Proof that the medians meet at the centroid

Proof

We compute the point two thirds of the way along each median.

1. Let $M_A = \frac{1}{2}(B + C)$, $M_B = \frac{1}{2}(C + A)$ and $M_C = \frac{1}{2}(A + B)$ be the midpoints of the three sides.
2. We have $A + \frac{2}{3}(M_A - A) = \frac{1}{3} A + \frac{2}{3} \cdot \frac{1}{2}(B + C) = \frac{1}{3}(A + B + C)$.
3. By the definition of the centroid, the right side of step 2 is $G$. So $G = A + \frac{2}{3}(M_A - A)$.
4. In the same way, $B + \frac{2}{3}(M_B - B) = \frac{1}{3} B + \frac{1}{3}(C + A) = G$ and $C + \frac{2}{3}(M_C - C) = \frac{1}{3} C + \frac{1}{3}(A + B) = G$.
5. The median from $A$ is the set of the points $A + t (M_A - A)$ with $t \in [0, 1]$. Since $\frac{2}{3} \in [0, 1]$, step 3 shows that $G$ lies on it.
6. In the same way, step 4 with $t = \frac{2}{3}$ shows that $G$ lies on the medians from $B$ and from $C$.
7. So $G$ lies on each of the three medians, and $G = A + \frac{2}{3}(M_A - A)$, with the same formula at $B$ and at $C$.