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20 13 problem-solving process of the fifth multiple-choice question of one model mathematics in Mentougou senior middle school
Science? Geometric solution: ∫ab is the diameter o of the circle, ∴ ∠ ACB = 90.

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.