## Point Mass Centroid (centre of gravity or balancing point) of Quadrilateral

Theorem
The respective centroids C', D', A' and B' of triangles ABD, ABC, BCD and CDA of quadrilateral ABCD form a quadrilateral A'B'C'D', similar to the original and scale factor -1/3 (a halfturn and reduction by 1/3), with lines AA', BB', CC' and DD' concurrent in G. Then this point of concurrency G (centre of similarity between ABCD and A'B'C'D') is defined as the (point mass) centroid of the quadrilateral.

#### .sketch_canvas { border: medium solid lightgray; display: inline-block; } Point Mass Centroid (centre of gravity) of Quadrilateral

Challenge
Can you explain why (prove that) the result is true?
If stuck, see my paper at Generalizations involving maltitudes.
This concurrency (centroid) result generalizes to any polygon as shown in Yaglom (1968), and is applied in another, interesting related generalization at Triangle Centroids of a Hexagon form a Parallelo-Hexagon as well as in Generalizing the Nagel line to Circumscribed Polygons by Analogy.

In the physical, real world context, this point mass centroid is the centre of gravity or balancing point of equal point masses placed at the vertices of a polygon. Click on the Show Varignon Parallelogram button to see why this is the case for a quadrilateral - for example the masses at the vertices can be replaced by point masses at the midpoints of the sides, which in turn will balance at the centre of the parallelogram (the intersection of its diagonals), which as you can see above, coincides with the point G.

Note: The centroid of a cardboard quadrilateral (a planar quadrilateral of uniform density), unlike the case for a triangle, does NOT always coincide with the point mass centroid illustrated above - click on the Show Cardboard centroid button. Also see the dynamic geometry sketch at Centroid of Cardboard Quadrilateral.

Parallelogram Theorem
Regarding the above Note, the following interesting & important theorem holds:
The point mass centroid G and the cardboard centroid of a quadrilateral coincides, if and only if, the quadrilateral is a parallelogram.
Challenge: First try to prove it yourself - before reading the translation from German of a neat proof by Arnold Kirsch in Humenberger (2023).

A 'bisect-diagonal' quadrilateral is a quadrilateral with at least one of its diagonals bisecting the other. The following interesting theorem in relation to a bisect-diagonal quadrilateral holds:
If ABCD is a bisect-diagonal quadrilateral with diagonal BD bisected (cut in half) by diagonal AC, then the cardboard (lamina) centroid and point mass centroid both lie on diagonal AC. More-over, if P is the midpoint of AC, then the distance between the cardboard centroid and the point P is twice that of the distance between the cardboard centroid and point mass centroid.
Challenge: First try to prove it yourself - before reading De Villiers (2021).

Explore Further
1) Is the point mass centroid of a quadrilateral always inside? Specifically check by dragging until the quadrilateral becomes concave.
2) Can you find another way of locating the point mass centroid of a quadrilateral with equal masses at the vertices using the coordinates of its vertices?
3) How would you locate the point mass centre in 2) above, if different masses are located at the vertices?
4) Can you figure out different ways of finding the point mass centroid of a pentagon, hexagon, etc. with equal or different masses at the vertices?
5) Where would you locate the balancing point of a 'perimeter' quadrilateral? For example, of a quadrilateral consisting of just sticks or rods forming its perimeter?

References
De Villiers, M. (2021). Some more properties of the bisect-diagonal quadrilateral. The Mathematical Gazette, Volume 105 , Issue 564 , November, pp. 474 - 480.
Humenberger, Hans. (2023). Centroids of Quadrilaterals and a Peculiarity of Parallelograms. At Right Angles, November, pp. 1-9.
Yaglom, I.M. (1968). Geometric Transformations II. Washington, DC: MAA, pp.24; 108-109.