Proving that the medians of a triangle are concurrent

Question

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!

http://jwilson.coe.uga.edu/EMT668/EMAT6680.F99/McCallum/Assignment%204/Median%20Concurrence.html — lab bhattacharjee, Jun 05 '13 at 06:19

score 13 · Answer 1 · edited Jul 14 '18 at 15:44

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

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score 6 · Accepted Answer · answered Jun 05 '13 at 06:54

6

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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answered Jun 05 '13 at 06:54

Mikasa

67,374

Paper is perfectly fine. Thank you for your time. – Sujaan Kunalan Jun 05 '13 at 06:56
You're Welcome. – Mikasa Jun 05 '13 at 06:57
5

original style of proof +1 – Adi Dani Jun 06 '13 at 16:08
1

This isn't a proof, but you might be interested in checking out Ceva's theorem. – Steven Alexis Gregory Aug 17 '15 at 06:35

score 2 · Answer 3 · edited Mar 13 '17 at 10:47

2

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.

edited Mar 13 '17 at 10:47

Glorfindel

3,955

answered Mar 13 '17 at 10:18

Julian Antonio

31

score 0 · Answer 4 · answered Oct 09 '16 at 04:04

0

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.

answered Oct 09 '16 at 04:04

SANTANU NANDI

301

Proving that the medians of a triangle are concurrent

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