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In geometry, the **Euler line**, named after Leonhard Euler (US /ˈɔɪlər/, * OY-lər* or UK /ˈɔɪlə/,

The concept of a triangle's Euler line extends to the Euler line of other shapes, such as the quadrilateral and the tetrahedron.

Euler showed in 1765 that in any triangle, the orthocenter, circumcenter and centroid are collinear.^{[2]} This property is also true for another triangle center, the nine-point center, although it had not been defined in Euler's time. In equilateral triangles, these four points coincide, but in any other triangle they are all distinct from each other, and the Euler line is determined by any two of them.

Other notable points that lie on the Euler line include the de Longchamps point, the Schiffler point, and the Exeter point.^{[1]} However, the incenter generally does not lie on the Euler line;^{[3]} it is on the Euler line only for isosceles triangles,^{[4]} for which the Euler line coincides with the symmetry axis of the triangle and contains all triangle centers.

The *tangential triangle* of a reference triangle is tangent to the latter's circumcircle at the reference triangle's vertices. The circumcenter of the tangential triangle lies on the Euler line of the reference triangle.^{[5]}^{:p.104,#211;p.242,#346}

Let *A*, *B*, *C* denote the vertex angles of the reference triangle, and let *x* : *y* : *z* be a variable point in trilinear coordinates; then an equation for the Euler line is

Another way to represent the Euler line is in terms of a parameter *t*. Starting with the circumcenter (with trilinear coordinates ) and the orthocenter (with trilinears , every point on the Euler line, except the orthocenter, is given by the trilinear coordinates

formed as a linear combination of the trilinears of these two points, for some *t*.

For example:

- The circumcenter has trilinears

- corresponding to the parameter value

- The centroid has trilinears , corresponding to the parameter value
- The nine-point center has trilinears corresponding to the parameter value
- The De Longchamps point has trilinears corresponding to the parameter value

In a Cartesian coordinate system, denote the slopes of the sides of a triangle as and and denote the slope of its Euler line as . Then these slopes are related according to^{[6]}^{:Lemma 1}

Thus the slope of the Euler line (if finite) is expressible in terms of the slopes of the sides as

Moreover, the Euler line is parallel to an acute triangle's side *BC* if and only if^{[6]}^{:p.173}

On the Euler line the centroid *G* is between the circumcenter *O* and the orthocenter *H* and is twice as far from the orthocenter as it is from the circumcenter:^{[5]}^{:p.102}

The center *N* of the nine-point circle lies along the Euler line midway between the orthocenter and the circumcenter:^{[1]}

Thus the Euler line could be repositioned on a number line with the circumcenter *O* at the location 0, the centroid *G* at 2*t*, the nine-point center at 3*t*, and the orthocenter *H* at 6*t* for some scale factor *t*.

Furthermore, the squared distance between the centroid and the circumcenter along the Euler line is less than the squared circumradius *R*^{2} by an amount equal to one-ninth the sum of the squares of the side lengths *a*, *b*, and *c*:^{[5]}^{:p.71}

In addition,^{[5]}^{:p.102}

In a right triangle, the Euler line contains the median on the hypotenuse—that is, it goes through both the right-angled vertex and the midpoint of the side opposite that vertex. This is because the right triangle's orthocenter, the intersection of its altitudes, falls on the right-angled vertex while its circumcenter, the intersection of its perpendicular bisectors of sides, falls on the midpoint of the hypotenuse.

The Euler line of an isosceles triangle coincides with the axis of symmetry. In an isosceles triangle the incenter falls on the Euler line.

A triangle's Kiepert parabola is the unique parabola that is tangent to the sides of the triangle and has the Euler line as its directrix.^{[7]}^{:Thm. 3}

Consider a triangle *ABC* with Fermat–Torricelli points *F*_{1} and *F*_{2}. The Euler lines of the 10 triangles with vertices chosen from *A, B, C, F*_{1} and *F*_{2} are concurrent at the centroid of triangle *ABC*.^{[8]}

The Euler lines of the four triangles formed by an orthocentric system (a set of four points such that each is the orthocenter of the triangle with vertices at the other three points) are concurrent at the nine-point center common to all of the triangles.^{[5]}^{:p.111}

In a convex quadrilateral, the quasiorthocenter *H*, the "area centroid" *G*, and the quasicircumcenter *O* are collinear in this order on the Euler line, and *HG* = 2*GO*.^{[9]}

A tetrahedron is a three-dimensional object bounded by four triangular faces. Seven lines associated with a tetrahedron are concurrent at its centroid; its six midplanes intersect at its Monge point; and there is a circumsphere passing through all of the vertices, whose center is the circumcenter. These points define the "Euler line" of a tetrahedron analogous to that of a triangle. The centroid is the midpoint between its Monge point and circumcenter along this line. The center of the twelve-point sphere also lies on the Euler line.

- ^
^{a}^{b}^{c}Kimberling, Clark (1998). "Triangle centers and central triangles".*Congressus Numerantium***129**: i–xxv, 1–295. **^**Euler, Leonhard (1767). "Solutio facilis problematum quorundam geometricorum difficillimorum" [Easy solution of some difficult geometric problems].*Novi Commentarii academiae scientarum imperialis Petropolitanae***11**: 103–123. E325. Reprinted in*Opera Omnia*, ser. I, vol. XXVI, pp. 139–157, Societas Scientiarum Naturalium Helveticae, Lausanne, 1953, MR 0061061. Summarized at: Dartmouth College.**^**Schattschneider, Doris; King, James (1997).*Geometry Turned On: Dynamic Software in Learning, Teaching, and Research*. The Mathematical Association of America. pp. 3–4. ISBN 978-0883850992.**^**Edmonds, Allan L.; Hajja, Mowaffaq; Martini, Horst (2008), "Orthocentric simplices and biregularity",*Results in Mathematics***52**(1-2): 41–50, doi:10.1007/s00025-008-0294-4, MR 2430410,It is well known that the incenter of a Euclidean triangle lies on its Euler line connecting the centroid and the circumcenter if and only if the triangle is isosceles

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^{a}^{b}^{c}^{d}^{e}Altshiller-Court, Nathan,*College Geometry*, Dover Publications, 2007 (orig. Barnes & Noble 1952). - ^
^{a}^{b}Wladimir G. Boskoff, Laurent¸iu Homentcovschi, and Bogdan D. Suceava, "Gossard’s Perspector and Projective Consequences",*Forum Geometricorum*, Volume 13 (2013), 169–184. [1] **^**Scimemi, Benedetto, "Simple Relations Regarding the Steiner Inellipse of a Triangle",*Forum Geometricorum*10, 2010: 55–77.**^**Beluhov, Nikolai Ivanov. "Ten concurrent Euler lines",*Forum Geometricorum*9, 2009, pp. 271–274. http://forumgeom.fau.edu/FG2009volume9/FG200924index.html**^**Myakishev, Alexei (2006), "On Two Remarkable Lines Related to a Quadrilateral",*Forum Geometricorum***6**: 289–295.

- Altitudes and the Euler Line and Euler Line and 9-Point Circle at cut-the-knot
- Triangle centers on the Euler line, by Clark Kimberling.
- An interactive applet showing several triangle centers that lies on the Euler line.
- Weisstein, Eric W., "Euler Line",
*MathWorld*. - "Euler Line" and "Non-Euclidean Triangle Continuum" at the Wolfram Demonstrations Project
- Nine-point conic and Euler line generalization and A further Euler line generalization at Dynamic Geometry Sketches
- The quasi-Euler line of a quadrilateral and a hexagon at Dynamic Geometry Sketches. This interactive sketch, with a link to a paper, shows a generalization of the Euler line to any quadrilateral and hexagon involving their so-called quasi-circumcenters, quasi-orthocenters and lamina centroids.