Medians of Triangle Meet at Centroid

Theorem

Let $\triangle ABC$ be a triangle.

Then the medians of $\triangle ABC$ meet at a single point.

This point is called the centroid of $\triangle ABC$.

Corollary

Let $AA'$ be produced beyond $BC$ to $X$, where $A'X = AG$.

Then the straight lines $B'X$ and $CX$ are parallel to $CC'$ and $BB'$ respectively, and $\dfrac 2 3$ of their length.

Proof 1

Let $A'$ be the midpoint of $BC$.

Let $B'$ be the midpoint of $AC$.

Let $C'$ be the midpoint of $AB$.

Hence $AA'$, $BB'$ and $CC'$ are the medians of $\triangle ABC$.

Let $AA'$ and $BB'$ intersect at $G$.

Hence $A'B'$ is a midline of $\triangle ABC$.

By the Midline Theorem, $A'B'$ is parallel to $AB$ and half the length of $AB$.

We have that $\triangle AGB$ and $\triangle A'GB'$ are similar.

Hence:

 $\ds A'G$ $=$ $\ds \dfrac {AG} 2$ $\ds B'G$ $=$ $\ds \dfrac {BG} 2$

Hence $BB'$ meets $AA'$ $\dfrac 1 3$ of the distance along $AA'$ from $A'$.

That is:

$A'G = \dfrac {AA'} 3$

Similarly, mutatis mutandis, it can be shown that $CC'$ also meets $AA'$ $\dfrac 1 3$ of the distance along $AA'$ from $A'$.

Hence we have shown that $BB'$ and $CC'$ both meet at the same point on $AA'$

That is, the medians of $\triangle ABC$ meet at a single point.

Proof 2

Let $\vec a, \vec b, \vec c$ be $\vec{OA}, \vec{OB}, \vec{OC}$ respectively.

Let the midpoint of $BC, AC, AB$ be $\vec d, \vec e, \vec f$ respectively.

Then:

 $\ds \vec d$ $=$ $\ds \frac {\vec b + \vec c} 2$ $\ds \vec e$ $=$ $\ds \frac {\vec a + \vec c} 2$ $\ds \vec f$ $=$ $\ds \frac {\vec a + \vec b} 2$

The three medians are $\vec{AD}, \vec{BE}, \vec{CF}$ respectively:

 $\ds \vec {AD}$ $=$ $\ds \vec d - \vec a$ $\ds$ $=$ $\ds \frac {\vec b + \vec c - 2 \vec a} 2$ $\ds \vec {BE}$ $=$ $\ds \vec e - \vec b$ $\ds$ $=$ $\ds \frac {\vec a + \vec c - 2 \vec b} 2$ $\ds \vec {CF}$ $=$ $\ds \vec f - \vec c$ $\ds$ $=$ $\ds \frac {\vec a + \vec b - 2 \vec c} 2$

Their equations:

 $\text {(1)}: \quad$ $\ds \vec {AD}: \ \$ $\ds \vec r$ $=$ $\ds \vec a + x \paren {\frac {\vec b + \vec c - 2\vec a} 2}$ $\text {(2)}: \quad$ $\ds \vec {BE}: \ \$ $\ds \vec r$ $=$ $\ds \vec b + y \paren {\frac {\vec a + \vec c - 2\vec b} 2}$ $\text {(3)}: \quad$ $\ds \vec {CF}: \ \$ $\ds \vec r$ $=$ $\ds \vec c + z \paren {\frac {\vec a + \vec b - 2\vec c} 2}$

It can be verified that $x = y = z = \dfrac 2 3$ produce the same point:

When $x = \dfrac 2 3$, from $(1)$:

$\vec r = \vec a + \dfrac 2 3 \paren {\dfrac {\vec b + \vec c - 2\vec a} 2} = \dfrac {\vec a + \vec b + \vec c} 3$

When $y = \dfrac 2 3$, from $(2)$:

$\vec r = \vec b + \dfrac 2 3 \paren {\dfrac {\vec a + \vec c - 2\vec b} 2} = \dfrac {\vec a + \vec b + \vec c} 3$

When $z = \dfrac 2 3$, from $(3)$:

$\vec r = \vec c + \dfrac 2 3 \paren {\dfrac {\vec a + \vec b - 2\vec c} 2} = \dfrac {\vec a + \vec b + \vec c} 3$

Therefore, the three medians meet at a single point, namely $\dfrac {\vec a + \vec b + \vec c} 3$.

$\blacksquare$