Answer 1

Given: ABCD is a parallelogram

To prove: BP × DQ = AB × BC

Proof: In ΔABP and ΔQDA

∠B = ∠D                             [Opposite angles of parallelogram]

∠BAP = ∠AQD                     [Alternate interior angles]

Then, ΔABP ~ ΔQDA              [By AA similarity]

`therefore"AB"/"QD"="BP"/"DA"`                [Corresponding parts of similar Δ are proportional]

But, DA = BC                     [Opposite sides of parallelogram]

Then, `therefore"AB"/"QD"="BP"/"BC"`

⇒ AB × BC = QD × BP