A Concurrency from Circumcircles of Subtriangles
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.
See Four Circles—a problem from the Canadian Mathematical Olympiad 2007.