Position of Centroid of Triangle on Median/Proof 2
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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:
and so:
The result follows.
$\blacksquare$