Theorem: Prove that the three medians of a triangle are concurrent and that the point of concurrence, the centroid, is two-thirds the distance from each vertex to the opposite side.

Proof:

Let ABC be a triangle and D and F be the midpoints of AB and AC respectively.

In triangles ADF and ABC, since <DAF = <BAC and AB = 2AD and AC = 2AF, they are congruent by SAS Congruency Theorem. It immediately follows that <ADF = <ABC and <AFD = <ACB; therefore, DF//BC.

In triangles GFD and GBC, since DF//BC, <GFD = <GBC, <GDF = <GCB (properties of alternate interior angles) and <DGF = <CGB (sum of the angles of a triangle=180). Therefore, it follows from AAA CongruencyTheorem that the triangles GFD and GBC are congruent.

So, we have DF/BC=GF/GB=1/2 or GF=1/2GB; therefore, GF=1/3BF.

Similarly, median CD intersects BF at G so that GD=1/3CD.

If we repeat the same calculations for the pairs of medians AE & CD, and AE & BF, we see they intersects at a point such that it cuts a median into two pieces so that the piece closest to the side is half of the piece closest to the vertex. That point could be one, and it is G.