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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
1 year, 5 months ago

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