Varignon's theorem: Difference between revisions

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Revision as of 03:08, 27 February 2020

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==Proof==

Referring to the diagram above, triangles ADC and HDG are similar by the side-angle-side criterion, so angles 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 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 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.

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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.

From the secondfirst proof, one can see that the sum of the diagonals is equal to the perimeter of the quadrilateralparallelogram 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.

A second proof requires less algebra. By drawing in the diagonals of the quadrilateral, one notices two triangles are created for each diagonal. And by the [[midpoint theorem]], the segment containing two midpoints of adjacent sides is both parallel and half the respective diagonal. Since two opposite sides are equal and parallel, the quadrilateral must be a parallelogram.

From the second proof, one 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.

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