In ΔPAC and ΔPDB ∠P=∠P (Common) ∠PAC=∠PBD Remember: Exterior angle of a cyclic quadrilateral is equal to the opposite interior angle. ΔPAC~ΔPDB (AA Similarity) ⇒
PC
PB
=
PA
PD
=
AC
BD
(Corresponding sides are proportional) ⇒PA×PB=PC×PD Hence, only statement 2 is correct