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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. - Mathematics

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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

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APPEARS IN

 RD Sharma Solution for Class 10 Maths (2018 (Latest))
Chapter 7: Triangles
Ex. 7.5 | Q: 17 | Page no. 75
 RD Sharma Solution for Class 10 Maths (2018 (Latest))
Chapter 7: Triangles
Ex. 7.5 | Q: 17 | Page no. 75
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.
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