是不是这个三角形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 = ...CD平分角ACB ==> AD：BD = AC：BC = 3：√3 --- (1)

怎么得到的？？∵AB为直径∴∠ADB=∠ACB=90°
∴AD：BD = AC：BC = 3：√3
求采纳