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.

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 = AB and FC = 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) = (AB * FG) [From (i)


3 Area of parallelogram AECF = AB * FG (ii)


And,


Area of parallelogram ABCD = AB * FG (iii)


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


3 Area of parallelogram AECF = Area of parallelogram ABCD


Area of parallelogram AECF = Area of parallelogram ABCD


Hence, proved



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