2018 AMC 8 Problema 20

Use similarity from the two parallel-line conditions.

Compare the areas of the two small corner triangles to the area of $\triangle ABC$.

Let the area of $\triangle ABC$ be equal to $t.$ Since $DE \parallel BC$ and $FE \parallel CA ,$ we can deduce that $$ADE \sim ABC$$ and $$EFB \sim ABC.$$ Since $AE = \dfrac{AB}3,$ the area of $ADE$ is equal to $\left(\dfrac13\right)^2 t = \dfrac{t}{9} .$ Since $EB = \dfrac{2AB}3,$ the area of $EFB$ is equal to $\left(\dfrac23\right)^2 t = \dfrac{4}{9}t .$ Finally, to find the area of $CDEF,$ we take the area of $ABC =t$ and subtract the areas of $ADE$ and $EFB.$ This is equivalent to the expression $$t- \frac{t}{9} - \frac{4t}{9} = \frac{4t}{9} .$$ Therefore, the ratio of the area of $CDEF$ and $ABC$ is $\dfrac{\left(\dfrac{4t}{9}\right)}{t} = \dfrac{4}{9} .$

Thus, A is the correct answer.