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  • In Fig. 6.62, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that PA . PB = PC . PD

 

Q8 (2)   In Fig. 6.62, two chords AB and CD of a circle intersect each other at the point P
             (when produced) outside the circle. Prove that
             PA . PB = PC . PD

            

Answers (1)

best_answer

In \Delta PAC \,and \,\Delta PDB,

\angle P=\angle P         (Common)

\angle PAC=\angle PDB  (Exterior angle of a cyclic quadrilateral is equal to opposite interior angle)

So,  \Delta PAC \sim \Delta PDB       ( By AA rule)

24440\frac{AP}{DP}=\frac{PC}{PB}=\frac{CA}{BD}             (Corresponding sides of similar triangles are proportional)

\Rightarrow \frac{AP}{DP}=\frac{PC}{PB}

\Rightarrow AP.PB=PC.DP

Posted by

seema garhwal

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