# Medians of Triangle Meet at Centroid/Corollary

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## 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:

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