# Position of Centroid on Euler Line

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## Theorem

Let $\triangle ABC$ be a triangle which is not equilateral.

Let $O$ be the circumcenter of $\triangle ABC$.

Let $G$ be the centroid of $\triangle ABC$.

Let $H$ be the orthocenter of $\triangle ABC$.

Then $G$ lies on the straight line connecting $O$ and $H$ such that:

$OG : GH = 1 : 2$

The line $OGH$ is the Euler line of $\triangle ABC$.

## Proof

First it is necessary to dispose of the case where $\triangle ABC$ is equilateral.

From Orthocenter, Centroid and Circumcenter Coincide iff Triangle is Equilateral, in that case $O$, $G$ and $H$ are the same point.

For all other triangles, $O$, $G$ and $H$ are distinct.

Let $A'$ be the midpoint of $BC$.

Let $B'$ be the midpoint of $AC$.

By definition of centroid, $G$ is the point at which $AA'$ and $BB'$ intersect.

By Circumscribing Circle about Triangle, $O$ is the point at which the perpendicular bisectors of $AC$ and $BC$ intersect.

By construction, the perpendicular bisectors of $BC$ and $AC$ pass through $A'$ and $B'$ respectively.

Let $OG$ be produced to $H$ such that $OG : HG = 1 : 2$.

It will be demonstrated that $H$ is the orthocenter of $\triangle ABC$.

Draw $AH$ and produce it to intersect $BC$ at $P$.

$A'G : AG = 1 : 2$

Consider $\triangle A'OG$ and $\triangle AHG$.

From above:

$OG : HG = A'G : AG$
$\angle OGA' = \angle HGA$
$\triangle A'OG$ and $\triangle AHG$ are similar.

Thus:

$\angle GHA = \angle GOA'$
$AH \parallel A'O$
$\angle APC = \angle OA'C$

As $OA'$ is perpendicular to $BC$, it follows that $AP$ is also perpendicular to $BC$.

Thus $AP$ is a straight line through $A$ perpendicular to $BC$.

So by definition $AP$ is an altitude of $\triangle ABC$ which passes through $H$.

Similarly, draw $BH$ and produce it to intersect $AC$ at $Q$.

By a similar analysis of the triangles $\triangle B'OG$ and $\triangle BHG$:

$BQ$ is an altitude of $\triangle ABC$ which passes through $H$.

From Altitudes of Triangle Meet at Point, the altitude of $\triangle ABC$ from $C$ to $AB$ likewise will pass through $H$.

Thus $H$ is by definition the orthocenter of $\triangle ABC$.

By construction, $OGH$ is a straight line such that:

$OG : GH = 1 : 2$

Hence the result.

$\blacksquare$

## Historical Note

It was Leonhard Paul Euler who published the result that the circumcenter, centroid and orthocenter of a triangle all lie on the same straight line.

He also determined the ratio of the distances between these points.

Hence this line is now known as the triangle's Euler line.

## Sources

This article incorporates material from Euler Line Proof on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.