Abcd is a parallelogram and bc is produced to a point q such that ad= cq. If aq intersects dc at p, show that area of bpc= area of dpq
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ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ (fig). If AQ intersects DC at P, show that ar(△BPC) = ar(△DPQ).
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answered Dec 26, 2017 by navnit40 (-2,641 points)
In ||gm ABCD ,
ar(△APC) = ar(△BCP) ---i)
[∵ triangles on the same base and between the same parallels have equal area]
Similarly, ar(△ADQ) = ar(△ADC) ---ii)
Now, ar(△ADQ) - ar(△ADP) = ar(△ADC) - ar(△ADP)
ar(△DPQ) = ar(△ACP) ---iii)
From (i) and (iii) , we have
ar(△BCP) = ar(△DPQ)
or ar(△BPC) = ar(△DPQ)
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answered Dec 15, 2018 by anandkthas gmail.com Expert (2,014 points)
even i want the answer
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