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Deepak Bora
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Asked: 2020-07-26T20:33:19+05:30 In:

Two right triangles ABC and DBC are drawn on the same hypotenuse BC and on the same side of BC. If AC and BD intersect at P, prove that AP × PC = BP × DP.

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CBSE 10th Class past year question paper

Year 2019

SET 1, Code Number- 30/2/1

Question number – 14 (a)

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    2 Answers

    1. Newbie
      Two right triangles ABC and DBC are drawn on the same hypotenuse BC and on the same side of BC. If AC and BD intersect at P, prove that AP × PC = BP × DP.

      ∆APB ~ ∆DPC [AA similarity]

      AP/DP =BP/PC

      AP × PC = BP × DP

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    2. Two right triangles ABC and DBC are drawn on the same hypotenuse BC and on the same side of BC. If AC and BD intersect at P, prove that AP × PC = BP × DP.

      Angle A = angle D……( 90 degrees given)

      Angle APB= angle DPC…..(VERTICALLY OPPOSITE ANGLES)

      Therefore, APB similar to triangle DPC…(by AA rule)

      Therefore, AP/DP=BP/PC

      So,AP×PC=BP×DP

      HENCE,PROVED.

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