Q : 14     In Fig. \small 9.28,  \small AP\parallel BQ\parallel CR. Prove that \small ar(AQC)=ar(PBR).

                

Answers (1)


We have, AP || BQ || CR
\DeltaBCQ and \Delta BQR lie on the same base (BQ) and between same parallels (BQ and CR)
Therefore, ar (\DeltaBCQ) = ar (\DeltaBQR)........(i)

Similarly, ar (\DeltaABQ) = ar (\DeltaPBQ)  [common base BQ and BQ || AP]............(ii)

Add the eq(i) and (ii), we get

ar (\DeltaAQC) = ar (\DeltaPBR)

Hence proved

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