Point $G_1$ represents the centroid of $\triangle ABD$ and point $G_2$ represents the centroid of $\triangle ABC$ in the trapezoid $ABCD$ $(AB||CD)$

Question

Point $G_1$ represents the centroid of $\triangle ABD$ and point $G_2$ represents the centroid of $\triangle ABC$ in the trapezoid $ABCD$ $(AB||CD)$. Show $G_1G_2 \parallel AB \parallel CD$ and find $G_1G_2$ if $CD=a$ $cm$.

$G_1$ is the centroid of $\triangle ABD$ iff $\vec{G_1A}+\vec{G_1B}+\vec{G_1D}=\vec{0}$, but this does not help here. I am trying to show $\vec{G_1G_2}=\dfrac{1}{3}\vec{DC}$. Can someone help me?

Answer 1

Since you wanted to see how to do it with vectors, The centroid of the triangle with vertices $\vec u, \vec v, \vec w$ is given by their average: $\frac 13(\vec u + \vec v + \vec w)$. Therefore (choosing any arbitrary point as the origin, and identifying all the points with the vector from the origin to it): $$G_1 = \frac{(A + B + C)}3,\quad G_2 = \frac{(A + B + D)}3$$ And therefore $$\vec{G_1G_2} = G_2 - G_1 = \frac D3 - \frac C3 = \frac13(D - C) = \frac13\vec{CD}$$

Answer 2

Let $M$ is the midpoint of $AB$.

$\vec{G_1G_2}=\vec{G_1D}+\vec{DC}+\vec{CG_2}=\dfrac{2}{3}\vec{MD}+\vec{DC}+\dfrac{2}{3}\vec{CM}=\dfrac{2}{3}(\vec{MD}+\vec{CM}) + \vec{DC}=\dfrac{2}{3}(\vec{MD}-\vec{MC})+\vec{DC}=\dfrac{2}{3}\vec{CD}+\vec{DC}=\vec{DC}-\dfrac{2}{3}\vec{DC}=\dfrac{1}{3}\vec{DC}$

$\Rightarrow G_1G_2 || AB || CD$ and $G_1G_2=\dfrac{1}{3}CD$