# Position of Centroid of Triangle on Median

## 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

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

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

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 \ \$ $\ds AG$ $=$ $\ds 2 \cdot GL$
$BG = 2 \cdot GM$
$CG = 2 \cdot GN$

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.

$\blacksquare$