**Proof That the Medians of a Triangle are
Concurrent**

**Michael McCallum**

The medians of a triangle are concurrent and the point of concurrence, the centroid, is one-third of the distance from the opposite side to the vertex along the median.

Proof:Given triangle ABC and medians AE, BD and CF. So F is the midpoint of AB, E is the mipoint of BC and D is the midpoint of AC by definition of the median.

First, draw medians AE and BD and segment DE.

Claim: Triangle ABC is similar to triangle DEC.

Proof: Angle ACB = angle DCE; AC = 2CD; BC = 2CE; so similar by Side-Angle-Side Similarity Theorem.

Claim: DE//AB

Proof: Angle CDE = Angle CAB and Angle CED = Angle CBA from similarity of triangles ACE and DCE.

Claim: Angle GED = angle GAB and angle GDE = angle GBA.

Proof: DE//AB

Claim: Angle DGE = angle AGB

Proof : Vertical interior angles are congruent.

Thus triangle ABG is similar to triangle EDG by the Angle-Angle-Angle Similarity Theorem.

Thus DE/AB = GE/GA =1/2, which implies GE = 1/2 GA, which in turn implies GE = 1/3 AE. Also GD = 1/2 GB, which implies GD = 1/3 BD by the above.

Repeating the above for AE paired with CE and BD paired with CE, 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.

For a Geometer Sketch Pad demonstration of this, click here.