two parallelograms ABCDand AEFB are on opposite side of AB.Prove
ar(ABCD)+ar(AEFB)=ar(EFCD)

Dear student,

Following is the answer to your query:


Extend AB and let it intersect ED at P and FC at Q.
Now in EPQF, 
EF || PQ      (since EF || AB)
EP || FQ      (since ED || FC)

Therefore, EPQF is a parallelogram.
similarly, PQCD is also a parallelogram.

Now ar(AEFB) = ar(EPQF) and ar(ABCD) = ar(PQCD)  (since parallelograms on the same base and between same parallel lines are equal in area)

Therefore, ​ar(AEFB) + ar(ABCD) = ar(EPQF) + ​ar(PQCD) = ar(EFCD)
Hence proved.