In a `/_\ABC `points D,E,F are taken on the sides BC,CA and AB respectively such that `(BD)/(DC)=(CE)/(EA)=(AF)/(FB)=n` prove that `/_\DEF= (n^2-n+1)/((n+1)^2)` /_\ABC`

Explanation

Take A as the origin and last the postion vectors of point B and C be `vecb and vecc`, Respectively.
Therefore, the position vectors, of D, E and F are, respectively, `(nvecc+vecb)/(n+1),vecc/(n+1)and (n vecb)/(n +1)`. Therefore,
`vec(ED)=vec(AD)-vec(AE)=((n-1)vecc+vecb)/(n+1)and vec(EF)=(nvecb-vecc)/(n+1)`
vector area of `DeltaABC=1/2(vecbxxvecc)`
vector area of `DeltaDEF=1/2(vec(EF)xxvec(ED))`
`1/(2(n+1)^(2))[(nvecb-vecc)xx{(n-1)vecc+vecb}]`
`= 1/(2(n+1)^(2))[(n^(2)-n)vecbxxvecc+vecbxxvecc]`
`1/(2(n+1)^(2))[(n^(2)-n+1)(vecbxxvecc)]=(n^(2)-n+1)/((n+1)^(2))Delta_(ABC)`