Position of Centroid of Triangle on Median

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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 1

CentroidOfTriangle.png


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.

$\blacksquare$


Proof 2

CentroidOfTriangle.png

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.

$\blacksquare$


Proof 3

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.

$\blacksquare$


Sources