（1）由已知得∠ACB = 90,∠ABC = 30,
∴ ∠Q = 30,∠BCO = ∠ABC = 30．
∵ CD是⊙O的切线,CO是半径,
∴ CD⊥CO,
∴ ∠DCQ =∠BCO = 30,
∴ ∠DCQ =∠Q,故△CDQ是等腰三角形．
（2）设⊙O的半径为1,则AB = 2,OC = 1,AC = AB∕2 = 1,BC =根号3 ．
∵ 等腰三角形CDQ与等腰三角形COB全等,∴ CQ = BC = 根号3．
于是 AQ = AC + CQ = 1 +根号3 ,进而 AP = AQ∕2 =（1 +根号3 ）∕2,
∴ BP = AB－AP = 2－（1 +根号3 ）∕2 =（3－根号3 ）∕2,
PO = AP－AO =（1 +根号3 ）∕2－1 =（ 根号3－1）∕2,
∴ BP:PO =根号3．