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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 = 2FC. 

Prove that AE CF is a parallelogram whose area is one third of the area of parallelogram ABCD.

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Best answer

Construction : 

Draw FG perpendicular to AB 

Proof : 

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 = \(\frac{1}{3}\) AB and FC = \(\frac{1}{3}\)DC  ...(i)

But, 

AB = DC 

Then, 

AE = FC 

(Opposite sides of a parallelogram) 

Thus, 

AE || FC 

Such that,

AE = FC 

Then, 

AECF is a parallelogram 

Now, 

Area of parallelogram (AECF) = \(\frac{1}{3}\)(AB x FG) [From (i)] 

3 Area of parallelogram AECF = AB x FG ..(ii) 

And,

Area of parallelogram ABCD = AB x FG ...(iii) 

Compare equation (ii) and (iii), we get 

3 Area of parallelogram AECF = Area of parallelogram ABCD 

Area of parallelogram AECF = \(\frac{1}{3}\)Area of parallelogram ABCD

Hence, proved

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