2013 USAMO Problems/Problem 1
In triangle , points lie on sides respectively. Let , , denote the circumcircles of triangles , , , respectively. Given the fact that segment intersects , , again at respectively, prove that
In this solution, all lengths and angles are directed.
Firstly, it is easy to see by that concur at a point . Let meet again at and , respectively. Then by Power of a Point, we have Thusly But we claim that . Indeed, and Therefore, . Analogously we find that and we are done.