Points $$C$$ and $$D$$ lie on opposite sides of the line $$AB$$. Let $$M$$ and $$N$$ be the centroids of the triangles $$ABC$$ and $$ABD$$ respectively. If $$AB = 25$$, $$BC = 24$$, $$AC = 7$$, $$AD = 20$$ and $$BD = 15$$, find $$MN$$.

Since $$M = \frac{A+B+C}{3}$$ and $$N = \frac{A+B+D}{3}$$, we get $$MN = \frac{CD}{3}$$. Both triangles are right angled, at $$C$$ and at $$D$$, so with $$A = (0,0)$$ and $$B = (25,0)$$ the feet give $$C = (1.96, 6.72)$$ and $$D = (16, -12)$$. Then $$CD = \sqrt{14.04^2 + 18.72^2} = 23.4$$ and $$MN = \frac{23.4}{3} = 7.8$$.