In Fig. 15.84, PSDA is a parallelogram in which PQ=QR=RS and AP||BQ||CR. Prove that

ar(Δ PQE) = ar(Δ CFD)

Question

In Fig. 15.84, PSDA is a parallelogram in which PQ=QR=RS and AP||BQ||CR. Prove that ar(&#x394; PQE) = ar(&#x394; CFD)

Philoid · Accepted Answer

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

In Fig. 15.84, PSDA is a parallelogram in which PQ=QR=RS and AP||BQ||CR. Prove thatar(Δ PQE) = ar(Δ CFD)