∠ bag =60.
And < b = < DOP = 90,
∴∠PDO=∠AOB=30，
∴PD=2PO=AO，
∴△POD≌△ABO
(2) BD is the tangent of ⊙ p.
Prove: ∫ from (1) △ pod △ ABO: do = bo,
∴△BOD is isosceles△,
∠ DOB =∠ Bao = 60。
∴△BOD is equilateral,
∴∠DBO=60
∠ PBO =∠ POB = 30。
∴∠DBP=∠DBO+∠PBO=60 +30 =90
∴ BD is the tangent of⊙ P.
(3) DO=BO=3，AO=2AB。 We can get: AB=√3, AO=2√3. Therefore,
The distance from b to AO is: 1.5,
The coordinate of ∴B is (-3√3/2, 1.5), then the analytical formula of straight line BD can be obtained as follows:
y=√3/3*x+3。