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''' holds that the midpoints of the sides of an arbitrary [[quadrilateral]] form a [[parallelogram]], called the '''Varignon parallelogram'''. It is named after [[Pierre Varignon]], whose proof was published posthumously in 1731.<ref>Peter N. Oliver: [http://www.maa.org/sites/default/files/images/upload_library/46/NCTM/mt2001-Varignon1.pdf ''Pierre Varignon and the Parallelogram Theorem'']{{Dead link|date=November 2024 |bot=InternetArchiveBot |fix-attempted=yes }}. Mathematics Teacher, Band 94, Nr. 4, April 2001, pp. 316-319</ref>

==Theorem==

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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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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.