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!

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. • Paper is perfectly fine. Thank you for your time. – Sujaan Kunalan Jun 5 '13 at 6:56
• You're Welcome. – mrs Jun 5 '13 at 6:57
• original style of proof +1 – Adi Dani Jun 6 '13 at 16:08
• This isn't a proof, but you might be interested in checking out Ceva's theorem. – steven gregory Aug 17 '15 at 6:35

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

Proof:

Since AG = GP and GD = GP/2, AG = 2GD.

AD = (AG + GD) = (2GD + GD) = 3GD. If you know Ceva's Theorem, apply it. Note that if $D,E$ and $F$ are the midpoints of sides, $AD/DB = BE/EC = CF/FA = 1$ 