Varignon's theorem: Difference between revisions

Revision as of 03:43, 24 September 2017

Latest revision as of 21:23, 1 May 2025

(34 intermediate revisions by 24 users not shown)

Line 1:

{{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.<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 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.

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;+&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.

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.

AFrom secondthe first proof, requiresone lesscan algebra.see By drawing inthat the diagonalssum of the quadrilateral,diagonals weis noticeequal twoto trianglesthe areperimeter createdof forthe eachparallelogram diagonalformed. AndAlso, bywe thecan [[midpointuse theorem]],vectors 1/2 the segment containing two midpointslength of adjacenteach sidesside isto bothfirst paralleldetermine andthe halfarea of the respectivequadrilateral, diagonal.and Sincethen twoto oppositefind sidesareas areof equalthe andfour parallel,triangles wedivided haveby thateach theside quadrilateralof must bethe ainner parallelogram.

From the second proof, we can see that the sum of the diagonals is equal to the perimeter of the quadrilateral formed. Also, 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"

|-

Line 33 ⟶ 35:

[[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==

Line 49 ⟶ 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>

Line 58 ⟶ 66:

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

Line 64 ⟶ 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 70 ⟶ 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 79 ⟶ 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 93 ⟶ 113:

| 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=Josefsson/>{{rp|p. 14}} <ref name=deV />{{rp|p. 169}}

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 is a line segment traversed twice.

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

| date = February 1901

| doi = 10.1017/s0013091500032892

| journal = Proceedings of the Edinburgh Mathematical Society

| pages = 70–72

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

| volume = 20| 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==

Line 105 ⟶ 134:

== 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 ''PierreConsequences of Varignon and the Parallelogram Theorem'']. Mathematics Teacher, Band 94, Nr. 45, AprilMai 2001, pp. 316406-319408

*Peter N. Oliver: [http://www.maa.org/sites/default/files/images/upload_library/46/NCTM/mt2001-Varignon1.pdf ''Consequences of Varignon Parallelogram Theorem'']. Mathematics Teacher, Band 94, Nr. 5, Mai 2001, pp. 406-408

==External links==

Line 112 ⟶ 140:

*{{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

[[Category:QuadrilateralsTheorems about quadrilaterals]]

[[Category:TheoremsEponymous intheorems of geometry]]

[[Category:Euclidean geometry]]

[[Category:Theorems in geometry]]

[[Category:Articles containing proofs]]