Question

In Fig. 15.84, PSDA is a parallelogram in which PQ=QR=RS and AP||BQ||CR. Prove that
ar(Δ PQE) = ar(Δ CFD)

Answer 1

Given : PSDA is a parallelogram in which PQ = QR = RS and AP || BQ || CR.  

 

To Prove : ar(Δ PQE) = ar(Δ CFD)

 

Proof :  

Since, AP ‖ BQ ‖ CR ‖ DS and AD ‖ PS

∴ PQ = CD …………....(i)

In ∆BED, C is the mid-point of BD and CF ‖ BE

∴ F is the mid-point of ED

EF = FD

Similarly, EF = PE  

∴ PE = FD ………...(ii)

In ∆PQE and ∆CFD , we have

PE = FD (from eq ii)

∠EPQ = ∠FDC (Alternate angle)

PQ = CD (from eq i)

So, by SAS congruence criterion, we have  

∆PQE  ≅ ∆DCF  

ar (Δ PQE) = ar (Δ CFD)

[∵ Congruent triangles have equal area]

Hence, proved

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Answer 2

Answer:-

$\:$

Given :

PSDA is a parallelogram in which PQ = QR = RS and AP || BQ || CR.  

 

To Prove :

ar(Δ PQE) = ar(Δ CFD)

 

Proof :  

Since, AP ‖ BQ ‖ CR ‖ DS and AD ‖ PS

∴ PQ = CD …………....(i)

In ∆BED, C is the mid-point of BD and CF ‖ BE

∴ F is the mid-point of ED

EF = FD

Similarly, EF = PE  

∴ PE = FD ………...(ii)

In ∆PQE and ∆CFD , we have

PE = FD (from eq ii)

∠EPQ = ∠FDC (Alternate angle)

PQ = CD (from eq i)

So, by SAS congruence criterion, we have  

∆PQE  ≅ ∆DCF  

ar (Δ PQE) = ar (Δ CFD)

[∵ Congruent triangles have equal area]

Hence, proved

$\:$

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