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!
Best Answer
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