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Euler line proof
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(Proof)
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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 $H$
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.
Repeating the same argument for the other medians proves that $P$ lies on the three heights and therefore it must be the orthocenter $H$
The ratio is $OG/GH=1/2$ since we constructed it that way.
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"Euler line proof" is owned by drini. [ full author list (2) | owner history (1) ]
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See Also: Euler line
| Keywords: |
Triangle, Euler Line, Orthocenter, Centroid |
This object's parent.
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Cross-references: ratio, lies on, argument, height, similar, triangles, midpoint, median, orthocenter, point, centroid, circumcenter
This is version 7 of Euler line proof, born on 2001-10-06, modified 2008-10-06.
Object id is 156, canonical name is EulerLineProof.
Accessed 12588 times total.
Classification:
| AMS MSC: | 51M99 (Geometry :: Real and complex geometry :: Miscellaneous) |
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Pending Errata and Addenda
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