In the sides $BC,CA,AB$ are taken three points $A',B',C'$ such that $BA':A'C=CB':B'A=AC':C'B=m:n$.Prove that if $AA',BB',CC'$ are joined they will form by their intersections a triangle whose area is to that of the triangle ABC as $\frac{(m-n)^2}{m^2+mn+n^2}$
I dont know how to find the area of inner $A'B'C'$ and how to relate with area of outer triangle.Please help.