Medians of Triangle Meet at Centroid/Corollary
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 $A'X = AG$.
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.
But as $\Box BGCX$ is a parallelogram:
- $BG = CX$
- $GC = BX$
Hence the result.