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>. | ||
+ | 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{HG}{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 02:44, 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
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