Medians of Triangle Meet at Centroid/Proof 2

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 $(1)$:
 * $\vec r = \vec a + \dfrac 2 3 \paren {\dfrac {\vec b + \vec c - 2\vec a} 2} = \dfrac {\vec a + \vec b + \vec c} 3$

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 $\dfrac {\vec a + \vec b + \vec c} 3$.