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Question

In the adjoining figure ,ABCD is a parallelogram. Any line through A cuts DC at a point P and BC produced at Q. Prove that : ar(ΔBPC)=ar(ΔDPQ).
1293878_f8df6de59fdb466294cd8e3124285b76.1293878-Q

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Solution

R.E.F image
Prove:- Area BPC = Area DPQ
form figure ---ACQ & DQC
CQ||AD
CQ=CQ same base
Area ACQ=areaDQC
(Between parallel lines & same base area of the triangle
should be equal)
Area ACQ = Area DQC
subtract the area CPQ form both side
Area ACQ - Area CPQ = Area DQC - Area CPQ.
Area APC=Area DPQ----(1)
sin a AB||PC ( PC is a part of DC)
APC & BCP are on the same base PC & between the same parallels PC||AB so they are equal in
Area ie Area (APC) = Aera (BCP)-----(ii)
form eq(i) & eq(ii)
Area (BCP) = Aera DPQ

1199672_1293878_ans_fc7c73b4be194b2689b859a9e5643c4a.jpg

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