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 \(\angle AXM = 90^{\circ} \).

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 \(\boxed{Q.E.D.}\)

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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 \(\boxed{Q.E.D.}\)

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