Let $C',B',A'$ be points respectively on sides $AB,AC,BC$ of $\triangle ABC$ satisfying $ \tfrac{AB'}{B'C}= \tfrac{BC'}{C'A}=\tfrac{CA'}{A'B}=k$. Prove that the ratio of the area of the triangle formed by the lines $AA',BB',CC'$ over the area of $\triangle ABC$ is $\tfrac{(k-1)^2}{(k^2+k+1)}$.