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CBSE Class 9 Answered

In the figure, PQ||DC||AB. Prove that ar(ACP) = ar(BDQ)
Asked by Topperlearning User | 04 Jun, 2014, 01:23: PM
answered-by-expert Expert Answer

PQ||DC

ar(PDC) = ar(QCD) ...(I)

[Triangle between same parallels and on the same base]

ACD and CDB are on same base CD and between same parallels, CD||AB

ar(ACD) = ar(CDB) …(II)

Adding (I) and (II),

ar(PDC) + ar(ACD) = ar(QCD) + ar(CDB)

ar(ACP) = ar(BDQ)

Answered by | 04 Jun, 2014, 03:23: PM

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