Solution: Connection proof: even OA, OB.

∫PA, PB is the tangent of circle o,

∴OA⊥PA,OB⊥PB

OA = OB，OP=OP。

∴Rt△OAP

≌Rt△OBP。

∴PA=PB

(2)

∠OPA=∠OPC。 (or PA=PC, or AB=CD, or the distance from the center o to Pb and PD is equal, or the arc AB is equal to the arc CD).