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Answer :
(i) In \(\triangle\) PAC and \(\triangle\) PDB,
\(\angle\) P = \(\angle\) P (Common Angles)
As we know, exterior angle of a cyclic quadrilateral is \(\angle\) PCA and \(\angle\) PBD is opposite interior angle, which are both equal.
\(\angle\) PAC = \(\angle\) PDB
Thus, \(\triangle\) PAC ~ \(\triangle\) PDB(AA similarity criterion)
(ii) We have already proved above,
\(\triangle\) APC ~ \(\triangle\) DPB
We know that the corresponding sides of similar triangles are proportional.
Therefore,
\(\frac{AP}{DP} = \frac{PC}{PB} = \frac{CA}{BD} \)
\(\frac{AP}{DP} = \frac{PC}{PB} \)
AP. PB = PC. DP