Given: AB || DC in the given figure.
To Prove: ar(BDE) = ar(ACED)
Proof: ∵ ΔADC and ΔBDC are on the same base DC and between the same parallels AB and DC ∴ ar(ΔADC) = ar(ΔBDC)
⇒ ar(ΔADC) + ar(ΔDCE)
= ar(ΔBDC) + ar(ΔDCE)
| Adding ar(ΔDCE) to both sides ⇒ ar(ACED) = ar(BDE)
⇒ ar(BDE) = ar(ACED)
In the given figure, ABED is a parallelogram in which DE = EC. Show that area (ABF) = area (BEC)
In the following figure, ABCD is a parallelogram and EFCD is a rectangle. Also, AL ⊥ DC. Prove that
(i) ar(ABCD) = ar(EFCD)
(ii) ar(ABCD) = DC x AL.