Clay Bennett

Asssignment 4.14

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.