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In an acute angled triangle ABC, let H be the orthocenter, and let D, E, F be the feet of altitudes from A, B, C to the opposite sides, respectively. Let L, M, N be midpoints of segments AH, EF, BC, respectively. Let X, Y be feet of altitudes from L, N on to the line DF. Prove that XM is perpendicular to MY.

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A circle having diameter BC passes through E and F. Also AEHF is a cycle quadrilateral having AH as diameter. So, EF is the common chord of two above mentioned circles which will be perpndicularily bisected by line joining their centres (which is LN). 

So L, M and N are collinear. 

Now, we claim that △LFN and △XMY are similar, because ∠MXY = ∠FLN (XLMF is cyclic quadrilateral) and ∠XYM = ∠LNF (MFNY is cyclic quadrilateral) 

Hence third angle of the two triangles will also be the same ∠XMY = ∠LFN = 90° (LN is the diameter of nine point circle passing through F) 

XM ⊥ MY

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