# Difference between revisions of "Euler line"

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==Another Proof== | ==Another Proof== | ||

Let <math>M</math> be the midpoint of <math>BC</math>. | Let <math>M</math> be the midpoint of <math>BC</math>. | ||

− | Extend <math>CG</math> to point <math>H'</math> such that <math>CG = \frac{1}{2} GH</math>. We will show <math>H</math> is the orthocenter. | + | Extend <math>CG</math> to point <math>H'</math> such that <math>CG = \frac{1}{2} GH</math>. We will show <math>H'</math> is the orthocenter. |

− | Consider triangles <math>MGO</math> and <math>AGH</math>. Since <math>\frac{MG}{GA}=\frac{ | + | Consider triangles <math>MGO</math> and <math>AGH'</math>. Since <math>\frac{MG}{GA}=\frac{H'G}{GC} = \frac{1}{2}</math>, and they both share a vertical angle, they are similar by SAS similarity. Thus, <math>AH' \parallel OM \perp BC</math>, so <math>H'</math> lies on the <math>A</math> altitude of <math>\triangle ABC</math>. We can analogously show that <math>H'</math> also lies on the <math>B</math> and <math>C</math> altitudes, so <math>H'</math> is the orthocenter. <math>\square</math> |

==Proof Nine-Point Center Lies on Euler Line== | ==Proof Nine-Point Center Lies on Euler Line== |

## Revision as of 02:45, 6 December 2018

In any triangle , the **Euler line** is a line which passes through the orthocenter , centroid , circumcenter , nine-point center and de Longchamps point . It is named after Leonhard Euler. Its existence is a non-trivial fact of Euclidean geometry. Certain fixed orders and distance ratios hold among these points. In particular, and

Euler line is the central line .

Given the orthic triangle of , the Euler lines of ,, and concur at , the nine-point circle of .

## Contents

## Proof Centroid Lies on Euler Line

This proof utilizes the concept of spiral similarity, which in this case is a rotation followed homothety. Consider the medial triangle . It is similar to . Specifically, a rotation of about the midpoint of followed by a homothety with scale factor centered at brings . Let us examine what else this transformation, which we denote as , will do.

It turns out is the orthocenter, and is the centroid of . Thus, . As a homothety preserves angles, it follows that . Finally, as it follows that Thus, are collinear, and .

## Another Proof

Let be the midpoint of . Extend to point such that . We will show is the orthocenter. Consider triangles and . Since , and they both share a vertical angle, they are similar by SAS similarity. Thus, , so lies on the altitude of . We can analogously show that also lies on the and altitudes, so is the orthocenter.

## Proof Nine-Point Center Lies on Euler Line

Assuming that the nine point circle exists and that is the center, note that a homothety centered at with factor brings the Euler points onto the circumcircle of . Thus, it brings the nine-point circle to the circumcircle. Additionally, should be sent to , thus and .

## Analytic Proof of Existence

Let the circumcenter be represented by the vector , and let vectors correspond to the vertices of the triangle. It is well known the that the orthocenter is and the centroid is . Thus, are collinear and

## See also

*This article is a stub. Help us out by expanding it.*