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10981Re: circumcevian reflections

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  • jpehrmfr
    Jan 6, 2005
      Dear Floor
      > First of all I wish you a very happy and fruitful 2005.

      Happy New Year to you and to each Hyacinthist.

      > To start the new year let me offer a little theorem:
      > If A'B'C' is a circumcevian triangle, and A"B"C" are the
      reflections of
      > A'B'C' through the sides of ABC, then the circles (ABC), (A"B"C),
      (A"BC")
      > and (AB"C") are concurrent in one point.

      Note that if A',B',C' lie on the circumcircle, the circles A"B"C,
      B"C"A, C"A"B have a common point.
      I suspect that this common point lies on the circumcircle if and
      only if the lines AA', BB', CC' concur.
      Any idea?
      Friendly. Jean-Pierre
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