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.
Given an arbitrary triangle, OBA.
Now I will construct the three medians of the triangle.
Notice that I have constructed P, Q, and R as distinct points. My goal is to show that P = Q = R are equal when Q is the point (2/3) of the way from O to M, when P is the point (2/3) of the way from B to L, and when R is the point (2/3) of the way from A to N. Before I start my proof I will let L, M, and N be the midpoints of OA, AB, and OB. I will also let them be vectors where point O is the origin and vector OA = x and vector OB = y.
OP = OB + BP = OB + (2/3)BL = y + (2/3)(.5x – y)
= (1/3)x + (1/3)y
OQ = (2/3)OM = (2/3)(.5(x + y)) = (1/3)(x + y)
OR = OA + AR
= OA + (2/3)AN
= x + (2/3)(.5y – x) = (1/3)x + (1/3)y.
From this we see that OP = OQ = OR, which implies P = Q = R.