 The Problem

Suppose ABC is any triangle. D, E, F are points on BC, CA, AB respectively such that $(\frac{BD}{DC} = \frac{CE}{EA} = \frac{AF}{FB})$. Prove that the centroids of triangles ABC and DEF coincide.

A little Complex Number

Let A, B, C be points on the Complex plane with complex coordinates a, b, c.

The origin of the plane is made to coincide with the centroid of triangle ABC (the complex coordinates of the vertices adjusted accordingly).

Hence the complex coordinate of the centroid G is 0. $(G = \frac{a+b+c}{3} = 0)$ implying a+b+c= 0.

Next we find the complex coordinates of D, E, F in terms of a, b, c and the given ratio $(\frac{BD}{DC} = \frac{CE}{EA} = \frac{AF}{FB} = k)$ (say)

Using the section formula we find that $(d = \frac{ck+b}{k+1})$ $(e = \frac{ak+c}{k+1})$ $(f = \frac{bk+a}{k+1})$

where d, e, f are the complex coordinates of D, E, F respectively.

Hence the centroid of triangle DEF is given by $(\frac{d+e+f}{3} = \frac{\frac{ck+b}{k+1} + \frac{ak+c}{k+1} + \frac{bk+a}{k+1} } {3} = 0)$ using (a+b+c) = 0.

Thus the centroid of triangle ABC and DEF are same.

A little Projective Geometry

We project triangle ABC into a plane T such that the projection is an equilateral triangle.

(to be continued)