Given In a parallelogram PSDA, points 0 and R are on PS such that PQ = QR = RS and PA || QB || RC. To prove ar (PQE) = ar (CFD) Proof In parallelogram PABQ, and PA||QB [given] So, PABQ is a parallelogram. PQ = AB …(i) Similarly, QBCR is also a parallelogram. QR = BC …(ii) and RCDS is a parallelogram. RS =CD …(iii) Now, PQ=QR = RS …(iv) From Eqs. (i), (ii) (iii) and (iv), PQ || AB [∴ in parallelogram PSDA, PS || AD] In ΔPQE and ΔDCF, ∠QPE = ∠FDC [since, PS || AD and PD is transversal, then alternate interior angles are equal] PQ=CD [from Eq. (v)] and ∠PQE = ∠FCD [∴ ∠PQE = ∠PRC corresponding angles and ∠PRC = ∠FCD alternate interior angles] ΔPQE = ΔDCF [by ASA congruence rule] ∴ ar (ΔPQE) = ar (ΔCFD) [since,congruent figures have equal area] Hence proved.