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 $\triangle ABC$.

Proof
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:

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

Their equations:

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

When $x = \dfrac 2 3$, from $\left({1}\right)$:
 * $\vec r = \vec a + \dfrac 2 3 \left({ \dfrac {\vec b + \vec c - 2\vec a} {2} }\right) = \dfrac {\vec a + \vec b + \vec c} {3}$

When $y = \dfrac 2 3$, from $\left({2}\right)$:
 * $\vec r = \vec b + \dfrac 2 3 \left({ \dfrac {\vec a + \vec c - 2\vec b} {2} }\right) = \dfrac {\vec a + \vec b + \vec c} {3}$

When $z = \dfrac 2 3$, from $\left({3}\right)$:
 * $\vec r = \vec c + \dfrac 2 3 \left({ \dfrac {\vec a + \vec b - 2\vec c} {2} }\right) = \dfrac {\vec a + \vec b + \vec c} {3}$

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

Also see

 * Definition:Centroid of Triangle