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(i) \(\triangle\) PAC ~ \(\triangle\) PDB

(ii) PA . PB = PC . PD.

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

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