Euler line

Builds on Circumcentre Centroid Orthocentre Collinear points

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.

Reading path

To understand this page from the ground up, read these 5 pages in order. The foundations come first.

Show the path as a graph

Each page uses the pages above it. The colour of the bar on a box shows how strong the evidence is, and the box also gives it as text. The list has the same pages.

no evidence level

Centroid concept
Circumcentre concept
Collinear points concept
Orthocentre concept
Euler line statement

Tag	`0VX0CPRQTZ`
URI	`m2://0VX0CPRQTZ`
Kind	statement, theorem
Writing	W1 stated
MSC	51M04
Status	This copy of the site has no build data. Run `m2 status 0VX0CPRQTZ`.
Uses	Circumcentre, Centroid, Orthocentre, Collinear points

Euler line

Statement

Remarks

History

Reading path

Further reading

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