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

such that DF = 2 FC. Prove that AE CF is a parallelogram whose area is one third of the

area of parallelogram ABCD.

#### Solution

Construction: Draw FG ⊥ AB

Proof: We have

BE = 2EA and DF = 2 FC

⇒ AB - AE = 2EA and DC - FC = 2FC

⇒ AB = 3EA and DC = 3FC

⇒ AE = `1/2` AB and FC = ` 1/3` CD .......... (1)

But AB = DC

Then, AE = DC [opposite sides of ||^{gm}]

Then, AE = FC

Thus, AE = FC and AE || FC.

Then, AECF is a parallelogram

Now ar (||^{gm} = AECF) = AE × FG

⇒ ar (||^{gm} =AECF) =`1/3 ABxx FG ` form ....... (1)

⇒ 3ar (||^{gm} =AECF) = AB × FG ........(2)

and area (||^{gm} =ABCD) = AB × FG ........... (3)

Compare equation (2) and (3)

⇒ 3 ar (||^{gm} =AECF) = area (||^{gm} =ABCD)

⇒ area (||^{gm} =AECF) = `1/3` area (||^{gm} =ABCD)