# Euler line

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

- 1 Proof Centroid Lies on Euler Line
- 2 Another Proof
- 3 Proof Nine-Point Center Lies on Euler Line
- 4 Analytic Proof of Existence
- 5 The points of intersection of the Euler line with the sides of the triangle
- 6 Angles between Euler line and the sides of the triangle
- 7 Distances along Euler line
- 8 Position of Kimberling centers on the Euler line
- 9 Triangles with angles of or
- 10 Euler lines of cyclic quadrilateral (Vittas’s theorem)
- 11 Concurrent Euler lines and Fermat points
- 12 Euler line of Gergonne triangle
- 13 Thebault point
- 14 Schiffler point
- 15 Euler line as radical axis
- 16 De Longchamps point X(20)
- 17 De Longchamps line
- 18 CIRCUMCENTER OF THE TANGENTIAL TRIANGLE X(26)
- 19 PERSPECTOR OF ORTHIC AND TANGENTIAL TRIANGLES X(25)
- 20 Exeter point X(22)
- 21 Far-out point X(23)
- 22 Symmetric lines
- 23 H–line Claim
- 24 See also

## 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 past 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

## The points of intersection of the Euler line with the sides of the triangle

**Acute triangle**

Let be the acute triangle where Denote

Let Euler line cross lines and in points and respectively.

Then point lyes on segment

Point lyes on segment

Point lyes on ray

**Proof**

Denote

We use the formulae (see Claim “Segments crossing inside triangle” in “Schiffler point” in “Euler line”).

Centroid lyes on median

Orthocenter lyes on altitude Therefore We use the signed version of Menelaus's theorem and get

**Obtuse triangle**

Let be the obtuse triangle where

Let Euler line cross lines and in points and respectively.

Similarly we get

ray

**Right triangle**

Let be the right triangle where Then Euler line contain median from vertex

**Isosceles triangle**

Let be the isosceles triangle where Then Euler line contain median from vertex

**Corollary: Euler line is parallel to side**

Euler line is parallel to side iff

**Proof**

After simplification in the case we get

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## Angles between Euler line and the sides of the triangle

Let Euler line of the cross lines and in points and respectively. Denote smaller angles between the Euler line and lines and as and respectively.

Prove that

**Proof**

WLOG,

Let be the midpoint be the circumcenter of

Symilarly, for other angles.

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## Distances along Euler line

Let and be orthocenter, centroid, circumcenter, and circumradius of the respectively.

Prove that

**Proof**

WLOG, is an acute triangle,

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## Position of Kimberling centers on the Euler line

Let triangle ABC be given. Let and are orthocenter, circumcenter, circumradius and inradius, respectively.

We use point as origin and as a unit vector.

We find Kimberling center X(I) on Euler line in the form of For a lot of Kimberling centers the coefficient is a function of only two parameters and

Centroid Nine-point center de Longchamps point Schiffler point Exeter point Far-out point Perspector of ABC and orthic-of-orthic triangle Homothetic center of orthic and tangential triangles Circumcenter of the tangential triangle

Midpoint of X(3) and
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## Triangles with angles of or

**Claim 1**

Let the in triangle be Then the Euler line of the is parallel to the bisector of

**Proof**

Let be circumcircle of

Let be circumcenter of

Let be the circle symmetric to with respect to

Let be the point symmetric to with respect to

The lies on lies on

is the radius of and translation vector to is

Let be the point symmetric to with respect to Well known that lies on Therefore point lies on

Point lies on

Let be the bisector of are concurrent.

Euler line of the is parallel to the bisector of as desired.

**Claim 2**

Let the in triangle be Then the Euler line of the is perpendicular to the bisector of

**Proof**

Let be circumcircle, circumcenter, orthocenter and incenter of the points are concyclic.

The circle centered at midpoint of small arc

is rhomb.

Therefore the Euler line is perpendicular to as desired.

**Claim 3**

Let be a quadrilateral whose diagonals and intersect at and form an angle of If the triangles PAB, PBC, PCD, PDA are all not equilateral, then their Euler lines are pairwise parallel or coincident.

**Proof**

Let and be internal and external bisectors of the angle .

Then Euler lines of and are parallel to and Euler lines of and are perpendicular to as desired.

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## Euler lines of cyclic quadrilateral (Vittas’s theorem)

**Claim 1**

Let be a cyclic quadrilateral with diagonals intersecting at The Euler lines of triangles are concurrent.

**Proof**

Let be the circumcenters (orthocenters) of triangles Let be the common bisector of and Therefore and are parallelograms with parallel sides.

bisect these angles. So points are collinear and lies on one straight line which is side of the pare vertical angles and Similarly, points are collinear and lies on another side of these angles. Similarly obtuse so points and are collinear and lies on one side and points and are collinear and lies on another side of the same vertical angles.

We use * Claim* and get that lines are concurrent (or parallel if or ).

**Claim 2 (Property of vertex of two parallelograms)**

Let and be parallelograms, Let lines and be concurrent at point Then points and are collinear and lines and are concurrent.

**Proof**

We consider only the case Shift transformation allows to generalize the obtained results.

We use the coordinate system with the origin at the point and axes

We use and get points and are colinear.

We calculate point of crossing and and and and get the same result: as desired (if then point moves to infinity and lines are parallel, angles or

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~minor edit by Yiyj1

## Concurrent Euler lines and Fermat points

Consider a triangle with Fermat–Torricelli points and The Euler lines of the triangles with vertices chosen from and are concurrent at the centroid of triangle We denote centroids by , circumcenters by We use red color for points and lines of triangles green color for triangles and blue color for triangles

**Case 1**

Let be the first Fermat point of maximum angle of which smaller then Then the centroid of triangle lies on Euler line of the The pairwise angles between these Euler lines are equal

**Proof**

Let and be centroid, circumcenter, and circumcircle of respectevely.

Let be external for equilateral triangle is cyclic.

Point is centroid of Points and are colinear, so point lies on Euler line of

**Case 2**

Let be the first Fermat point of

Then the centroid of triangle lies on Euler lines of the triangles and The pairwise angles between these Euler lines are equal

**Proof**

Let be external for equilateral triangle, be circumcircle of is cyclic.

Point is centroid of

Points and are colinear, so point lies on Euler line of as desired.

**Case 3**

Let be the second Fermat point of Then the centroid of triangle lies on Euler lines of the triangles and

The pairwise angles between these Euler lines are equal

**Proof**

Let be internal for equilateral triangle, be circumcircle of

Let and be circumcenters of the triangles and Point is centroid of the is the Euler line of the parallel to

is bisector of is bisector of is bisector of is regular triangle.

is the inner Napoleon triangle of the is centroid of this regular triangle.

points and are collinear as desired.

Similarly, points and are collinear.

**Case 4**

Let and be the Fermat points of Then the centroid of point lies on Euler line is circumcenter, is centroid) of the

**Proof**

* Step 1* We will find line which is parallel to

Let be midpoint of Let be the midpoint of

Let be point symmetrical to with respect to

as midline of

* Step 2* We will prove that line is parallel to

Let be the inner Napoleon triangle. Let be the outer Napoleon triangle. These triangles are regular centered at

Points and are collinear (they lies on bisector

Points and are collinear (they lies on bisector

Points and are collinear (they lies on bisector angle between and is

Points and are concyclic Points and are concyclic

points and are concyclic

Therefore and are collinear or point lies on Euler line

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## Euler line of Gergonne triangle

Prove that the Euler line of Gergonne triangle of passes through the circumcenter of triangle

Gergonne triangle is also known as the contact triangle or intouch triangle. If the inscribed circle touches the sides of at points and then is Gergonne triangle of .

Other wording: Tangents to circumcircle of are drawn at the vertices of the triangle. Prove that the circumcenter of the triangle formed by these three tangents lies on the Euler line of the original triangle.

**Proof**

Let and be orthocenter and circumcenter of respectively. Let be Orthic Triangle of

Then is Euler line of is the incenter of is the incenter of

Similarly,

where is the perspector of triangles and

Under homothety with center P and coefficient the incenter of maps into incenter of , circumcenter of maps into circumcenter of are collinear as desired.

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## Thebault point

Let and be the altitudes of the where

a) Prove that the Euler lines of triangles are concurrent on the nine-point circle at a point T (Thebault point of )

b) Prove that if then else

**Proof**

**Case 1 Acute triangle**

a) It is known, that Euler line of acute triangle cross AB and BC (shortest and longest sides) in inner points.

Let be circumcenters of

Let and be centroids of

Denote is the circle (the nine-points circle).

is the midpoint where is the orthocenter of

Similarly

is the midline of

Let cross at point different from

spiral similarity centered at maps onto

This similarity has the rotation angle acute angle between Euler lines of these triangles is

Let these lines crossed at point Therefore points and are concyclic

Similarly, as desired.

b) Point lies on median of and divide it in ratio 2 : 1.

Point lies on Euler line of

According the **Claim,**

Similarly

**Case 2 Obtuse triangle**

a) It is known, that Euler line of obtuse cross AC and BC (middle and longest sides) in inner points.

Let be circumcenters of

Let and be centroids of

Denote is the circle (the nine-points circle).

is the midpoint where is the orthocenter of

Similarly

is the midline of

Let cross at point different from

spiral similarity centered at maps onto

This similarity has the rotation angle acute angle between Euler lines of these triangles is

Let these lines crossed at point Therefore points and are concyclic

Similarly, as desired.

b)

Point lies on median of and divide it in ratio

Point lies on Euler line of
According the **Claim,**

Similarly

**Claim (Segment crossing the median)**

Let be the midpoint of side of the

Then

**Proof**

Let be (We use sign to denote the area of

Denote
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## Schiffler point

Let and be the incenter, circumcenter, centroid, circumradius, and inradius of respectively. Then the Euler lines of the four triangles and are concurrent at Schiffler point .

**Proof**

We will prove that the Euler line of cross the Euler line of at such point that .

Let and be the circumcenter and centroid of respectively.

It is known that lies on circumcircle of

Denote

It is known that is midpoint point lies on median points belong the bisector of

Easy to find that ,

We use sigh [t] for area of t. We get

Using * Claim* we get
Therefore each Euler line of triangles cross Euler line of in the same point, as desired.

**Claim (Segments crossing inside triangle)**

Given triangle GOY. Point lies on

Point lies on

Point lies on

Point lies on Then

**Proof**

Let be (We use sigh for area of
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## Euler line as radical axis

Let with altitudes and be given.

Let and be circumcircle, circumcenter, orthocenter and circumradius of respectively.

Circle centered at passes through and is tangent to the radius AO. Similarly define circles and

Then Euler line of is the radical axis of these circles.

If is acute, then these three circles intersect at two points located on the Euler line of the

**Proof**

The power of point with respect to and is

The power of point with respect to is

The power of point with respect to is

The power of point with respect to is

It is known that

Therefore points and lies on radical axis of these three circles as desired.

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## De Longchamps point X(20)

**Definition 1**

The De Longchamps’ point of a triangle is the radical center of the power circles of the triangle. Prove that De Longchamps point lies on Euler line.

We call A-power circle of a the circle centered at the midpoint point with radius The other two circles are defined symmetrically.

**Proof**

Let and be orthocenter, circumcenter, and De Longchamps point, respectively.

Denote power circle by power circle by WLOG,

Denote the projection of point on

We will prove that radical axes of power and power cicles is symmetric to altitude with respect Further, we will conclude that the point of intersection of the radical axes, symmetrical to the heights with respect to O, is symmetrical to the point of intersection of the heights with respect to

Point is the crosspoint of the center line of the power and power circles and there radical axis. We use claim and get:

and are the medians, so

We use Claim some times and get: radical axes of power and power cicles is symmetric to altitude with respect

Similarly radical axes of power and power cicles is symmetric to altitude radical axes of power and power cicles is symmetric to altitude with respect Therefore the point of intersection of the radical axes, symmetrical to the heights with respect to is symmetrical to the point of intersection of the heights with respect to lies on Euler line of

**Claim (Distance between projections)**

**Definition 2**

We call circle of a the circle centered at with radius The other two circles are defined symmetrically. The De Longchamps point of a triangle is the radical center of circle, circle, and circle of the triangle (Casey – 1886). Prove that De Longchamps point under this definition is the same as point under **Definition 1.**

**Proof**

Let and be orthocenter, centroid, and De Longchamps point, respectively. Let cross at points and The other points are defined symmetrically. Similarly is diameter

Therefore is anticomplementary triangle of is orthic triangle of So is orthocenter of

as desired.

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## De Longchamps line

The de Longchamps line of is defined as the radical axes of the de Longchamps circle and of the circumscribed circle of

Let be the circumcircle of (the anticomplementary triangle of

Let be the circle centered at (centroid of ) with radius where

Prove that the de Longchamps line is perpendicular to Euler line and is the radical axes of and

**Proof**

Center of is , center of is where is Euler line. The homothety with center and ratio maps into This homothety maps into and there is two inversion which swap and

First inversion centered at point Let be the point of crossing and

The radius of we can find using

Second inversion centered at point We can make the same calculations and get as desired.

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## CIRCUMCENTER OF THE TANGENTIAL TRIANGLE X(26)

Prove that the circumcenter of the tangential triangle of (Kimberling’s point lies on the Euler line of

**Proof**

Let and be midpoints of and respectively.

Let be circumcircle of It is nine-points circle of the

Let be circumcircle of Let be circumcircle of

and are tangents to inversion with respect swap and Similarly, this inversion swap and and Therefore this inversion swap and

The center of and the center of lies on Euler line, so the center of lies on this line, as desired.

After some calculations one can find position of point on Euler line (see Kimberling's point

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## PERSPECTOR OF ORTHIC AND TANGENTIAL TRIANGLES X(25)

Let be the orthic triangle of Let be the circumcenter of Let be the tangencial triangle of Let be the circumcenter of

Prove that lines and are concurrent at point, lies on Euler line of

**Proof**

and are antiparallel to BC with respect

Similarly,

Therefore homothetic center of and is the point of concurrence of lines and Denote this point as

The points and are the corresponding points (circumcenters) of and so point lies on line

Points and lies on Euler line, so lies on Euler line of

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## Exeter point X(22)

Exeter point is the perspector of the circummedial triangle and the tangential triangle By another words, let be the reference triangle (other than a right triangle). Let the medians through the vertices meet the circumcircle of triangle at and respectively. Let be the triangle formed by the tangents at and to (Let be the vertex opposite to the side formed by the tangent at the vertex A). Prove that the lines through and are concurrent, the point of concurrence lies on Euler line of triangle the point of concurrence lies on Euler line of triangle where - circumcenter, - orthocenter, - circumradius.

**Proof**

At first we prove that lines and are concurrent. This follows from the fact that lines and are concurrent at point and * Mapping theorem*.

Let and be the midpoints of and respectively. The points and are collinear. Similarly the points and are collinear.

Denote the inversion with respect It is evident that

Denote

The power of point with respect is

Similarly the power of point with respect is

lies on radical axis of and

Therefore second crosspoint of and point lies on line which is the Euler line of Point lies on the same Euler line as desired.

Last we will find the length of as desired.

**Mapping theorem**

Let triangle and incircle be given. Let be the point in the plane Let lines and crossing second time at points and respectively.

Prove that lines and are concurrent.

**Proof**

We use Claim and get: Similarly,

We use the trigonometric form of Ceva's Theorem for point and triangle and get We use the trigonometric form of Ceva's Theorem for triangle and finish proof that lines and are concurrent.

**Claim (Point on incircle)**

Let triangle and incircle be given. Prove that

**Proof**

Similarly

We multiply and divide these equations and get:

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## Far-out point X(23)

Let be the tangential triangle of

Let and be the centroid, circumcircle, circumcenter, circumradius and orthocenter of

Prove that the second crosspoint of circumcircles of and is point Point lies on Euler line of

**Proof**

Denote the inversion with respect midpoints of

It is evident that

The inversion of circles are lines which crosses at point

Therefore point lies on Euler line of as desired.

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## Symmetric lines

Let triangle having the circumcircle be given.

Prove that the lines symmetric to the Euler line with respect and are concurrent and the point of concurrence lies on

**Proof**

The orthocenter lies on the Euler line therefore the Euler line is We use * H-line Clime* and finish the proof.

## H–line Claim

Let triangle having the orthocenter and circumcircle be given. Denote any line containing point

Let and be the lines symmetric to with respect and respectively.

Prove that and are concurrent and the point of concurrence lies on

**Proof**

Let and be the crosspoints of with and respectively.

WLOG Let and be the points symmetric to with respect and respectively.

Therefore

Let be the crosspoint of and is cyclic

Similarly is cyclic the crosspoint of and is point

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## See also

- Kimberling center
- Kimberling’s point X(20)
- Kimberling’s point X(21)
- Kimberling’s point X(22)
- Kimberling’s point X(23)
- Kimberling’s point X(24)
- Kimberling’s point X(25)
- Kimberling’s point X(26)
- De Longchamps point
- Gossard perspector
- Evans point
- Double perspective triangles
- Steiner line
- Miquel's point
- Spieker center
- Simson line
- Complete Quadrilateral
- Gauss line
- Isogonal conjugate
- Barycentric coordinates
- Symmetry
- Symmedian and Antiparallel
- Central line
- Gergonne line
- Gergonne point

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