Euler line proof

Let $O$ the circumcenter of $\mathrm{△}ABC$ 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 $A{A}^{\prime }$ where $A^{\prime }$ is the midpoint of $BC$. Triangles $OG{A}^{\prime }$ and $PGA$ are similar, since $GP=2GO$, $AG=2A^{\prime }G$ and $\mathrm{\angle }OG{A}^{\prime }=\mathrm{\angle }PGA$. Then $\mathrm{\angle }O{A}^{\prime }G=\mathrm{\angle }PGA$ and $O{A}^{\prime }\parallel AP$. But $O{A}^{\prime }⟂BC$ so $AP⟂BC$, 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.

Title	Euler line proof
Canonical name	EulerLineProof
Date of creation	2013-03-22 11:44:29
Last modified on	2013-03-22 11:44:29
Owner	drini (3)
Last modified by	drini (3)
Numerical id	15
Author	drini (3)
Entry type	Proof
Classification	msc 51M99
Classification	msc 55U10
Classification	msc 18E30
Classification	msc 18-00
Classification	msc 55U35
Classification	msc 46-01
Classification	msc 47B25
Classification	msc 81-01
Related topic	EulerLine