Question

In the given figure, ABCD and AEFG
are two parallelograms. Prove that
ar(//gm ABCD) = ar(//AEFC
G).​

Answer 1

I have to understand that

Answer 2

Answer:

Step-by-step explanation:

const: join bg

proof: llgm AEFG and triangle ABG are on the same base ag and between same ll lines.

so, ar(ABG) = 1/2ar(AEFG)

similarly,

ar(ABG) = 1/2ar(ABCD)

from above statements

ar(AEFG)=ar(ABCD)