Let \(D,E,F\) be the midpoints of the sides \(BC\), \(CA\) and \(AB\) respectively. It is easy to prove that \(O\), the circumcenter of \(\Delta ABC\), is the orthocenter of \(\Delta DEF\). And now we can achieve triangle \(\Delta DEF\) from \(\Delta ABC\) by making a transformation about centroid, i.e. squeezing the length \(AG\) to half and rotating \(180^0\) gives \(DG\). Similarly we do this for \(BG\) and \(CG\) to get \(EG\) and \(FG\). So, what happens the.... length of the line segment joining orthocenter and a vertex of triangle \(\Delta ABC\) is twice the line joining orthocenter and a vertex of triangle \(\Delta DEF\). But first corresponds to \(AH\) and second corresponds to \(OD\). Therefore, \(AH=2OD\).

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TopNewestLet \(D,E,F\) be the midpoints of the sides \(BC\), \(CA\) and \(AB\) respectively. It is easy to prove that \(O\), the circumcenter of \(\Delta ABC\), is the orthocenter of \(\Delta DEF\). And now we can achieve triangle \(\Delta DEF\) from \(\Delta ABC\) by making a transformation about centroid, i.e. squeezing the length \(AG\) to half and rotating \(180^0\) gives \(DG\). Similarly we do this for \(BG\) and \(CG\) to get \(EG\) and \(FG\). So, what happens the.... length of the line segment joining orthocenter and a vertex of triangle \(\Delta ABC\) is twice the line joining orthocenter and a vertex of triangle \(\Delta DEF\). But first corresponds to \(AH\) and second corresponds to \(OD\). Therefore, \(AH=2OD\).

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Thanx a lot!

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