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.

Answer 1

SOLUTION :

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

To prove : BP x DQ = AB x BC

Proof :

In ∆ABP and ∆QDA

∠B = ∠D [Opposite angles of parallelogram]

∠BAP = ∠AQD [Alternate interior angles]

Then, ∆ABP ~ ∆QDA [By AA similarity]

AB/QD = BP/DA

[Since, the corresponding sides of similar triangles are proportional]

DA = BC  

[opposite sides of a parallelogram]

AB/QD = BP/BC

AB × BC = QD × BP

Hence proved

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