      Generalization (Locus):

      Let ABC be a triangle, P a point and A'B'C'
      the pedal triangle of P.

      The line PA' intersects BA, CA at Ba,Ca, resp.

      The line PB' intersects CB, AB at Cb,Ab, resp.

      The line PC' intersects AC, BC at Ac,Bc, resp.

      Denote:

      A# = AbAc /\ BcCb [=BC]

      B# = BcBa /\ CaAc [=CA]

      C# = CaCb /\ AbBa [=AB]


      The triangles ABC, A#B#C# are orthologic.
      (Orthologic centers: P, ??)


      Greetings from Athens

      Antreas

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