Question

Abcd is a parrallogram,a line through a cuts dc at point p and bc produced at q.Prove that trianglebcp is equal in area to triangle dpq

Answer 1

Construction: Join A & C. Proof: BC extends upto point Q. ∴ CQ is also parallel to AD. △ ACQ and △ DQC are on the same base CQ and between the same parallels CQ ∥ AD, so they are equal in area ,i.e., Area (△ ACQ) = Area (△ DQC) Subtracting Area (△ CPQ) from both the triangles, Area (△ ACQ) - Area (△ CPQ) = Area (△ DQC) - Area (△ CPQ) ⇒ Area (△ APC) = Area (△ DPQ) .....(i) Since, AB ∥ DC (∵ Opposite sides of a parallelogram are equal) ∴ AB ∥ PC (∵ PC is a part of DC) △ APC and △ BCP are on the same base PC and between the same parallels PC ∥ AB, so they are equal in area ,i.e., Area (△ APC) = Area (△ BCP) .....(ii) From (i) and (ii), Area (△ BCP) = Area (△ DPQ)


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Answer 2

Answer:

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Step-by-step explanation:

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