Problem: Let ABC be an acute triangle with orthocenter H, and let M and N denote the midpoints of AB and AC. Rays MH and NH intersect the circumcircle of ABC again at points X and Y. Prove that the four points M,N,X,Y lie on a circle.
Solution
Solution: Let Ω be the circumcircle of △ABC. Let P and Q be the reflections of H across M and N. Since ∠APB=∠AHB=180∘−∠C, point P lies on Ω; thus so does Q. Moreover, MN∥PQ, so ∠XMN=∠XPQ=∠XYQ and we are done.
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