In Fig., ABCD and AEFD are two parallelograms. Prove that ar(Δ PEA) = ar(Δ QFD).

Proof:

In triangles PEA and QFD, we have

∠APE = ∠DQF (Corresponding angles)

AE = DF (Opposite sides of II" AEFD)

∠AEP = ∠DFQ (Corresponding angles)

∴ △PEA ≅ △QFD (AAS congruence criterion)

As congruent triangles have equal area.

∴ ar (△PEA) ≅ ar (△QFD)