Question

PSDA is a parallelogram. Point Q and R are taken on PS such that PQ =QR =RS and PA is parallel to QB and QB is parallel to RC.Prove that area of PQE = area of CFD

Answer 1

PA║QB║RC║SD & PQ=QR=RS
Equal intercept theorem states that if three or more parallel lines makes equal intercept on traversal,then they make equal intercept on any other form of traversal.
Therefore PE=EF=FD& AB=BC=CD
In ΔPQE & ΔDCF
∠PEQ=∠DFC
PE=DF
∠QPE=∠CDF
Therefore ΔPQE≈ΔDCF
ΔPQE=ΔDCF(Congruent figures have equal areas) 

Answer 2

PSDA is a parallelogram. PA ║ QB║RC║SD and PQ = QR = RS.
Equal Intercept Theorem states that if three or more parallel lines makes equal intercepts on one transversal,then they make equal intercepts on any other transversal.
∴ PE = EF = FD and AB = BC = CD
In Δ PQE and Δ DCF,
∠ PEQ = ∠ DFC             (Exterior opposite angles are equal)
PE = DF               (Proved)
∠QPE = ∠ CDF       (Alternate angles)
∴ Δ PQE ≈  Δ DCF           (ASA Congruence Theorem)
⇒ ar (ΔPQE) = ar (Δ DCF)      (Congurent figures have equal areas)