In figure, PSDA is a parallelogram in which PQ = QR = RS and AP ∥ BQ ∥ CR. Prove that ar (ΔPQE) = ar(ΔCFD).

Solution:

Given that PSDA is a parallelogram

Since, AP ∥ BQ ∥ CR ∥ DS and AD ∥ PS

Therefore, PQ = CD       (equ. 1)

In triangle BED, C is the midpoint of BD and CF ∥ BE

Therefore, F is the midpoint of ED

⇒ EF = PE

Similarly,

EF = PE

Therefore, PE = FD             (equ. 2)

In triangles PQE and CFD, we have

PE = FD

Therefore, ∠ EPQ = ∠FDC     [Alternate angles]

So, by SAS criterion, we have

ΔPQE ≅ ΔDCF

⇒ ar(ΔPQE) = ar(ΔDCF)