Question

ABCD is a parallelogram E is a point on AB such that BE=2EA and F is a point on DC such that DF =2FC . Prove that AEFC is a parallelogram whose area is one third of parallelogram ABCD.

Answer 1

ANSWER:-

Given:

ABCD is a parallelogram E is the point on AB such that BE= 2EA & F is a point on DC such that DF= 2FC.

To prove:

AEFC is a parallelogram whose area is one third of a parallelogram ABCD.

Construction:

Draw FG perpendicular to AB.

Proof:

We have,

BE= 2EA &

DF= 2FC [given]

⚫AB - AE= 2AE

&

⚫DC - FC= 2FC

AB= 3AE & DC= 3FC

$= > AE = \frac{1}{3} AB \: and \: FC = \frac{1}{3} DC...........(1)$

AB= DC

then,

AE= FC [opposite side of a ||gm]

So,

AE||FC

such that AE = FC

Therefore,

AECF is a parallelogram.

Now,

Area of parallelogram(AECF):

$= > \frac{1}{3} (AB \times FG) \: \: \: \: \: \: \: \: \: [From \: (1)]$

=)3 area of ||gm AECF=AB × FG..........(2)

&

Area of ||gm ABCD= AB × FG...........(3)

Comparing equation (2) & (3), we get;

=)3 Ar. of ||gm (AECF)= Ar.of ||gm ABCD

=)Ar. of ||gm AECF=1/3Ar. of ||gm ABCD

Hence,

Proved.

Hope it helps ☺️