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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 $CD=a$ $cm$.

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$G_1$ 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?

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  • $\begingroup$ I have a geometric method. $\endgroup$ Sep 6 '19 at 11:17
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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 $$\vec{G_1G_2} = G_2 - G_1 = \frac D3 - \frac C3 = \frac13(D - C) = \frac13\vec{CD}$$

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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 $G_1G_2=\dfrac{1}{3}CD$

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  • $\begingroup$ What is $M$? And why is $\vec{G_1D} = \frac23\vec{MD}$ and $\vec{CG_2} = \frac23\vec{CM}$? $\endgroup$ Sep 8 '19 at 15:31
  • $\begingroup$ @PaulSinclair, $M$ is the midpoint of $AB$. $\vec{G_1D}=\frac{2}{3}\vec{MD}$, because $G_1D=\frac{2}{3}MD$ and they lie on the same line. $\endgroup$ Sep 8 '19 at 16:14
  • $\begingroup$ My point is that this information has to be part of your solution. Your solution is wrong without it. And how do you know that $G_1D = \frac 23 MD$ and $G_1, M$ and $D$ are colinear? If this has already been proven, then you need to explicitly reference that. If it hasn't, then the proof needs to be part of your solution as well. $\endgroup$ Sep 8 '19 at 16:27
  • $\begingroup$ @PaulSinclair, that's obvious. $G_1$ represents the centroid of $\triangle ABD$; thus, $G_1$ lies on every median. $M$ is the midpoint of $AB$, so $DM$ is a median and $DG_1=\frac{2}{3}DM$ (the centroid divides the length of each median in $2:1$ ratio). $\endgroup$ Sep 8 '19 at 23:19
  • $\begingroup$ How do you know that the centroid lies on every median? How do you know that that the centroid divides the length of each median in a 2:1 ratio? If you drew a picture of this and showed it to someone intelligent but who has not studied the geometry of triangles in any depth, would they agree that these two things are "obvious"? As a rule of thumb, anything that you cannot prove in 5 lines no wider than your hand does not qualify as "obvious". $\endgroup$ Sep 9 '19 at 0:29

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