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{{for|the theorem about the moment of a force|Varignon's theorem (mechanics)}}
[[Image:Varignon parallelogram convex.svg|thumb|300px|Area(''EFGH'') = (1/2)Area(''ABCD'')]]
In [[Euclidean geometry]], '''Varignon's theorem'''
==Theorem==
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Referring to the diagram above, [[triangle]]s ''ADC'' and ''HDG'' are similar by the side-angle-side criterion, so [[angle]]s ''DAC'' and ''DHG'' are equal, making ''HG'' parallel to ''AC''. In the same way ''EF'' is parallel to ''AC'', so ''HG'' and ''EF'' are parallel to each other; the same holds for ''HE'' and ''GF''.
Varignon's theorem can also be proved as a theorem of [[affine geometry]] organized as [[linear algebra]] with the linear combinations restricted to coefficients summing to 1, also called affine or [[Barycentric coordinates (mathematics)|barycentric coordinates]]. The proof applies even to skew quadrilaterals in spaces of any dimension.
Any three points ''E'', ''F'', ''G'' are completed to a parallelogram (lying in the plane containing ''E'', ''F'', and ''G'') by taking its fourth vertex to be ''E'' − ''F'' + ''G''. In the construction of the Varignon parallelogram this is the point (''A'' + ''B'')/2 − (''B'' + ''C'')/2 + (''C'' + ''D'')/2 = (''A'' + ''D'')/2. But this is the point ''H'' in the figure, whence ''EFGH'' forms a parallelogram.
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:<math>m=\tfrac{1}{2}\sqrt{-a^2+b^2-c^2+d^2+p^2+q^2}</math>
where ''p'' and ''q'' are the length of the diagonals.<ref>
:<math>n=\tfrac{1}{2}\sqrt{a^2-b^2+c^2-d^2+p^2+q^2}.</math>
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The lengths of the bimedians can also be expressed in terms of two opposite sides and the distance ''x'' between the midpoints of the diagonals. This is possible when using Euler's quadrilateral theorem in the above formulas. Whence<ref>{{citation
| last = Josefsson
| first = Martin | journal = Forum Geometricorum
| pages = 155–164
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| url = http://forumgeom.fau.edu/FG2011volume11/FG201116.pdf
| volume = 11
| year = 2011
| access-date = 2016-04-05
| archive-date = 2020-01-05
| archive-url = https://web.archive.org/web/20200105031952/http://forumgeom.fau.edu/FG2011volume11/FG201116.pdf
| url-status = dead
}}.</ref>
:<math>m=\tfrac{1}{2}\sqrt{2(b^2+d^2)-4x^2}</math>
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In a convex quadrilateral, there is the following [[Duality (mathematics)|dual]] connection between the bimedians and the diagonals:<ref name=Josefsson>{{citation
| last = Josefsson
| first = Martin | journal = Forum Geometricorum
| pages = 13–25
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| url = http://forumgeom.fau.edu/FG2012volume12/FG201202.pdf
| volume = 12
| year = 2012
| access-date = 2012-12-28
| archive-date = 2020-12-05
| archive-url = https://web.archive.org/web/20201205213638/http://forumgeom.fau.edu/FG2012volume12/FG201202.pdf
| url-status = dead
}}.</ref>
* The two bimedians have equal length [[if and only if]] the two diagonals are [[perpendicular]].
* The two bimedians are perpendicular if and only if the two diagonals have equal length.
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== References and further reading ==
*H. S. M. Coxeter, S. L. Greitzer: ''Geometry Revisited''. MAA, Washington 1967, pp. 52-54
*Peter N. Oliver: [
==External links==
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[[Category:Theorems about quadrilaterals]]
[[Category:Eponymous theorems of geometry]]
[[Category:Euclidean geometry]]
[[Category:Articles containing proofs]]
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