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In the figure, PQ||DC||AB. Prove that ar(ACP) = ar(BDQ)

Asked by Topperlearning User 4th June 2014, 1:23 PM
Answered by 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 Expert 4th June 2014, 3:23 PM
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