The centroid of a triangle is two-thirds the distance from each vertex to the midpoint of the opposite side.

If $\overline{AE},$ $\overline{BD},$ and $\overline{CF}$ are the medians of $△ABC,$ then

This can be proven using midpoints and parallel lines.
Consider $△ABC.$ The points $D$ and $E$ are midpoints on their respective side. Thus, $\overline{AE}$ and $\overline{BD}$ are medians.

The two medians intersect at the point $F.$

Now, two new points are introduced — the midpoints of $\overline{AF}$ and $\overline{BF}.$ Call them $G$ and $H.$

Since $G$ and $H$ are midpoints of $\overline{AF}$ and $\overline{BF},$ $\overline{GH}$ is a midsegment of $△AFB.$ Thus, by the Triangle Midsegment Theorem, $\overline{GH}$ is parallel to and half the length of $\overline{AB}.$

Similarly, $\overline{DE}$ is a midsegment of $△ABC$ since $D$ and $E$ are the midpoints of $\overline{AC}$ and $\overline{BC}.$ Therefore, $\overline{DE}$ is also parallel to and half the length of $\overline{AB}.$ It follows that

Since $\overline{DE}$ and $\overline{GH}$ are parallel and congruent, $DEHG$ are the vertices of a parallelogram. By the Parallelogram Diagonals Theorem, the diagonals of a parallelogram bisect each other. Therefore, $\overline{DF}\cong \overline{FH}$ and $\overline{GF}\cong \overline{FE}.$

Thus, the median $\overline{DB}$ intersects $\overline{AE}$ at two-thirds of the distance from $A.$ Now, by applying the same reasoning for the third median, it also intersects $\overline{AE}$ at two-thirds from $A.$

The median from point $C$ also intersects $\overline{AE}$ at two-thirds from point $A.$ Therefore, the centroid of a triangle is two-thirds the distance from each vertex to the midpoint of the opposite side.