# Wittenbauer's Parallelogram: What is it?

A Mathematical Droodle

What if applet does not run? |

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny### Wittenbauer's Parallelogram

Divide the sides of a quadrilateral ABCD into the thirds. Draw straight lines through the pairs of the division points next to each vertex. The four lines form a parallelogram bearing the name of F. Wittenbauer (1857-1922). The sides of the parallelogram are parallel to the diagonals of ABCD (which in fact explains why there's a parallelogram in the first place), exactly as the sides of Varignon's parallelogram.

Wittenbauer's theorem states that the center of Wittenbauer's parallelogram is exactly the *barycenter* - the *center of gravity* - of the quadrilateral ABCD. In other words, if ABCD is thought of as a physical thin piece of a uniformly distributed material, then the shape could be balanced at that point on a pointed stick or a needle.

The drawing is suggestive in the case where ABCD is convex. In geometric terms the statement is as follows:

Let M, N, P, Q be the points of intersection of the medians of triangles ACD, ABC, BCD, and ABD, respectively. Then the point of intersection of MN and PQ coincides with the center - the point of intersection of the diagonals - of Wittenbauer's parallelogram.

The mechanical proof is the most natural. Consider triangles ACD and ABC. Their masses are proportional to their areas. Since they share the base AC, their masses are in the ratio of their altitudes to AC:

The vertical distance, i.e., the distance in the direction perpendicular to AC, between M and N is

Now let's have a look at the Wittenbauer's parallelogram. The vertical distance between XW and YZ is

When the quadrilateral is concave or even self-intersecting the theorem still holds, however the proof must obviously be modified.

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny63718379 |