Position of Centroid of Triangle on Median/Proof 2

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.

Proof

Let $\triangle ABC$ be a triangle.

Let $AL$, $BM$ and $CN$ be the medians of $\triangle ABC$.

Let the medians be concurrent at the centroid, $G$.

By the definition of median, the sides of $\triangle ABC$ are bisected.

\(\ds BL\)

\(=\)

\(\ds LC\)

by hypothesis

\(\ds BC\)

\(=\)

\(\ds BL + LC\)

addition

\(\text {(1)}: \quad\)

\(\ds \leadsto \ \ \)

\(\ds \frac {BC} {LC}\)

\(=\)

\(\ds 2\)

\(\ds AN\)

\(=\)

\(\ds NB\)

by hypothesis

\(\text {(2)}: \quad\)

\(\ds \leadsto \ \ \)

\(\ds \frac {AN} {NB}\)

\(=\)

\(\ds 1\)

\(\ds \dfrac {LG} {GA} \cdot \dfrac {AN} {NB} \cdot \dfrac {BC} {CL}\)

\(=\)

\(\ds -1\)

Menelaus's Theorem

\(\ds CL\)

\(=\)

\(\ds -LC\)

Definition of Directed Line Segment

\(\ds \leadsto \ \ \)

\(\ds \dfrac {LG} {GA} \cdot \dfrac {AN} {NB} \cdot \dfrac {BC} {LC}\)

\(=\)

\(\ds 1\)

\(\ds \dfrac {LG} {GA} \cdot 2\)

\(=\)

\(\ds 1\)

substitute from $(1)$ and $(2)$

\(\ds \dfrac {GA} {LG}\)

\(=\)

\(\ds 2\)

rearranging

Hence:

the distance from the vertex to the centroid is $2/3$ of the whole length of the median.

and so:

the distance from the centroid to the side is $1/3$ of the whole length of the median.

The result follows.

$\blacksquare$