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Question
In the below fig. ABCD and AEFD are two parallelograms. Prove that
(1) PE = FQ
(2) ar (Δ APE) : ar (ΔPFA) = ar Δ(QFD) : ar (Δ PFD)
(3) ar (ΔPEA) = ar (ΔQFD)
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Solution
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 11gm]
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)
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