(1)a2-b2+c2-2ac=(a-c)2-b2=(a-c+b)(a-c-b)∵a、b、c为△ABC三边的长,∴(a-c+b)>0,(a-c-b)<0,∴a2-b2+c2-2ac<0.(2)由a2+2b2+c2=2b(a+c)得:a2-2ab+b2+b2-2bc+c2=0配方得:(a-b)2+(b-c)2=0∴a=b=c∴△ABC为等边三角形.