In figure, ABCD and AEFD are two parallelograms. Prove that

Solution:

Given that, ABCD and AEFD are two parallelograms

(i) In triangles, EPA and FQD

∠PEA = ∠QFD         [corresponding angles]

∠EPA=∠FQD           [corresponding angles]

PA = QD                   [opposite sides of parallelogram]

Then, ΔEPA ≅ ΔFQD           [By AAS condition]

Therefore, EP = FQ               [C.P.C.T]

(ii) Since triangles, PEA and QFD stand on equal bases PE and FQ lies between the same parallels EQ and AD

Therefore, ar(ΔPEA) = ar(ΔQFD)  ... (1)

Since, triangles PEA and PFD stand on the same base PF and between same parallels PF and AD

Therefore, ar(ΔPFA) = ar(ΔPFD) .... (2)

Divide the equation 1 by equation 2

ar(ΔPEA) ar(ΔPFA) = ar(ΔQFD) ar(ΔPFD)

(iii) From part (i), 

ΔEPA ≅ ΔFQD

Then, ar(ΔPEA) =  ar(ΔQFD).