Centroid of Triangle is Centroid of Medial

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:

$BF$ bisects $DE$

$DC$ bisects $EF$.

Hence the result.

$\blacksquare$