Q15 of 168 Page 5

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

Given :- ABCD is a parallelogram


To prove :- BP x DQ = AB x BC


Proof:- In ΔABP and ΔQDA


<B = <D (Opposite angles of parallelogram)


<BAP = <AQD (Alternative interior angle)


Then, ΔABP ~ ΔQDA


SO, (Corresponding parts of similar triangle area proportion) But, DA = BC (Opposite side of parallelogran)


Then,


Or, AB x BC = QD X BP


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