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In triangle A B C ,points D , Ea n dF ar...

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

Text Solution

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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)`
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