Centroid of Triangle is Centroid of Medial
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Theorem
Let $\triangle ABC$ be a triangle.
Let $\triangle DEF$ be the medial triangle of $\triangle ABC$.
Let $G$ be the centroid of $\triangle ABC$.
Then $G$ is also the centroid of $\triangle DEF$.
Proof
By definition of centroid and medial triangle, the lines $AE$, $BF$ and $CD$ intersect at $G$.
It remains to be shown that $AE$, $BF$ and $CD$ bisect the sides of $DF$, $DE$ and $EF$ respectively.
Without loss of generality, let $AE$ intersect $DF$ at $H$.
From the working of Triangle is Medial Triangle of Larger Triangle, we have that:
- $DE \parallel AF$
- $EF \parallel AD$
and so $\Box ADEF$ is a parallelogram whose diagonals are $DF$ and $AE$.
From Diameters of Parallelogram Bisect each other, $AE$ bisects $DF$ at $H$.
Similarly:
Hence the result.
$\blacksquare$