Sarthaks Test
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In the below fig. ABCD and AEFD are two parallelograms. Prove that 

(i) PE = FQ 

(ii) ar (∆ APE) : ar (∆PFA) = ar ∆(QFD) : ar (∆ PFD) 

(iii) ar (∆ APE) = ar ∆(QFD)

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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

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