Question

In Fig. 15.88, ABCD and AEFD are two parallelograms. Prove that:
(i) PE=FQ
(ii) ar(ΔAPE):ar(Δ PFA)= ar(Δ QFD)=ar(Δ PFD)
(iii) ar(Δ PEA) = ar(Δ QFD)

Answer 1

Given : ABCD and AEFD are two parallelograms.

 

To Prove:

(i) PE = FQ

(ii) ar(ΔAPE): ar(Δ PFA) = ar(Δ QFD) : ar(Δ PFD)

(iii) ar(Δ PEA) = ar(Δ QFD)

 

 

Proof :

(i) In ∆APE and ∆DQF

∠APE = ∠DQF [Corresponding angles]

∠AEP = ∠DFQ [Corresponding angles]

PA = QD [Opposite sides of a ||ᵍᵐ]

Then,  ∆APE ≅ ∆DQF [By AAS congruence criterion]

∴ PE = QF

[CPCT]

And ar(∆APE ) = ar (∆DQF) ………..(1)

[Congruent triangles have equal areas]

This proves (i) and (iii)  

(ii) Clearly, ∆PFA and ∆PFD stand on equal base PF and lies between the same parallel PQ and AD.

∴ ar(∆PFA) = ar (PFD) …………….(2)

Divide the equation (1) by (2), we get :  

ar(∆APE ) / ar(∆PFA) = ar (∆DQF)/ar (PFD)

Hence, ar(∆APE ) :  ar(∆PFA) = ar (∆DQF) : ar (PFD)

HOPE THIS ANSWER WILL HELP YOU…..

 

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Answer 2

Answer:

Step-by-step explanation:

In ∆APE and ∆DQF

∠APE = ∠DQF [Corresponding angles]

∠AEP = ∠DFQ [Corresponding angles]

PA = QD [Opposite sides of a ||ᵍᵐ]

Then,  ∆APE ≅ ∆DQF [By AAS congruence criterion]

∴ PE = QF

[CPCT]

And ar(∆APE ) = ar (∆DQF) ………..(1)

[Congruent triangles have equal areas]

This proves (i) and (iii)  

(ii) Clearly, ∆PFA and ∆PFD stand on equal base PF and lies between the same parallel PQ and AD.

∴ ar(∆PFA) = ar (PFD) …………….(2)

Divide the equation (1) by (2), we get :  

ar(∆APE ) / ar(∆PFA) = ar (∆DQF)/ar (PFD)