Euler line proofEulerLineProof2001-10-06 18:38:152008-10-06 17:17:04ProofEuler line0TriangleEuler LineOrthocenterCentroid\usepackage{amssymb}
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Let $O$ the circumcenter of $\triangle ABC$ and $G$ its centroid. Extend $OG$ until a point $P$ such that $OG/GP=1/2$. We'll prove that $P$ is the orthocenter $H$.
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Draw the median $AA'$ where $A'$ is the midpoint of $BC$. Triangles $OGA'$ and $PGA$ are similar, since $GP=2GO$, $AG=2A'G$ and $\angle OGA'=\angle PGA$. Then $\angle OA'G =\angle PGA$ and $OA'\parallel AP$. But $OA'\perp BC$ so $AP\perp BC$, that is, $AP$ is a height of the triangle.\smallskip
Repeating the same argument for the other medians proves that $P$ lies on the three heights and therefore it must be the orthocenter $H$.\smallskip
The ratio is $OG/GH=1/2$ since we constructed it that way.\medskip