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:
(i) \(\triangle\) PAC ~ \(\triangle\) PDB
(ii) PA . PB = PC . PD.

(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