Varignon's theorem: Difference between revisions

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{{short description|The midpoints of the sides of an arbitrary quadrilateral form a parallelogram}}

{{for|the theorem about the momentummoment 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]], whowhose proof was published itposthumously 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==

The midpoints of the sides of an arbitrary quadrilateral form a parallelogram. If the quadrilateral is [[convex polygon|convex]] or [[concave polygon|reentrantconcave]], (the quadrilateral is not a [[Quadrilateral#Self-intersecting quadrilateralsComplex_quadrilaterals|crossing quadranglecomplex]]), then the [[area]] of the parallelogram is half the area of the quadrilateral.

If one introduces the concept of oriented areas for [[Polygon|''n''-gons]], then thethis area equality above also holds for crossedcomplex quadrilaterals as well.<ref name=Coxeter>[[Coxeter|Coxeter, H. S. M.]] and Greitzer, S. L. "Quadrangle; Varignon's theorem" §3.1 in Geometry Revisited. Washington, DC: Math. Assoc. Amer., pp. 52&ndash;54, 1967.</ref>

The Varignon parallelogram exists even for a [[Quadrilateral#More_quadrilateralsSkew_quadrilaterals|skew quadrilateral]], and is planar whether the quadrilateral is planar or not. The theorem can be generalized to the [[midpoint polygon]] of an arbitrary polygon.

==Proof==

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 iscan easilyalso 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;&minus;−&nbsp;''F''&nbsp;+&nbspthinsp;''G''. In the construction of the Varignon parallelogram this is the point (''A''&nbsp;+&nbsp;''B'')/2&nbsp;&minus;−&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.

In short, the [[centroid]] of the four points ''A'', ''B'', ''C'', ''D'' is the midpoint of each of the two diagonals ''EG'' and ''FH'' of ''EFGH'', showing that the midpoints coincide.

A second proof requires less algebra. By drawing inFrom the diagonalsfirst of the quadrilateralproof, weone noticecan twosee triangles are created for each diagonal. And by the [[Midline Theorem]], the segment containing two midpoints of adjacent sides is both parallel and half the respective diagonal. Therefore,that the sum of the diagonals is equal to the perimeter of the quadrilateralparallelogram formed. SecondlyAlso, we can use vectors 1/2 the length of each side to first determine the area of the quadrilateral, and then to find areas of the four triangles divided by each side of the inner parallelogram.

{| class="wikitable"

|-

! convex quadrilateral

! concave quadrilateral

! crossed quadrilateral

|-

||

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

||

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

||

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

|}

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

|An arbitrary quadrilateral and its diagonals.

|Bases of similar triangles are parallel to the blue diagonal.

|Ditto for the red diagonal.

|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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*Each pair of opposite sides of the Varignon parallelogram are parallel to a diagonal in the original quadrilateral.

*A side of the Varignon parallelogram is half as long as the diagonal in the original quadrilateral it is parallel to.

*The area of the Varignon parallelogram equals half the area of the original quadrilateral. This is true in convex, concave and crossed quadrilaterals provided the area of the latter is defined to be the difference of the areas of the two triangles it is composed of.<ref>H. S. M. name=Coxeter and S. L. Greitzer, Geometry Revisited, MAA, 1967, pp. 52-53.</ref>

*The [[perimeter]] of the Varignon parallelogram equals the sum of the diagonals of the original quadrilateral.

*The diagonals of the Varignon parallelogram are the bimedians of the original quadrilateral.

Line 25 ⟶ 57:

:<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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This is also a [[corollary]] to the [[parallelogram law]] applied in the Varignon parallelogram.

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 name=Josefsson3>{{citation

| last = Josefsson
| first = Martin

| journal = Forum Geometricorum

| pages = 155–164

Line 40 ⟶ 73:

| 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>

Line 46 ⟶ 84:

:<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

Line 55 ⟶ 94:

| 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.

===Special cases===

The Varignon parallelogram is a [[rhombus]] if and only if the two diagonals of the quadrilateral have equal length, that is, if the quadrilateral is an [[equidiagonal quadrilateral]].<ref name=deV>{{citation

| last = de Villiers | first = Michael

| isbn = 9780557102952

Line 66 ⟶ 110:

| publisher = Dynamic Mathematics Learning

| title = Some Adventures in Euclidean Geometry

| url = httphttps://books.google.com/books?id=R7uCEqwsN40C&pg=PA58

| year = 2009}}.</ref>

The Varignon parallelogram is a [[rectangle]] if and only if the diagonals of the quadrilateral are [[perpendicular]], that is, if the quadrilateral is an [[orthodiagonal quadrilateral]].<ref name=Josefsson2Josefsson/>{{citationrp|p. 14}} <ref name=deV />{{rp|p. 169}}

| last = Josefsson | first = Martin

| journal = Forum Geometricorum

| pages = 13–25

| title = Characterizations of Orthodiagonal Quadrilaterals

| url = http://forumgeom.fau.edu/FG2012volume12/FG201202.pdf

| volume = 12

| year = 2012}}.</ref>{{rp|p.14}}

For a [[list of self-intersecting polygons|self-crossing]] quadrilateral, the Varignon parallelogram can degenerate to four collinear points, forming a line segment traversed twice. This happens whenever the polygon is formed by replacing two parallel sides of a [[trapezoid]] by the two diagonals of the trapezoid, such as in the [[antiparallelogram]].<ref>{{citation

If a crossing quadrilateral is formed from either pair of opposite parallel sides and the diagonals of a parallelogram, the Varignon parallelogram has a side of length zero and is a line segment.{{Citation needed|date=April 2015}}

| last = Muirhead | first = R. F. | author-link = Robert Franklin Muirhead

| date = February 1901

==Proof==

| doi = 10.1017/s0013091500032892

Varignon's theorem is easily 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.

| journal = Proceedings of the Edinburgh Mathematical Society

| pages = 13–2570–72

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;&minus;&nbsp;''F''&nbsp;+&nbsp;''G''. In the construction of the Varignon parallelogram this is the point (''A''&nbsp;+&nbsp;''B'')/2&nbsp;&minus;&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.

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

| volume = 20| doi-access = free

In short, the [[centroid]] of the four points ''A'', ''B'', ''C'', ''D'' is the midpoint of each of the two diagonals ''EG'' and ''FH'' of ''EFGH'', showing that the midpoints coincide.

}}</ref>

A second proof requires less algebra. By drawing in the diagonals of the quadrilateral, we notice two triangles are created for each diagonal. And by the [[Midline Theorem]], the segment containing two midpoints of adjacent sides is both parallel and half the respective diagonal. Therefore, the sum of the diagonals is equal to the perimeter of the quadrilateral formed. Secondly, we can use vectors 1/2 the length of each side to first determine the area of the quadrilateral, and then to find areas of the four triangles divided by each side of the inner parallelogram.

{| class="wikitable"

|-

! convex quadrilateral

! concave quadrilateral

! crossed quadrilateral

|-

||

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

||

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

||

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

|}

==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==

<references/>

== References and further reading ==

*H. S. M. Coxeter, S. L. Greitzer: ''Geometry Revisited''. MAA, Washington 1967, pp. 52-54

*Peter N. Oliver: [https://old.maa.org/sites/default/files/images/upload_library/46/NCTM/mt2001-Varignon2.pdf ''Consequences of Varignon Parallelogram Theorem'']. Mathematics Teacher, Band 94, Nr. 5, Mai 2001, pp. 406-408

==External links==

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*{{MathWorld|urlname=VarignonsTheorem|title= Varignon's theorem}}

*[http://www.vias.org/comp_geometry/geom_quad_varignon.html Varignon Parallelogram in Compendium Geometry]

* [http://frink.machighwaydynamicmathematicslearning.com/~dynamicm/octagoncentroids.html A generalization of Varignon's theorem to 2n-gons and to 3D] at [http://dynamicmathematicslearning.com/JavaGSPLinks.htm Dynamic Geometry Sketches], interactive dynamic geometry sketches.

*[http://www.cut-the-knot.org/Curriculum/Geometry/Varignon.shtml Varignon parallelogram] at cut-the-knot-org

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