Re: [math-fun] Triangle 1/7th of generating ABC triangle

Given triangle ABC, A' divides side BC in the ratio of m:1, B' divides CA in the ratio n:1, and C' divides AB in the ratio p:1. Now draw cevians AA', BB', and CC'. The ratio of the area of the center triangle to the area of the original triangle is: f(m,n,p) = (mnp-1)^2/((mn+n+1)(mp+m+1)(np+p+1))

Lovely. Is there an equally lovely proof? -- g