in the given figure , abcd is a parallelogram.prove that ar(triangle bcp) = ar(triangle dpq), if bc=cq

given: ABCD is a parallelogram.

TPT:area(ΔBCP)=area(ΔDPQ) 

proof:

area(ΔBCP)=area(ΔAPC).............(1)  [since PC is parallel to AB and the triangles formed between same set of parallel line and with same base are equal in areas]

similarly area(ΔADC)=area(ΔADQ)  [the triangles have the same base AD and between the parallel lines of BQ and AD]

⇒area(ΔADP)+area(ΔAPC)=area(ΔADP)+area(ΔDPQ)

therefore area(ΔAPC)=area(ΔDPQ).......(2)

from (1) and (2)

area(ΔBCP)=area(ΔDPQ)

hope this helps you.

cheers!!

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