已知AB=AC,AB的中垂线MN交AC于点D,交AB于点M,若BD平分角ABC,求AD/AC

已知,DM垂直平分AB,可得：AD = BD ；
则有：∠A = ∠ABD = ∠CBD ,∠BDC = ∠A+∠ABD = ∠CBD+∠ABD = ∠ABC = ∠BCD ；
可得：BC = BD = AD ；
因为,在△BDC和△ABC中,∠CBD = ∠CAB ,∠C为公共角,
所以,△BDC ∽ △ABC ,
可得：BC/DC = AC/BC ；
令 AD/AC = k（k > 0）,
则有：BC = AD = kAC ,DC = AC-AD = (1-k)AC ；
可得：BC/DC = k/(1-k) ,AC/BC = 1/k ,
所以,k/(1-k) = 1/k ,
解得：k = (√5-1)/2（舍去负值）,
即有：AD/AC = (√5-1)/2 .