Varignon's theorem: Difference between revisions

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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''' isholds athat statementthe inmidpoints [[Euclideanof geometry]],the thatsides dealsof withan thearbitrary construction[[quadrilateral]] ofform a particular [[parallelogram]], called the '''Varignon parallelogram''', from an arbitrary [[quadrilateral]] (quadrangle). It is named after [[Pierre Varignon]], whose proof was published posthumously in 1731.<ref>Peter N. Oliver: [https://web.archive.org/web/20150906161939/http://www.maa.org/sites/default/files/images/upload_library/46/NCTM/mt2001-Varignon1.pdf ''Pierre Varignon and the Parallelogram Theorem'']. Mathematics Teacher, Band 94, Nr. 4, April 2001, pp. 316-319</ref>

==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&nbsp;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&nbsp;''G'') by taking its fourth vertex to be ''E''&nbsp;−&nbsp;''F''&nbsp;+&thinsp;''G''. In the construction of the Varignon parallelogram this is the point (''A''&nbsp;+&nbsp;''B'')/2&nbsp;−&nbsp;(''B''&nbsp;+&nbsp;''C'')/2&nbsp;+&nbsp;(''C''&nbsp;+&nbsp;''D'')/2 =&nbsp;(''A''&nbsp;+&nbsp;''D'')/2. But this is the point ''H'' in the figure, whence ''EFGH'' forms a parallelogram.

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[[Image:Varignon parallelogram crossed.svg|300px]]

|}

[[File:varignon_parallelogram.svg|thumb|[[Proof without words]] of Varignon's theorem: {{olist

<br />1. |An arbitrary quadrilateral and its diagonals.

<br />2. |Bases of similar triangles are parallel to the blue diagonal.

<br />3. |Ditto for the red diagonal.

<br />4. |The base pairs form a parallelogram with half the area of the quadrilateral, ''A_q'', as the sum of the areas of the four large triangles, ''A_l'' is 2 ''A_q'' (each of the two pairs reconstructs the quadrilateral) while that of the small triangles, ''A_s'' is a quarter of ''A_l'' (half linear dimensions yields quarter area), and the area of the parallelogram is ''A_q'' minus ''A_s''.]]

}}]]

==The Varignon 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>[{{Cite web |url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=363253 |title=Mateescu Constantin, Answer to ''Inequality Of Diagonal''] |access-date=2016-04-05 |archive-date=2014-10-24 |archive-url=https://web.archive.org/web/20141024134419/http://www.artofproblemsolving.com/Forum/viewtopic.php?t=363253 |url-status=dead }}</ref> The length of the bimedian that connects the midpoints of the sides ''b'' and ''d'' is

:<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}}.</ref>

| 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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:<math>n=\tfrac{1}{2}\sqrt{2(a^2+c^2)-4x^2}.</math>

Note that theThe two opposite sides in these formulas are not the two that the bimedian connects.

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}}.</ref>

| 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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| pages = 70–72

| title = Geometry of the isosceles trapezium and the contra-parallelogram, with applications to the geometry of the ellipse

| volume = 20}}</ref>| doi-access = free

}}</ref>

==See also==

*[[Perpendicular bisector construction of a quadrilateral]], a different way of forming another quadrilateral from a given quadrilateral

*[[Morley's trisector theorem]], a related theorem on triangles

==Notes==

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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: [httphttps://wwwold.maa.org/sites/default/files/images/upload_library/46/NCTM/mt2001-Varignon1Varignon2.pdf ''Consequences of Varignon Parallelogram Theorem'']. Mathematics Teacher, Band 94, Nr. 5, Mai 2001, pp. 406-408

==External links==

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[[Category:Theorems about quadrilaterals]]

[[Category:Eponymous theorems of geometry]]

[[Category:Euclidean geometry]]

[[Category:Articles containing proofs]]