190, 我总算做出来了。

找到正确的辅助线用了约两天时间。剩下的工作用了大约半天。整理到目前的条理用了一个多小时。最近几天的早锻炼（走路约50分钟）都在考虑这道题目。

Let ∆ABC = S.

Choose F in AB so that ED//CF. In ∆ACF,

sin ∠A/CF = sin(∠ACF)/AF

i.e.

CF/sin(45°) = AF/sin(75°)

Because ∠B = 30° and ∠CFB = 120°, CF = BF。

BF/AF = sin(45°)/ sin(75°) = sqrt(3) – 1

∆BCF/∆ACF = BF/AF = sqrt(3) – 1

∆ACF = S/sqrt(3) (by ∆ACF+∆BCF=S)

∆AED/∆ACF = sqrt(3)/2 = (AD/AF)^2 (by ∆AED=S/2)

AD/AB = AD/(AF+BF) = (AD/AF)/(1+BF/AF)

= (AD/AF)/sqrt(3) =12^(-1/4)