Q14 of 25 Page 9

In Fig given below PSDA is a parallelogram in which PQ=QR=RS and AP||BQ||CR. Prove that ar(Δ PQE) = ar(Δ CFD).

Given that,

PSDA is a parallelogram


Since,


AP ‖ BQ ‖ CR ‖ DS and AD ‖ PS


Therefore,


PQ = CD (i)


In


C is the mid-point of BD and CF ‖ BE


Therefore,


F is the mid-point of ED


EF = PE


Similarly,


PE = FD (ii)


In PQE and , we have


PE = FD


EPQ = FDC (Alternate angle)


And,


PQ = CD


So, by SAS theorem, we have



Area (Δ PQE) = Area (Δ CFD)


Hence, proved


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In figure, 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).

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