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Question

ABCD is a parallelogram and APQ is a straight line meeting BC at Pend DC produced at Q. prove that the rectangle obtained by BP and DQ is equal to the rectangle contained by AB and BC. 


Solution

ABCD is a parallelogram and APQ is a straight line meeting BC at P and DC produced at Q.

We need to prove, the rectangle obtained by BP and DQ is equal to the rectangle contained by AB and BC. We need to prove that BP x DQ = AB x BC

8

In    ABP and    QCP,

   ABP  =   QCP   (alternate angles as AB  DC)

  BPA =     (VOA)

   ABPsim  QCP   (AA similarity)

We know that corresponding sides of similar triangles are proportional
fraction numerator A B over denominator Q C end fraction equals fraction numerator B P over denominator C P end fraction equals fraction numerator P A over denominator P Q end fraction
BP x QC = AB x CP
BP x (DQ -DC) = AB x (BC - BP)
BP x (DQ - AB) = AB x (BC - BP)
BP x DQ - BP x AB  = AB x BC - AB x BP
BP x DQ = AB x BC 
Hence Proved

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