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Geometry Proof 12

In triangle \(ABC\), denote \(M\) as the midpoint of \(BC\) , \(H\) as the orthocenter, and \(\omega\) as the cirumcircle. Ray \(\vec{MH}\) intersects \(\omega\) at \(X\). Prove that \(\angle AXM = 90^{\circ} \).

Note by Alan Yan
11 months, 4 weeks ago

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