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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, 9 months ago

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· 4 months, 1 week ago

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 $$\boxed{Q.E.D.}$$ · 1 month, 3 weeks ago