answer me th following question pl........
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hey mate here is your answer....
given abcd is a parallelogram.p and q lie on the two points lying in the side DC and ad
to prove .ar (apb)=ar(bqc)
proof. first of all join ap and bp to form triangle apb now ,as triangle apb and parallelogram abcd lie on the same base ab and between same parallels DC and ab {ar (apb)=1/2ar (abcd)
this is the first equation
now join qc and qb to form triangle qbc.
as here also the triangle qbc and the parallelogram abcd lie on the same base abcd lie on the same base bc and between same parallels bc and ad so ,ar (qbc)=1/2ar(abcd)....take this is as second equation...........from the above two equation we consider that..... ar (qbc)=ar (abp)....
I hope it may help you....
given abcd is a parallelogram.p and q lie on the two points lying in the side DC and ad
to prove .ar (apb)=ar(bqc)
proof. first of all join ap and bp to form triangle apb now ,as triangle apb and parallelogram abcd lie on the same base ab and between same parallels DC and ab {ar (apb)=1/2ar (abcd)
this is the first equation
now join qc and qb to form triangle qbc.
as here also the triangle qbc and the parallelogram abcd lie on the same base abcd lie on the same base bc and between same parallels bc and ad so ,ar (qbc)=1/2ar(abcd)....take this is as second equation...........from the above two equation we consider that..... ar (qbc)=ar (abp)....
I hope it may help you....
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