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Answer :
Firstly, let us join CB, in the given figure.
(i) In \(\triangle\) APC and \(\triangle\) DPB,
\(\angle\) APC = \(\angle\) DPB (Vertically opposite angles)
\(\angle\) CAP = \(\angle\) BDP (Angles in the same segment for chord CB)
Therefore,
\(\triangle\) APC ~ \(\triangle\) DPB (AA similarity criterion)
(ii) In the above, we have proved that \(\triangle\) APC ~ \(\triangle\) DPB
We know that the corresponding sides of similar triangles are proportional.
\(\frac{AP}{DP} = \frac{PC}{PB} = \frac{CA}{BD} \)
\(\frac{AP}{DP} = \frac{PC}{PB} \)
AP. PB = PC. DP
Hence, proved.