A Concurrency from Circumcircles of Subtriangles
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Let ABC be a triangle and let the incircle intersect BC, CA, and AB at A', B', and C', respectively. Let the circumcircles of AB'C', A'BC', and A'B'C intersect the circumcircle of ABC (apart from A, B, and C) at A'', B'', and C'', respectively. Then A'A'', B'B'', and C'C'' are concurrent.
Contributed by: Jay Warendorff (March 2011)
Open content licensed under CC BY-NC-SA
See Four Circles—a problem from the Canadian Mathematical Olympiad 2007.
"A Concurrency from Circumcircles of Subtriangles"
Wolfram Demonstrations Project
Published: March 7 2011