How can we prove this as theorem:

Parallelogram with same base and having equal areas lie between the same parallel.

Please let me know the answer as fast as possible

Dear Student!

Given: ABCD and ABEF are parallelograms on the same base AB. Area of parallelogram ABCD = Area of parallelogram ABEF

To prove: AB || DE

Construction: Draw DP ⊥ AB and FQ ⊥ AB

Proof: 

Area of parallelogram ABCD = AB × DP  ...(1)  (Area of parallelogram = Base × Corresponding attitude)

Area of parallelogram ABEF = AB × FQ  ...(2)

Given, Area of parallelogram ABCD = Area of parallelogram ABEF

∴ AB × DP = AB × FQ

⇒ DP = FQ

Hence, the distance between the sides AB and CD and AB and FE is equal.

∴ AB || DE

Thus, the parallelograms ABCD and ABEF lie between the same parallel lines AB and DE.

Cheers!