The medians of a triangle are concurrent

In case anyone reading this does not know, a median is a line connecting a vertex of a triangle to the midpoint of the opposite edge. The theorem is that the three medians of a triangle are concurrent (i.e. they meet in a single point). Here are four proofs.

Proof 1: (Transformation geometry)

Let E and F be the midpoints as shown and let BE and CF intersect at G. Consider the dilation about A with factor 2. It sends E to C, F to B and G to Q (this is the definition of Q). Then EG and CQ are parallel, as are FG and BQ. Thus BGCQ is a parallellogram and the diagonals of a parallelogram bisect each other QED.

Proof 2: (Vectors. Efficient and boring). Write {\bf a}, {\bf b} and {\bf c} for A, B and C respectively. Let G be ({\bf a}+{\bf b}+{\bf c})/3. It is easy then to check that the midpoint D of AB is ({\bf a}+{\bf b})/2 and that A, D and G are concurrent.

Proof 3: (Why not just prove Ceva’s Theorem)

Ceva’s Theorem states that in the situation shown, AD, BE and CF are concurrent if and only if

\displaystyle \frac{BD}{DC}\frac{CE}{EA}\frac{AF}{FB}=1.

To prove this, note that in the case of concurrence

\displaystyle \frac{BD}{DC}=\frac{|ABG|}{|ACG|}.

The rest of the proof is routine.

Proof 4: (my favourite) WLOG the triangle is equilateral. Now the statement is obvious (e.g. by symmetry).

Perhaps some elaboration should be made to the WLOG. An affine transformation of \mathbb{R}^2 is a map of the form {\bf{v}}\mapsto A{\bf{v}}+{\bf{b}} where A is an invertible matrix and {\bf{b}} is a vector. The affine transformations act transitively on the set of (nondegenerate) triangles and the property of having concurrent medians is clearly invariant under these transformations.

Acknowledgements: Thanks to Inna Lukyanenko for the first proof and tikz files, and to pdftoppm for converting the .pdf output to .png.