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Question -

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 AB CD.



Answer -

Given:
(1) ABCD is a parallelogram.
(2) E is a point on BA such that BE = 2EA 
(3) F is a point on DC such that DF = 2FC.
To find:
Area of parallelogram 
Proof: We have,
BE = 2EA and DF = 2FC
 AB − AE = 2AE and DC − FC = 2FC
 AB = 3AE and DC = 3FC
 AE =  AB and FC =  DC
 AE = FC [since AB = DC]
Thus, AE || FC such that AE = FC
Therefore AECF is a parallelogram.
Clearly, parallelograms ABCD and AECF have the same altitude and
AE =  AB.
Therefore
 
Hence proved that  

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