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 : $ar\left(\Delta BPC\right)=ar\left(\Delta DPQ\right).$

Answer 1

R.E.F image

Prove:- Area $△BPC$ = Area $△DPQ$

form figure ---$△ACQ$ & $△DQC$

$CQ\left|\right|AD$

CQ=CQ same base

Area $△ACQ=area△DQC$

(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 $AB\left|\right|PC$ $(\because$ PC is a part of DC)

$△APC$ & $△BCP$ are on the same base PC & between the same parallels $PC\left|\right|AB$ so they are equal in

Area ie Area $(△APC)$ = Aera $(△BCP)$-----(ii)

form eq(i) & eq(ii)

Area $(△BCP)$ = Aera $△DPQ$