Content deleted Content added
Re: the "proof without words", a suggested improvement to an already nice explanation (I won't protest a revert, or perhaps some of my changes will be kept but not others) to (a) use more consistent mathematical language once you get to the area variables (largely to make (b) easier) and (b) to tie things together a bit more clearly. Tag: Reverted |
Annawwanna (talk | contribs) Link suggestions feature: 2 links added. |
||
| (4 intermediate revisions by 4 users not shown) | |||
Line 2:
{{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: [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==
Line 14:
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.
Line 39:
|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<sub>q</sub>'', as the sum of the areas of the four large triangles, ''A<sub>l</sub>'' is 2 ''A<sub>q</sub>'' (each of the two pairs reconstructs the quadrilateral) while that of the small triangles, ''A<sub>s</sub>'' is a quarter of ''A<sub>l</sub>''
}}]]
Line 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>
:<math>n=\tfrac{1}{2}\sqrt{a^2-b^2+c^2-d^2+p^2+q^2}.</math>
Line 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 72 ⟶ 73:
| 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>
Line 81 ⟶ 87:
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 87 ⟶ 94:
| 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.
Line 122 ⟶ 134:
== References and further reading ==
*H. S. M. Coxeter, S. L. Greitzer: ''Geometry Revisited''. MAA, Washington 1967, pp. 52-54
*Peter N. Oliver: [
==External links==
| |||