Medial triangle

 ( Elementary Geometry ) 

In Euclidean geometry, the medial triangle or midpoint triangle of a triangle is the triangle with vertices at the of the triangle's sides . It is the case of the midpoint polygon of a polygon with sides. The medial triangle is not the same thing as the median triangle, which is the triangle whose sides have the same lengths as the medians of .

Each side of the medial triangle is called a midsegment (or midline). In general, a midsegment of a triangle is a line segment which joins the midpoints of two sides of the triangle. It is parallel to the third side and has a length equal to half the length of the third side.

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Properties [[File:Mittendreieck.svg|thumb|upright=1.6|

circumcenter of , orthocenter of

incenter of , Nagel point of

centroid of and

]] The medial triangle can also be viewed as the image of triangle transformed by a homothety centered at the centroid with ratio -1/2. Thus, the sides of the medial triangle are half and parallel to the corresponding sides of triangle ABC. Hence, the medial triangle is inversely similar and shares the same centroid and medians with triangle . It also follows from this that the perimeter of the medial triangle equals the semiperimeter of triangle , and that the area is one quarter of the area of triangle . Furthermore, the four triangles that the original triangle is subdivided into by the medial triangle are all mutually congruent by SSS, so their areas are equal and thus the area of each is 1/4 the area of the original triangle.Posamentier, Alfred S., and Lehmann, Ingmar. The Secrets of Triangles, Prometheus Books, 2012.

The orthocenter of the medial triangle coincides with the circumcenter of triangle . This fact provides a tool for proving Euler line. The medial triangle is the pedal triangle of the circumcenter. The nine-point circle circumscribes the medial triangle, and so the nine-point center is the circumcenter of the medial triangle.

The Nagel point of the medial triangle is the incenter of its reference triangle.Altshiller-Court, Nathan. College Geometry. Dover Publications, 2007.

A reference triangle's medial triangle is congruent to the triangle whose vertices are the midpoints between the reference triangle's orthocenter and its vertices.

The incenter of a triangle lies in its medial triangle. Franzsen, William N.. "The distance from the incenter to the Euler line", Forum Geometricorum 11 (2011): 231–236.

A point in the interior of a triangle is the center of an inellipse of the triangle if and only if the point lies in the interior of the medial triangle.Chakerian, G. D. "A Distorted View of Geometry." Ch. 7 in Mathematical Plums (R. Honsberger, editor). Washington, DC: Mathematical Association of America, 1979.

The medial triangle is the only inscribed figure for which none of the other three interior triangles has smaller area. Torrejon, Ricardo M. "On an Erdos inscribed triangle inequality", Forum Geometricorum 5, 2005, 137–141. http://forumgeom.fau.edu/FG2005volume5/FG200519index.html

The reference triangle and its medial triangle are orthologic triangles.

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Coordinates Let $a\; =\; |BC|,\; b\; =\; |CA|,\; c\; =\; |AB|$ be the sidelengths of triangle $\backslash triangle\; ABC.$ Trilinear coordinates for the vertices of the medial triangle $\backslash triangle\; EFD$ are given by

$\backslash begin\{alignat\}\{3\}$

E &={} \, 0 &&: \frac{1}{b} &&: \frac{1}{c}, \\5mu F &={} \frac{1}{a} &&: \,0 &&: \frac{1}{c}, \\5mu D &={} \frac{1}{a} &&: \frac{1}{b} &&: \, 0. \end{alignat}

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Anticomplementary triangle If $\backslash triangle\; EFD$ is the medial triangle of $\backslash triangle\; ABC,$ then $\backslash triangle\; ABC$ is the anticomplementary triangle or antimedial triangle of $\backslash triangle\; EFD.$ The anticomplementary triangle of $\backslash triangle\; ABC$ is formed by three lines parallel to the sides of the parallel to $AB$ through $C,$ the parallel to $AC$ through $B,$ and the parallel to $BC$ through $A.$

Trilinear coordinates for the vertices of the triangle $\backslash triangle\; E\text{'}F\text{'}D\text{'}$ anticomplementary to $\backslash triangle\; ABC$ are given by

$\backslash begin\{alignat\}\{3\}$

E' &= -\frac{1}{a} &&: \phantom{-}\frac{1}{b} &&: \phantom{-}\frac{1}{c}, \\5mu F' &= \phantom{-}\frac{1}{a} &&: -\frac{1}{b} &&: \phantom{-}\frac{1}{c}, \\5mu D' &= \phantom{-}\frac{1}{a} &&: \phantom{-}\frac{1}{b} &&: -\frac{1}{c}. \end{alignat}

The name "anticomplementary triangle" corresponds to the fact that its vertices are the anticomplements of the vertices $A,\; B,\; C$ of the reference triangle. The vertices of the medial triangle are the complements of $A,\; B,\; C.$

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See also

• Middle hedgehog, an analogous concept for more general convex sets

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External links

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