Position of Centroid of Triangle on Median/Proof 3

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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

CentroidOfTriangle.png

By Medians of Triangle Form Six Triangles of Equal Area‎:

these six triangles formed by the medians of $\triangle ABC$ have equal area:
  • $\triangle AGN$
  • $\triangle BGN$
  • $\triangle BGL$
  • $\triangle CGL$
  • $\triangle CGM$
  • $\triangle AGM$

Without loss of generality consider one of the three medians of $\triangle ABC$, $AGL$.

The following triangles both have their base on $AGL$ and share vertex $C$:

  • $\triangle CGA$
  • $\triangle CGL$

Since $\triangle CGA$ contains two of the small triangles:

\(\ds \leadsto \ \ \) \(\ds \AA \paren { \triangle CGA }\) \(=\) \(\ds \AA \paren { \triangle AGM } + \AA \paren { \triangle CGM }\)
\(\ds \dfrac { \AA \paren { \triangle CGA } } { \AA \paren { \triangle CGL } }\) \(=\) \(\ds \dfrac { \AA \paren { \triangle AGM } + \AA \paren { \triangle CGM } } { \AA \paren { \triangle CGL } }\)

Since

$\AA \paren { \triangle AGM } = \AA \paren { \triangle CGM } = \AA \paren { \triangle CGL }$
\(\ds \leadsto \ \ \) \(\ds \dfrac 2 1\) \(=\) \(\ds \dfrac { \AA \paren { \triangle AGM } + \AA \paren { \triangle CGM } } { \AA \paren { \triangle CGL } }\)
\(\ds \dfrac { \AA \paren { \triangle CGA } } { \AA \paren { \triangle CGL } }\) \(=\) \(\ds \dfrac 2 1\) Common Notion $1$
\(\ds \dfrac { \AA \paren { \triangle CGA } } { \AA \paren { \triangle CGL } }\) \(=\) \(\ds \dfrac { AG } { GL }\) Areas of Triangles and Parallelograms Proportional to Base
\(\ds \dfrac { AG } { GL }\) \(=\) \(\ds \dfrac 2 1\) Common Notion $1$
\(\ds \leadsto \ \ \) \(\ds AG\) \(=\) \(\ds 2 \cdot GL\)

mutatis mutandis:

$BG = 2 \cdot GM$
$CG = 2 \cdot GN$

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.

$\blacksquare$

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