ABCD is a parallelogram. E is a point on BA such that BE = 2EA and F is point on DC such that DF = 2FC.

Solution:

Draw FG ⊥ AB

We have,

BE = 2 EA and DF = 2FC

⇒ AB - AE = 2 AE and DC - FC = 2 FC

⇒ AB = 3 AE and DC = 3 FC

⇒ AE = (1/3) AB and FC = (1/3)DC ⋅⋅⋅⋅ (1)

But AB = DC

Then, AE = FC     [opposite sides of ∥^gm]

Thus, AE = FC and AE ∥ FC

Then, AECF is a parallelogram

Now, area of parallelogram AECF = AE × FG

⇒ ar(∥^gm AECF) = 1/3 AB × FG       from(1)

⇒ 3ar (∥^gm AECF) = AB × FG            ⋅⋅⋅ (2)

And ar(∥^gm ABCD) = AB × FG ⋅⋅⋅ (3)

Compare equation 2 and 3

⇒ 3ar(∥^gm AECF) = ar(∥^gm ABCD)

⇒ ar(∥^gm AECF) = 1/3 ar(∥^gm ABCD)