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!

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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.

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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).


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.


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

AD = (AG + GD) = (2GD + GD) = 3GD.

Hence GD = AD/3. QED

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.

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.

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