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Euler line
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==In special triangles== ===Right triangle=== In a [[right triangle]], the Euler line coincides with the [[median (triangle)|median]] to 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 [[Altitude (triangle)|altitudes]], falls on the right-angled vertex while its circumcenter, the intersection of its [[Bisection#Perpendicular bisectors|perpendicular bisectors]] of sides, falls on the midpoint of the hypotenuse. ===Isosceles triangle=== The Euler line of an [[isosceles triangle]] coincides with the [[axis of symmetry]]. In an isosceles triangle the [[incenter]] falls on the Euler line. ===Automedian triangle=== The Euler line of an [[automedian triangle]] (one whose [[median (geometry)|medians]] are in the same proportions, though in the opposite order, as the sides) is perpendicular to one of the medians.<ref name="parry">{{citation | last = Parry | first = C. F. | issue = 472 | journal = The Mathematical Gazette | jstor = 3620241 | pages = 151–154 | title = Steiner–Lehmus and the automedian triangle | volume = 75 | year = 1991| doi = 10.2307/3620241 }}.</ref> ===Systems of triangles with concurrent Euler lines=== Consider a triangle ''ABC'' with [[Fermat–Torricelli point]]s ''F''<sub>1</sub> and ''F''<sub>2</sub>. The Euler lines of the 10 triangles with vertices chosen from ''A, B, C, F''<sub>1</sub> and ''F''<sub>2</sub> are [[concurrent lines|concurrent]] at the centroid of triangle ''ABC''.<ref>Beluhov, Nikolai Ivanov. "Ten concurrent Euler lines", ''Forum Geometricorum'' 9, 2009, pp. 271–274. http://forumgeom.fau.edu/FG2009volume9/FG200924index.html</ref> 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.<ref name=ac/>{{rp|p.111}}
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