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