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  "body": "\n## Statement\n\nLet $ABC$ 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]].\nThen\n$$H = O + 3\\,(G - O).$$\nIn particular, the points $O$, $G$ and $H$ are [[collinear|0AFNNTRTJE]].\nIf the triangle is not equilateral, then $O \\neq G$, and the line through $O$ and $G$ is the Euler line of the triangle.\n\n## Remarks\n\nThe formula says that $G$ lies between $O$ and $H$, with $d(G, H) = 2\\, d(O, G)$.\nAn equivalent form is $H - O = (A - O) + (B - O) + (C - O)$.\nThe centre of the nine-point circle is the midpoint of $O$ and $H$, so it is also on the Euler line.\nIn an equilateral triangle, the points $O$, $G$ and $H$ are the same point.\n\n## History\n\nEuler showed in 1765 that the orthocentre, the circumcentre and the centroid of a triangle are collinear.\n\n## Explanations\n\n### L1 human\n\nThree 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.\n",
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