Euler line

Statement

Let $ABC$ be a triangle in the Euclidean space $\mathbb{R}^n$. Let $O$ be its circumcentre, $G = \frac{1}{3}(A + B + C)$ its centroid and $H$ its orthocentre. Then $$H = O + 3\,(G - O).$$ In particular, the points $O$, $G$ and $H$ are collinear. 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.