三角形abc的三个顶点都在圆o上,ab为直径 ==> ∠C = 90°；==>RtΔCAB
∠CAB = 30° ==> BC = 1/2 *AB；
AC = √3/2 *AB
∴ AB = AC/(√3/2) = 2√3 ；BC =1/2 *AB = √3
CD平分角ACB ==> AD：BD = AC：BC = 3：√3 --- (1)
而：AD+BD = AB = 2√3 --- (2)
(1)(2)联立,解得：
AD = 3√3-3；BD=3-√3