Question

ABCD is a ||gm and APQ is a straight line meeting BC at P and DC produced at Q. Prove that the rectangle obtained by BC and DQ is equal to the rectangle contained by AB n BC.

Answer 1

Answer:

The diagram shows parallelogram ABCD and the line APQ.

ΔABP and ΔCQP are similar, as the corresponding sides are parallel. Hence, their ratios: 

             CQ / AB = PQ / AP = CP / BP

As AB = CD,  Adding 1 to each of the above ratios, we get  

    (CD+CQ) / AB  = (PQ + AP) / AP = (CP + BP) / BP

         DQ / AB = AQ / AP = BC / BP

=>  BP * DQ  =  AB * BC 

Answer 2

Answer:

The diagram shows parallelogram ABCD and the line APQ.

ΔABP and ΔCQP are similar, as the corresponding sides are parallel. Hence, their ratios: 

             CQ / AB = PQ / AP = CP / BP

As AB = CD,  Adding 1 to each of the above ratios, we get  

        (CD+CQ) / AB  = (PQ + AP) / AP = (CP + BP) / BP

         DQ / AB = AQ / AP = BC / BP

=>  BP * DQ  =  AB * BC