Editing Euler line (section)

===A vector proof===
Let <math>ABC</math> be a triangle. A proof of the fact that the [[circumscribed circle|circumcenter]] <math>O</math>, the [[centroid]] <math>G</math> and the [[altitude (triangle)#Orthocenter|orthocenter]] <math>H</math> are '''collinear''' relies on [[euclidean vector|free vectors]]. We start by stating the prerequisites. First, <math>G</math> satisfies the relation

:<math>\vec{GA}+\vec{GB}+\vec{GC}=0.</math>

This follows from the fact that the [[barycentric coordinate system|absolute barycentric coordinates]] of <math>G</math> are <math>\frac{1}{3}:\frac{1}{3}:\frac{1}{3}</math>. Further, the [[Sylvester's triangle problem|problem of Sylvester]]<ref name=Dorrie>Dörrie, Heinrich, "100 Great Problems of Elementary Mathematics. Their History and Solution". Dover Publications, Inc., New York, 1965, {{ISBN|0-486-61348-8}}, pages 141 (Euler's Straight Line) and 142 (Problem of Sylvester)</ref> reads as

:<math>\vec{OH}=\vec{OA}+\vec{OB}+\vec{OC}.</math>

Now, using the vector addition, we deduce that

:<math>\vec{GO}=\vec{GA}+\vec{AO}\,\mbox{(in triangle }AGO\mbox{)},\,\vec{GO}=\vec{GB}+\vec{BO}\,\mbox{(in triangle }BGO\mbox{)},\,\vec{GO}=\vec{GC}+\vec{CO}\,\mbox{(in triangle }CGO\mbox{)}.</math>

By adding these three relations, term by term, we obtain that

:<math>3\cdot\vec{GO}=\left(\sum\limits_{\scriptstyle\rm cyc}\vec{GA}\right)+\left(\sum\limits_{\scriptstyle\rm cyc}\vec{AO}\right)=0-\left(\sum\limits_{\scriptstyle\rm cyc}\vec{OA}\right)=-\vec{OH}.</math>

In conclusion, <math>3\cdot\vec{OG}=\vec{OH}</math>, and so the three points <math>O</math>, <math>G</math> and <math>H</math> (in this order) are collinear.

In Dörrie's book,<ref name=Dorrie /> the '''Euler line''' and the [[Sylvester's triangle problem|problem of Sylvester]] are put together into a single proof. However, most of the proofs of the problem of Sylvester rely on the fundamental properties of free vectors, independently of the Euler line.