The three medians concur at the centroid

Step 1 (midpoint coordinates). Place $\triangle ABC$ in a coordinate plane with vertices $A$, $B$, $C$. The midpoint $M$ of side $BC$ is the average of $B$ and $C$: $M = \left(\dfrac{x_B+x_C}{2},\, \dfrac{y_B+y_C}{2}\right)$.

Step 2 (point two-thirds along $AM$). Move from $A$ toward $M$ by $\dfrac{2}{3}$ of the segment $AM$: the resulting point is $A + \dfrac{2}{3}(M - A) = \dfrac{1}{3}A + \dfrac{2}{3}\cdot\dfrac{B+C}{2} = \dfrac{A+B+C}{3}$, giving coordinates $G = \left(\dfrac{x_A + x_B + x_C}{3},\, \dfrac{y_A + y_B + y_C}{3}\right)$.

Step 3 (symmetry forces concurrency). Notice that the final formula $G = \left(\dfrac{x_A + x_B + x_C}{3},\, \dfrac{y_A + y_B + y_C}{3}\right)$ is completely symmetric in $A$, $B$, $C$ — swapping the roles of $A$ and $B$ (to find the point $\dfrac{2}{3}$ of the way along median $BN$) or of $A$ and $C$ (along $CP$) produces the exact same point $G$. Therefore all three medians pass through $G$, and $G$ lies $\dfrac{2}{3}$ of the way from each vertex to the opposite midpoint ($AG = \dfrac{2}{3} AM$).