Concurrency of the Medians of a Triangle

by

Mike Rosonet

This assignment will prove that the three medians of a triangle are concurrent at a point called the **Centroid** and that the **Centroid** is two-thirds the distance from each vertex to the opposite side.

Proof:

Let ABC be a triangle, and let the points D and E be the midpoints of AB and AC, respectively.

Construct the segment DE. Also, construct the medians BE and CD, and let them meet at the point G.

Since any segment joining the midpoints of two sides of a triangle is parallel to the base (which is the remaining side of the triangle) and equal to one-half of the base, then segment DE is parallel to segment BC.

Thus, it is true that and because they are alternate interior angles.

Also, segment DE is one-half of segment BC. Therefore, BC = 2DE.

Thus, it is true that by Double-Angle-Side-Angle.

In particular, BG = 2GE. Thus, G is the point on the median BE that is 2/3 of the way from B to E.

Likewise, CG = 2 GD. Thus, G is the point on the median CD that is 2/3 of the way from C to D.

Similarly, if F is the midpoint of segment BC, then the third median AF will pass through the point G and G will be 2/3 of the way from A to F.

Thus, all three medians meet at the point G, which is the

Centroid, and theCentroidlies on each median 2/3 of the way from the vertex to the midpoint of the opposite side.