证明:连接OA
∵OC = BC,AC =1/2OB
∴OA = OC = AC = OB
∴△AOC是等边三角形,△ABC是等腰三角形
∴∠OAC = ∠O =60°,∠CAB = ∠B = 30°
∴∠OAB = ∠OAC + ∠CAB = 90°
∴AB是○O的切线
计算:延长CO交○O于点E,连接DE
∴△CDE是直角三角形
∴CD = OC/sin15° = 7.7274
证明:连接OA
∵OC = BC,AC =1/2OB
∴OA = OC = AC = OB
∴△AOC是等边三角形,△ABC是等腰三角形
∴∠OAC = ∠O =60°,∠CAB = ∠B = 30°
∴∠OAB = ∠OAC + ∠CAB = 90°
∴AB是○O的切线
计算:延长CO交○O于点E,连接DE
∴△CDE是直角三角形
∴CD = OC/sin15° = 7.7274