Position of Centroid of Triangle on Median/Proof 1

Theorem

Let $\triangle ABC$ be a triangle.

Let $AL$, $BM$ and $CN$ be the medians of $\triangle ABC$ meeting at the centroid $G$ of $\triangle ABC$.

Then $G$ is $\dfrac 1 3$ of the way along $AL$ from $L$, and similarly for the other medians.

Let $\triangle ABC$ be embedded in a Cartesian plane such that $A = \tuple {x_1, y_1}$, $B = \tuple {x_2, y_2}$ and $C = \tuple {x_3, y_3}$.

The coordinates of $L$ are $\tuple {\dfrac {x_2 + x_3} 2, \dfrac {y_2 + y_3} 2}$.

Let $G$ be the point dividing $AL$ in the ratio $2 : 1$.

The coordinates of $G$ are $\tuple {\dfrac {x_1 + \paren {x_2 + x_3} } {1 + 2}, \dfrac {y_1 + \paren {y_2 + y_3} } {1 + 2} }$.

By similarly calculating the coordinates of $M$ and $N$, we get:

$\ds M$

$=$

$\ds \tuple {\dfrac {x_1 + x_3} 2, \dfrac {y_1 + y_3} 2}$

$\ds N$

$=$

$\ds \tuple {\dfrac {x_1 + x_2} 2, \dfrac {y_1 + y_2} 2}$

Similarly:

calculating the position of the point $G'$ dividing $BM$ in the ratio $2 : 1$

calculating the position of the point $G$ dividing $CN$ in the ratio $2 : 1$

we find that:

$G = G' = G = \tuple {\dfrac {x_1 + x_2 + x_3} 3, \dfrac {y_1 + y_2 + y_3} 3}$

and the result follows.

$\blacksquare$

1933: D.M.Y. Sommerville: Analytical Conics (3rd ed.) ... (previous) ... (next): Chapter $\text I$. Coordinates: $12$. Mean points and centres of gravity
1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text {III}$. Analytical Geometry: The Straight Line: Centre of gravity of the triangle $ABC$