Centroid of a quadrilateral $ABCD$: $\overrightarrow{GA} + \overrightarrow{GB} + \overrightarrow{GC} + \overrightarrow{GD} = \overrightarrow{0}$

Question

Let $ABCD$ be a quadrilateral and let $G_1$, $G_2$, $G_3$, $G_4$ be he centroids of triangles $ABC$, $BCD$, $CDA$ and $DAB$, respectively. Assume without proof that $G_1G_3$ and $G_2G_4$ always intersect at a unique point $G$ (this can be shown by verifying that $\overrightarrow{G_1G_3}$ and $\overrightarrow{G_2G_4}$ are not collinear). Is it true that $\overrightarrow{GA} + \overrightarrow{GB} + \overrightarrow{GC} + \overrightarrow{GD} = \overrightarrow{0}$?

I am well aware of properties like $3\overrightarrow{GG_1} = \overrightarrow{GA} + \overrightarrow{GB} + \overrightarrow{GC}$ but I cannot find a way to combine them to show the above.

Any help appreciated!

Answer 1

That property is false. Below you can find a counterexample: $G'$ is the unique point such that $\overrightarrow{G'A} + \overrightarrow{G'B} + \overrightarrow{G'C} + \overrightarrow{G'D} = \overrightarrow{0}$ (as you may easily check), but it is not the same as point $G$ as defined in the question.

The confusion might arise from the fact that the centroid of four points $ABCD$ ($G'$ in the figure) is not the same as the centroid of quadrilateral $ABCD$ (i.e. the centroid of its surface), which is in fact $G$. On the other hand, for a triangle both centroids are the same.

Answer 2

Here is an other counterexample. The involved points are (also identified with their affixes and with the corresponding vectors in $\Bbb R^2$): $$ \begin{aligned} A &= (-6,0)\ ,\\ B &= (0,-6)\ ,\\ C &= (12,0)\ ,\\ D &= (0, 18)\ ,\\[2mm] G_1 &=\frac 13(A+B+C)=(2,-2)\ ,\\ G_2 &=\frac 13(B+C+D)=(4,4)\ ,\\ G_3 &=\frac 13(C+D+A)=(2,6)\ ,\\ G_4 &=\frac 13(D+A+B)=(-2,4)\ ,\\[2mm] G &=(2,4)\ ,\\ A+B+C+D &= (6,12)\ ,\\ 4G &=(8,16)\ . \end{aligned} $$ In a picture:

One more observation that may explain the difference.

Let $\Gamma$ be the centroid of $ABCD$. Then it is also the centroid of $G_1G_2G_3G_4$. So a counterexample to the vectorial relation $$ 4G\Gamma =GA+GB+GC+GD=0 $$ is obtained by choosing the points $A,B,C,D$ so that for the associated quadrilateral $G_1G_2G_3G_4$ the intersection of its diagonals $G_1G_3$ and $G_2G_4$ is not its centroid.