Existence of the centroid. Prove that the medians of a triangle are concurrent.
Given triangle ABC and medians AD, BE and CF. So, F is themidpoint of AB, E is the midpoint of BC and D is the midpoint of AC by definition of the median. First, draw medians AD and BE and segment CF. Claim: Triangle ABC is similar to triangle ECD. Proof: Angle ACB = angle ECD; AC = 2CE; BC = 2CD; so similar by Side-Angle-Side Similarity Theorem. Claim: DE//AB Proof: Angle CED = Angle CAB and Angle CDE = Angle CBA from similarity of triangles ABC and ECD. Claim: Angle GDE = angle GAB and angle GED = angle GBA. Proof: DE//AB Angle DGE = angle AGB (vertical interior angles are congruent). Thus, triangle ABG is similar to triangle EDG by the Angle-Angle-Angle Similarity Theorem. Thus, DE/AB = GD/GA =1/2, which implies GD = 1/2 GA, which in turn implies GD = 1/3 AD. Also, GE = 1/2 GB, which implies GE = 1/3 BE by the above. Repeating the above for AD paired with CD and BE paired with CD, we see that each pair intersects at a point that cuts each median into two pieces at a point such that the piece closest to a side has 1/2 the length of the piece closest to the vertex. That point can be only one point and that is G, the centroid.