Question

Given a triangle ABC with A(6b,6c) B(0,0) and C (6a,0), prove that the medians of the triangle are concurrent at a point that is two thirds of the way from any vertex to the midpoint of the opposite side.

I'm not sure how to prove this. I tried and when I got to finding the centroid I got a very weird equation. Help?

Answer 1

Not just for your particular triangle, but for any triangle at all, the intersection of the medians is 2/3 of the way from each vertex to the opposite side midpoint. For proof see:
https://www.khanacademy.org/math/geometry/triangle-properties/medians_centroids/v/proving-that-the-centroid-is-2-3rds-along-the-median