# Position of Centroid on Euler Line

## 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 $P$.

From Medians of Triangle Meet at Centroid:

- $A'G : AG = 1 : 2$

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

From above:

- $OG : HG = A'G : AG$

From Two Straight Lines make Equal Opposite Angles:

- $\angle OGA' = \angle HGA$

From Triangles with One Equal Angle and Two Sides Proportional are Similar:

- $\triangle A'OG$ and $\triangle AHG$ are similar.

Thus:

- $\angle GHA = \angle GOA'$

From Equal Alternate Angles implies Parallel Lines:

- $AH \parallel A'O$

From Parallelism implies Equal Corresponding Angles:

- $\angle APC = \angle OA'C$

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

- 1986: David Wells:
*Curious and Interesting Numbers*... (previous) ... (next): $3$ - 1997: David Wells:
*Curious and Interesting Numbers*(2nd ed.) ... (previous) ... (next): $3$ - 2008: David Nelson:
*The Penguin Dictionary of Mathematics*(4th ed.) ... (previous) ... (next):**Euler line** - 2014: Christopher Clapham and James Nicholson:
*The Concise Oxford Dictionary of Mathematics*(5th ed.) ... (previous) ... (next):**Euler line** - 2014: Christopher Clapham and James Nicholson:
*The Concise Oxford Dictionary of Mathematics*(5th ed.) ... (previous) ... (next):**orthocentre**

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