The centroid of a triangle

Identify points with their position vectors. Let $M=\frac12(B+C)$ be the midpoint of $BC$, so $AM$ is the median from $A$.

Define $G=\dfrac{A+B+C}3$. Then
$$A+\frac23(M-A)=\frac A3+\frac{2M}3=\frac A3+\frac{B+C}3=G.$$
So $G$ lies on the median $AM$, at two thirds of the way from $A$ to $M$.

The expression $\frac{A+B+C}3$ is symmetric in $A,B,C$, so the same computation shows $G$ lies on the other two medians, also at two thirds from the vertex. Hence the three medians are concurrent at $G$.