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 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