Given that, ABCD and AEFD are two parallelograms
To prove:
(i) PE = FQ
(ii) ar (∆ APE) : ar (∆PFA) = ar ∆(QFD) : ar (∆ PFD)
(iii) ar (∆ APE) = ar ∆(QFD
RP = FQ
(ii) Since, ∆PEA and ∆QFD stand on the same base PE and FQ lie between the same parallels EQ and AD