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

ABCD is a parallelogram. E is a point on BA such that BE = 2EA and F is point on DC such that DF = 2FC. Prove that AECF is a parallelogram whose area is one third of the area of parallelogram ABCD.

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)

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