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Found problems: 1

2025 Sharygin Geometry Olympiad, 20
Let $H$ be the orthocenter of a triangle $ABC$, and $M$, $N$ be the midpoints of segments $BC$, $AH$ respectively. The perpendicular from $N$ to $MH$ meets $BC$ at point $A^{\prime}$. Points $B^{\prime}$ and $C^{\prime}$ are defined similarly. Prove that $A^{\prime}$, $B^{\prime}$, $C^{\prime}$ are collinear. Proposed by: F.Ivlev
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