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  "body": "\n## Proof\n\nWe argue with vectors in $\\mathbb{R}^n$.\n\n1. Let $ABC$ be a [[triangle|0CTNFGYMM2]] in $\\mathbb{R}^n$, and let $P$ be its plane, the [[affine span|0YMB1676R1]] of $A$, $B$, $C$. Let $W$ be the span of $B - A$ and $C - A$. Then $P = A + W$.\n2. The points $A$, $B$, $C$ are not collinear, so $B - A$ and $C - A$ are linearly independent, and $W$ has dimension $2$.\n3. Let $O$ be the [[circumcentre|0F98FZK14A]]. It lies in $P$ and satisfies $\\norm{A - O} = \\norm{B - O} = \\norm{C - O} = R$ for some $R \\geq 0$.\n4. Define the point $K = A + B + C - 2O$. It is the [[affine combination|0KJH6SRFZK]] of $A$, $B$, $C$, $O$ with the weights $1, 1, 1, -2$. These four points lie in $P$, so $K \\in P$.\n5. With the [[dot product|0S52TNNTH9]], we compute $(K - A) \\cdot (C - B) = ((B - O) + (C - O)) \\cdot ((C - O) - (B - O))$.\n6. This equals $\\norm{C - O}^2 - \\norm{B - O}^2 = R^2 - R^2 = 0$ by step 3.\n7. In the same way, $(K - B) \\cdot (A - C) = \\norm{A - O}^2 - \\norm{C - O}^2 = 0$ and $(K - C) \\cdot (B - A) = \\norm{B - O}^2 - \\norm{A - O}^2 = 0$.\n8. Let $H_A$ be the foot of the [[altitude|0C9MZY2RNA]] from $A$, so $H_A = B + k\\,(C - B)$ with $k = \\frac{(A - B) \\cdot (C - B)}{(C - B) \\cdot (C - B)}$.\n9. Then $(H_A - A) \\cdot (C - B) = (B - A) \\cdot (C - B) + (A - B) \\cdot (C - B) = 0$. Also $H_A \\neq A$, since $H_A$ is on the line $BC$ and $A$ is not.\n10. The map $\\phi \\colon W \\to \\mathbb{R}$, $\\phi(w) = w \\cdot (C - B)$, is linear. It is not zero, because $C - B \\in W$ and $\\phi(C - B) = \\norm{C - B}^2 \u003e 0$.\n11. By the [[rank–nullity theorem|0W4FSW1TVB]], the kernel of $\\phi$ has dimension $2 - 1 = 1$. By step 9, it contains the nonzero vector $H_A - A$, so it is the span of $H_A - A$.\n12. By steps 4 and 6, $K - A \\in W$ and $\\phi(K - A) = 0$. So $K - A$ is a multiple of $H_A - A$, and $K$ lies on the line through $A$ and $H_A$.\n13. That line contains the altitude from $A$. In the same way, steps 7 to 12 show that $K$ lies on the lines of the altitudes from $B$ and from $C$.\n14. The altitude lines from $A$ and $B$ are distinct. Otherwise this line is $AB$, so $B - A$ is orthogonal to $C - B$ and to $C - A$.\n   Then $B - A$ is orthogonal to $(C - A) - (C - B) = B - A$, so $A = B$, which is false. Two distinct lines meet in at most one point.\n   By the definition of the [[orthocentre|0WB619CA0V]], $H$ is the common point of the three altitude lines. Since $K$ is also a common point, $K = H$, that is, $H - O = (A - O) + (B - O) + (C - O)$.\n15. The [[centroid|0X7EEWES5G]] is $G = \\frac{1}{3}(A + B + C)$, so $3\\,(G - O) = A + B + C - 3O = H - O$. This gives $H = O + 3\\,(G - O)$.\n16. Each difference of two points of $\\set{O, G, H}$ is a multiple of $G - O$, because $H - O = 3(G - O)$ and $H - G = 2(G - O)$. So the span of these differences has dimension at most $1$, and $O$, $G$, $H$ are [[collinear|0AFNNTRTJE]].\n17. Now suppose that $O = G$. Put $a = A - G$, $b = B - G$ and $c = C - G$. Then $\\norm{a} = \\norm{b} = \\norm{c} = R$ and $a + b + c = 0$.\n18. From $c = -(a + b)$ we get $R^2 = \\norm{a + b}^2 = 2R^2 + 2\\, a \\cdot b$. Hence $a \\cdot b = -R^2/2$.\n19. Then the [[distance|0NPPGHQNXQ]] satisfies $d(A, B)^2 = \\norm{a - b}^2 = 2R^2 - 2\\, a \\cdot b = 3R^2$.\n20. The same computation for the pairs $(b, c)$ and $(c, a)$ gives $d(B, C)^2 = d(C, A)^2 = 3R^2$. So the triangle is equilateral.\n21. By contraposition, if $ABC$ is not equilateral, then $O \\neq G$. Then there is exactly one line through $O$ and $G$, and by step 16 it contains $H$. This line is the Euler line.\n\n## Idea\n\nThe point $A + B + C - 2O$ is easy to test against each altitude. The dot product with a side becomes a difference of two squared radii, which is zero. So this point is the orthocentre, and comparing it with $3(G - O)$ gives the collinearity at once.\n",
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