Proving that the medians of a triangle are concurrent I was wondering how to prove Euclid's theorem: The medians of a triangle are concurrent.
My work so far:
First of all my interpretation of the theorem is that if a line segment is drawn from each of the 3 side's medians to the vertex opposite to it, they intersect at one point.
Since a triangle has three sides and each side must have a median, I figure that at least 2 of them have to intersect as the lines can't be parallel. 
May anyone explain further? Thank you!
 A: I could do that by using Thales's Theorem. Sorry if I did it on a paper. It is really hard to do on this page.

A: This is my short proof from 1963:
In the triangle ABC draw medians BE, and CF, meeting at point G.
Construct a line from A through G, such that it intersects BC at point D.
We are required to prove that D bisects BC, therefore AD is a median, hence medians are concurrent at G (the centroid).
Proof:
Produce AD to a point P below triangle ABC, such that AG = GP.
Construct lines BP and PC.
Since AF = FB, and AG = GP, FG is parallel to BP. (Euclid)
Similarly, since AE = EC, and AG = GP, GE is parallel to PC
Thus BPCG is a parallelogram.
Since diagonals of a parallelogram bisect one another (Euclid),
therefore BD = DC.
Thus AD is a median.  QED
Corollary:  GD = AD/3.
Proof:
Since AG = GP and GD = GP/2, AG = 2GD.
AD = (AG + GD) = (2GD + GD) = 3GD.
Hence GD = AD/3.   QED

A: I have been trying to answer this question for an AS question, using only methods already shown in the course book. I'm sorry for the lousy handwriting, hope it's readable. 

A: If you know Ceva's Theorem, apply it. Note that if $ D,E$ and $F$ are the midpoints of sides, $AD/DB = BE/EC = CF/FA = 1$
Therefore proved by converse of Ceva's Theorem.
