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[parent] Euler line proof (Proof)

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 $ PAO$ are similar, since $ GP=2OG$, $ AG=2GA'$ and $ \angle OGA'=\angle PGA$. Then $ \angle OA'G =\angle GAP$ 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 $ H$ lies on the three heights and therefore it must be the orthocenter.

The ratio is $ OG/GH=1/2$ since we constructed it that way.




"Euler line proof" is owned by drini. [ owner history (1) ]
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See Also: Euler line

Keywords:  Triangle, Euler Line, Orthocenter, Centroid

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Cross-references: ratio, lies on, argument, height, similar, triangles, midpoint, median, orthocenter, point, centroid, circumcenter

This is version 5 of Euler line proof, born on 2001-10-06, modified 2006-06-15.
Object id is 156, canonical name is EulerLineProof.
Accessed 8654 times total.

Classification:
AMS MSC: 51M99 (Geometry :: Real and complex geometry :: Miscellaneous)
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