Loading ...
Sorry, an error occurred while loading the content.

10972circumcevian reflections

Expand Messages
  • Floor en Lyanne van Lamoen
    Jan 5, 2005
      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,
    • Show all 8 messages in this topic