Understand the problem
In an acute angled triangle ABC, let H be the orthocenter, and let D, E, F be the feet of the altitudes from A, B, C to the opposite sides, respectively. Let L, M, N be the midpoints of the 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
Source of the problem
Start with hints
Chase three cyclic quadrilaterals to understand DH is the internal angle bisector of \( \angle EDF \). Since \( \angle BFE = \angle BEC = 90 ^o \) hence BFEC is cyclic hence \(\angle FBE = \angle FCE \) (angle subtended by the segment FE). Since \( \angle ADC = \angle AFC = 90 ^o \) hence AFDC is cyclic hence \(\angle FDA = \angle FCA or \angle FCE \) (angle subtended by the segment FA). Since \( \angle BDA = \angle BEA = 90 ^o \) hence BDEA is cyclic hence \(\angle ADE = \angle ABE \) (angle subtended by the segment AE). Combining we have \( \angle FDA = \angle EDA \) implying DA bisects \( \angle FDE \) Can you show L-M-N are collinear and perpendicular to EF?
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