10981Re: circumcevian reflections
- Jan 6, 2005Dear 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:reflections of
> If A'B'C' is a circumcevian triangle, and A"B"C" are the
> 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.
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