Abcd Is Parallelogram And Bc Is Produced To Point Q Such That Ad Cq

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) =…


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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