When you draw two medians of a triangle they will intersect at some point. It turns out that when you draw a third median, it will intersect the other two at the same point (that is, they are concurrent). This also happens when you draw perpendicular bisectors (problem 5.1), altitudes (problem 5.2) and angle bisectors (problem 5.21). The first goal of this problem is to prove this fact.

When the problem says "further, show that the centroid divides the medians into a ratio of $2:1$", it is asking you to also show that the point where two medians intersect (so the centroid) divides each of the medians in a $2:1$ ratio. That is, if in $\triangle ABC$, $D$, $E$ and $F$ are the midpoints of $AB$, $BC$ and $CD$, respectively, then the medians $AD$, $BE$, and $CF$ intersect at a point $G$, and $$AG:GD = BG : GE = CG:GF= 2:1.$$