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Prove that <HMA=<GNS

Source: Serbia NMO 2010 problem 2

March 11, 2011
geometrycircumcircleratiogeometric transformationreflectiongeometry unsolved

Problem Statement

In an acute-angled triangle ABCABC, MM is the midpoint of side BCBC, and D,ED, E and FF the feet of the altitudes from A,BA, B and CC, respectively. Let HH be the orthocenter of ΔABC\Delta ABC, SS the midpoint of AHAH, and GG the intersection of FEFE and AHAH. If NN is the intersection of the median AMAM and the circumcircle of ΔBCH\Delta BCH, prove that ∠HMA=∠GNS\angle HMA = \angle GNS.
Proposed by Marko Djikic
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