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 AB and BC.
Solution
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
Is there an error in this question or solution?
Solution 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 Ab and Bc. Concept: Criteria for Similarity of Triangles.
