Geometry Polygons, polyhedra and solid geometry Theorem

Varignon's theorem

Statement

Let $ABCD$ be a simple quadrilateral, that is a simple polygon with four vertices, in the Euclidean plane. Let $K$ , $L$ , $M$ , $N$ be the midpoints of $AB$ , $BC$ , $CD$ , $DA$ . Then $KLMN$ is a parallelogram. Its sides are parallel to the diagonals $AC$ and $BD$ and half as long, and its area is half the area of $ABCD$ .

Remarks

With vectors, $L - K = \frac{1}{2}(C - A) = M - N$ , so $KL$ and $NM$ are parallel and equal, which makes $KLMN$ a parallelogram. In the same way $KN$ and $LM$ equal $\frac{1}{2}(D - B)$ . The area of $ABCD$ is $\frac{1}{2}\abs{(C - A) \times (D - B)}$ , and the area of $KLMN$ is $\abs{\frac{1}{2}(C - A) \times \frac{1}{2}(D - B)}$ , which is half of it. Example: for the square with vertices $(0, 0), (2, 0), (2, 2), (0, 2)$ , the midpoints form a square of side $\sqrt 2$ and area $2$ , half of $4$ . The parallelogram part needs no simplicity; for a self-crossing quadrilateral the area relation fails, and the parallelogram can even degenerate.

Going deeper

For a student

The proof is the midpoint theorem for triangles applied twice, once in each triangle cut off by a diagonal. It is a good first example of how vectors turn a geometric statement into one line of algebra.

Reading path

To understand this page from the ground up, read these 3 pages in order. The foundations come first.

Show the path as a graph

Each page uses the pages above it. The colour of the bar on a box shows how strong the evidence is, and the box also gives it as text. The list has the same pages.

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Midpoint concept
Polygon concept
Varignon's theorem statement

Tag	`B2ZJW0YBQ7`
URI	`m2://B2ZJW0YBQ7`
Kind	statement, theorem
Writing	W0 stub
MSC	51M04
Status	This copy of the site has no build data. Run `m2 status B2ZJW0YBQ7`.
About	Polygon
Uses	Polygon, Midpoint
Same as	Wikidata Q1896455 Wikipedia Varignon's theorem