**Question:**

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.

**Solution:**

Given:

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

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×DQ=AB×BC

Proof:

In ∆ABP and ∆QCP, we have

∠ABP = ∠QCP (Alternate angles as AB || DC)

∠BPA = ∠QPC ( Vertically opposite angles)

By AA similarity, we get

∆ABP ~ ∆QCP

We know that corresponding sides of similar triangles are proportional.

⇒ABQC=BPCP=APQP

⇒ABQC=BPCP⇒AB×CP=QC×BP

Adding AB×BP in both sides, we get

⇒AB×CP+AB×BP=QC×BP+AB×BP⇒AB×CP+BP=QC+AB×BP⇒AB×CP+BP=QC+CD×BP

ABCD is a parallelogram, AB = CD⇒AB×BC=DQ×BP⇒BP×DQ=AB×BC