Once again ∵AC=

1

2

AB, ∴∠ B = 30, and ∠ CAB = 60, BC=

three

2

ab blood type

PA = AC =

1

2

AB，

△ PAC uses cosine theorem,

PC=

PA2+AC2？ 2PA？ Accounting 120

=

three

AC=

three

2

AB，

That is BC=PC, a is correct;

∵PA=AC,BC=PC,∴PC？ AC=PA？ BC, get b correctly;

Available connection OC

∵ isosceles △PAC, ∠ PCA = 30 and equilateral △ACO, ∠ ACO = 60.

∴∠ OCP = 90, you can get PC⊥OC, so PC is the tangent of circle O, so C is correct;

According to secant theorem, BC2=PC2=PA? PB≠BA？ BP, so d is incorrect.

So choose: D.