---
tag: 0VX0CPRQTZ
kind: statement
role: theorem
name: Euler line
msc: [51M04]
uses: [0F98FZK14A, 0X7EEWES5G, 0WB619CA0V, 0AFNNTRTJE]
sources:
  - url: https://en.wikipedia.org/wiki/Euler_line
    locator: Individual centers
  - url: https://en.wikipedia.org/wiki/Euler_line
    locator: A vector proof
  - url: https://en.wikipedia.org/wiki/Euler_line
    locator: Distances between centers
---

## Statement

Let $ABC$ be a triangle in the Euclidean space $\mathbb{R}^n$. Let $O$ be its [[circumcentre|0F98FZK14A]], $G = \frac{1}{3}(A + B + C)$ its [[centroid|0X7EEWES5G]] and $H$ its [[orthocentre|0WB619CA0V]].
Then
$$H = O + 3\,(G - O).$$
In particular, the points $O$, $G$ and $H$ are [[collinear|0AFNNTRTJE]].
If the triangle is not equilateral, then $O \neq G$, and the line through $O$ and $G$ is the Euler line of the triangle.

## Remarks

The formula says that $G$ lies between $O$ and $H$, with $d(G, H) = 2\, d(O, G)$.
An equivalent form is $H - O = (A - O) + (B - O) + (C - O)$.
The centre of the nine-point circle is the midpoint of $O$ and $H$, so it is also on the Euler line.
In an equilateral triangle, the points $O$, $G$ and $H$ are the same point.

## History

Euler showed in 1765 that the orthocentre, the circumcentre and the centroid of a triangle are collinear.

## Explanations

### L1 human

Three famous centres of a triangle always lie on one straight line. The centroid is on this line between the other two. It is twice as far from the orthocentre as from the circumcentre.