Answer 1

Given that, ABCD and AEFD are two parallelograms
To prove:   (1) PE = FQ

(2) `"ar (ΔAPE)"/ "ar (ΔPFA)"` = `"ar (ΔQFD)"/"ar (ΔPED)"`

(3) ar (ΔPEA) = ar (ΔQFD)

Proof:  (1) In  ΔEPA and ΔFQD

∠PEA  = ∠QFD                 [ ∴ Corresponding angles]
∠EPA  = ∠FQD                 [Corresponding angles]

 PA = QD                    [opp .sides of 11^gm]

Then,  ΔEPA  ≅  ΔFQD      [By. AAS condition]

∴ EP = FQ                       [c. p. c.t]

(2)  Since, ΔPEA and ΔQFD stand on the same base PEand FQlie between the same
parallels EQ and AD

∴  ar  (ΔPEA ) = ar (ΔQFD)  →  (1) 

AD  ∴ ar (ΔPFA) = ar (PFD)        .....(2)

Divide the equation (1) by equation (2)

`"area of (ΔPEA)"/"area of (ΔPFA)"` = `"ar Δ(QFD)"/"ar Δ(PFD)"`

 (3) From (1) part ΔEPA  ≅ FQD

Then, ar (ΔEDA) = ar (ΔFQD)