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in Parallelograms by (65.5k points)

In Fig., AP || BQ || CR. 

Prove that ar.(AQC) = ar.(PBR).

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Data: AP || BQ || CR. 

To Prove: ar.(∆AQC) = ar.(∆PBR) 

Proof: AP || BQ. CR is a line segment. 

∆ABQ and ∆PBQ are on base BQ and in between 

AP || BQ ar.(∆ABQ) = ar.(∆PBQ) ………….. (i) 

∆BQC and ∆BQR are on base BQ and in between BQ || CR. 

∴ ar.(∆BQC) = ar.(∆BQR) ………….. (ii) 

Adding (i) and (ii),

 ar.(∆ABQ) + ar.(∆BQC) = ar.(∆PBQ) + ar.(∆BQR) 

∴ ar.(∆AQC) = ar.(∆PBR).

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