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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) - Mathematics

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)

  Is there an error in this question or solution?
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APPEARS IN

RD Sharma Mathematics for Class 9
Chapter 14 Areas of Parallelograms and Triangles
Exercise 14.3 | Q 26 | Page 47
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