In ∆ABC,P is a point on AB ,PR is drawn parallel to BC and PR=BC.PR curd AC at Qprove that ar(∆AQR)=ar(∆BPQ).

Since,PR।।BC and AP = BP =>AQ=QC Now, triangles PQB and PQC are on the same base and b/w same parallels PQ and BC => ar(PQB) = ar(PQC)Now,In triangles PQC and ARQ PQ. = RQ AnglesPQC= ARQ [vertically opposite angles] QA=QC Hence, by SAS congruency rule => Triangles PQC and AQR are congruent to each other Since,ar(PQC) = ar (PQB) and also ar(PQC) = ar(ARQ) => ar(PQB) = ar(ARQ)