Medians of Triangle Meet at Centroid/Corollary

Theorem
Let $\triangle ABC$ be a triangle.

Let $A'$ be the midpoint of $BC$.

Let $B'$ be the midpoint of $AC$.

Let $C'$ be the midpoint of $AB$.

Let $G$ be the centroid of $\triangle ABC$ Let $AA'$ be produced beyond $BC$ to $X$, where $A'X = AG$.

Then the straight lines $B'X$ and $CX$ are parallel to $CC'$ and $BB'$ respectively, and $\dfrac 2 3$ of their length.

Proof

 * Medians-meet-at-Centroid.png

Consider the quadrilateral $\Box BGCX$.

Its diagonals are $GX$ and $BC$.

By construction, they bisect each other.

From Quadrilateral with Bisecting Diagonals is Parallelogram, $\Box BGCX$ is a parallelogram.

From Position of Centroid of Triangle on Median:
 * $BG$ is $\dfrac 2 3$ the length of $BB'$
 * $GC$ is $\dfrac 2 3$ the length of $CC'$

But as $\Box BGCX$ is a parallelogram:
 * $BG = CX$

and:
 * $GC = BX$

Hence the result.