In triangle `A B C ,poin t sD , Ea n dF` are taken on the sides `B C ,C Aa n dA B ,` respectigvely, such that `(B D)/(D C)=(C E)/(E A)=(A F)/(F B)=ndot` Prove that `_(D E F)=(n^2-n+1)/((n+1)^2)_(A B C)dot`

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)`