Problem with medians in triangles

Question

It would appear that the segment passing through the centroid and being parallel to one of a triangle's sides is 2/3 the side to which it is parallel. I would be grateful if someone could help me out in comprehending why this is so. I have the feeling that the medians theorem should or might be resorted to in this demonstration, but I cannot find the path. Please do not consider anything trivial. And be as much as explicative as possible.

Answer 1

Yes. This is due to the classical result that the centroid of a triangle divides each median in two segments, the longest of which is twice the shortest. More precisely, if $M$ is the centroid of $\triangle ABC$ and $AA'$ is a median, then $\overline{AM}=2\overline{MA'}$ and so forth. Now, if $D\in AB$ and $E\in AC$ are such that $ED\parallel BC$ and $M\in ED$, then by similarity of $\triangle ADE$ and $\triangle ABC$ we have $$ED:BC=AD:AB$$ and, by the parallel lines theorem $$AD:AB=AM:AA'=2:3$$