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:
and so:
The result follows.
$\blacksquare$
Proof 3
By Medians of Triangle Form Six Triangles of 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:
and so:
The result follows.
$\blacksquare$
Sources
- 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): centroid
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): centroid