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  • 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 <img alt="AP\times PC = BP \times DP" src="https://entrancecorner.codecogs.com/gif.latex?AP%5Ctimes%20PC%20%3D%20BP%

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\times PC = BP \times DP

 

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In \Delta BAP & \Delta CDP

\angle BAP = \angle PDC = 90\degree

\angle BPA = \angle CPD \quad (\text{vertically opposite angles})

\Delta BAP \sim \Delta DPC\quad [\text{By AA - Similarity}]

So,    \frac{AP}{DP} = \frac{BP}{PC}

\Rightarrow AP \times PC = BP \times DP

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