(1)由题意知：sinA≠0,2C≠B(否则B+C>π,自己试证一下).
因为sin2C/(sinA－sin2C)=b／（a－b）=sinB/(sinA-sinB)
所以sin2C*(sinA-sinB)=sinB*(sinA－sin2C),整理并化简得：
sin2C=sinB,则有2C+B=π.而A+C+B=π,所以A=C,a=c.
因此△ABC是等腰三角形.
(2)由(1)中的a=c及平行四边形法则(自己画出图来)可知：sinA=1/c.
所以AB*向量BC
=|向量AB|*|向量BC|*cos(π-B)
=-c*a*cosB
=-c^2*cos(π-2C)
=(1/sinA)^2*cos2C
=cos2C/(sinA)^2
=cos2C/(sinC)^2
=cos2C/[(1-cos2C)/2]
=2cos2C/(1-cos2C)
=(2cos2C-2+2)/(1-cos2C)
=(2cos2C-2)/(1-cos2C)+2/(1-cos2C)
=-2+2/(1-cos2C)
因为π/3