∴∠POC=∠FDP, ∣∠p is a common angle.
∴△POC∽△PDF,
∴PD? PC=PF? Po,
∵PD? PC=PB? Dad,
∴PF? PO=PB? Dad.
(2)∫PB = 2BF,
∴ Let PB=x, then BF= 12x, PF=32x,
And ∵⊙O radius is 2, ∴PO=x+2, PA=x+4,
How do you know PF from (1)? PO=PB? Dad,
∴32x(x+2)=x(x+4),
X=2, x=0 (give up),
∴PB=2.