从A做AG⊥BC交BC于G,则AG=(√3)/2*BC=(√3)/2*AB
设对折后A落在BC上的点是A',∠A'AG=x(π/6≥x≥0)
因是对折,DE⊥AA'且交点M是AA'中点
对于直角△DMA,AD=(AA'/2)/cos(π/6-x)
对于直角△A'GA,AA'=AG/cos(x)=(√3)/2*AB/cos(x)
∴AD=[(√3)/4]*AB/[cos(x)*cos(π/6-x)]
当x=π/6-x,即x=π/12时,AD取最小值
AD:AB=[(√3)/4]/[cos(π/12)^2]=2√3-3