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!