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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

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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)

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

$\Rightarrow AP.PB=PC.DP$

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