- Jan 5, 2005Dear 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.
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