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.

Answer 1

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

In $△$ ABP and $△$ QCP,

$\angle$ ABP = $\angle$QCP (alternate angles as AB DC)

$\angle$BPA = $\angle$ (VOA)

$△$ ABP$sim$ $△$ QCP (AA similarity)

We know that corresponding sides of similar triangles are proportional

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