Question

ABCD is a parallelogram E is a point on BA such that BE=2EA and F is a point on DC such bthat DF=2FC prove that AECF is a parallelogram whose area is one third of the area of parallelogram ABCD

Answer 1

Here is your answer I hope you got that.....

Answer 2

Proved below.

Step-by-step explanation:

Given:

We have, E is a point on BA

⇒BE = 2 EA

And,  F is a point on DC

⇒DF = 2FC

AB - AE = BE

AB - AE =2 AE            [∵BE = 2 AE]

AB = 3 AE  

$AE=\frac{1}{3}AB$               [1]

And,

DC - FC = DF

DC - FC = 2 FC            [∵DF = 2FC]

DC = 3 FC

$FC= \frac{1}{3}DC$                [2]

But,

AB = DC                         [given]

Then,

AE = FC (Opposite sides of a parallelogram)

Thus,

AE || FC such that AE = FC  

Then,  AECF is a parallelogram.

Draw FG perpendicular to AB.

Now, Area of parallelogram $(AECF) = \frac{1}3} (AB \times FG)$          [from 1 and 2]

⇒$3(AECF) = (AB \times FG)$                              [3]

And, area of parallelogram $ABCD = AB \times FG$   [4]

Compare equation (3) and (4), we get

3 Area of parallelogram AECF = Area of parallelogram ABCD

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

Hence, proved.