# 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
2 years, 7 months ago

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## Comments

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- 1 year, 1 month ago

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By the lemma, we know that in any $$\Delta$$ the reflection of the orthocenter in the midpoint of a side lies on the circumcircle and is the end of the diameter from the opposite vertex. So $$\boxed{Q.E.D.}$$

- 11 months, 3 weeks ago

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