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

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.

$\blacksquare$

Sources

• 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text {IV}$. Pure Geometry: Plane Geometry: The centre of gravity