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Question:

In fig, ABCD is a parallelogram and BC is produced to a point Q such that AD=CQ. If AQ intersect DC at P, show that ar(BPC)=ar(DPQ).

Solution:

JoinAC.As we know that area of triangles on the same base and between the same parallels lines are equalTherefore ,ar(APC)=ar(BPC)... (1)Now In quadrilateralACQD, we haveAD=CQand ,AD∥CQ[Given]Therefore, this quadrilateralADQCwith one pair of opposite sides is equal and parallel is parallelogram.ThereforeADQCis a parallelogram.⇒AP=PQandCP=DP[Since diagonals of a|| gm bisect each other]InΔs APCandDPQwe haveAP=PQ[Proved above]∠APC=∠DPQ[Vertically opp.∠s]and ,PC=PD[Proved above]Therefore by SAS criterion of congruence,ΔAPC≅ΔDPQ⇒ar(APC)=ar(DPQ)... (2)[Since congruentΔshave equal area]Thereforear(BPC)=ar(DPQ)[From (1)]Hence ,ar(BPC)=ar(DPQ)