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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: (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