In Fig. 15.84, PS

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