The medians meet at the centroid

Statement

Let $ABC$ be a triangle in the Euclidean space $\mathbb{R}^n$, and let $G = \frac{1}{3}(A + B + C)$ be its centroid. Then $G$ lies on each of the three medians of $ABC$. More precisely, let $M_A$ be the midpoint of $B$ and $C$. Then $G = A + \frac{2}{3}(M_A - A)$, and the same holds for the vertices $B$ and $C$.

Remarks

The formula shows that $G$ divides each median in the ratio $2 : 1$, counted from the vertex. The proof is one line: $A + \frac{2}{3}\left(\frac{B + C}{2} - A\right) = \frac{1}{3}(A + B + C)$. The statement is true in each affine space over a field in which $2$ and $3$ are invertible. Mathlib states it for each simplex.

Explanations

L1 human

Draw the three medians of a paper triangle. They always cross at one point. This point is two thirds of the way from each vertex to the middle of the opposite side.