Question
Parallelograms on the same base and between same parallels are equal in area. Prove this.
Solution
Given: Two parallelograms ABCD and EFCD, on the same base DC and between the same parallels AF and DC.
To Prove: ar(ABCD) = ar(EFCD)
Proof: In ΔADE and ΔBCF,
∠DAE = ∠CBF ...(1)
Corresponding angles
(∵ AD || BC and a transversal AF intersects them)
∠AED = ∠BFC ...(2)
Corresponding angles (∵ ED || FC and a transversal AF intersects them)
∴ ∠ADE = ∠BCF ...(3)
| Angle sum property of a triangle Also AD = BC ...(4)
Opposite sides of the parallelogram ABCD
∴ ΔADE ≅ ΔBCF | By ASA congruence rule, using (1), (3) and (4) ∴ ar(ΔADE) = ar(ΔBCF)
| ∵ Congruent figures have equal areas ⇒ ar(ΔADE) + ar(EDCB)
= ar(ΔBCF) + ar(EDCB)
| Adding ar(EDCB) to both sides ⇒ ar(ABCD) = ar(EFCD)
So, parallelograms ABCD and EFCD are equal in area.
