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.