# Euler line 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 $AA^{\prime}$ where $A^{\prime}$ is the midpoint of $BC$. Triangles $OGA^{\prime}$ and $PGA$ are similar, since $GP=2GO$, $AG=2A^{\prime}G$ and $\angle OGA^{\prime}=\angle PGA$. Then $\angle OA^{\prime}G=\angle PGA$ and $OA^{\prime}\parallel AP$. But $OA^{\prime}\perp BC$ 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.

 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