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10972circumcevian reflections

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  • Floor en Lyanne van Lamoen
    Jan 5, 2005
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      Dear friends,

      First of all I wish you a very happy and fruitful 2005.

      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.

      Kind regards,
      Floor.
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