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) =…
Answers:
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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