M为三角形ABC中边AB上的一点，且AM^2+BM^2+CM^2=2AM+2BM+2CM-3,则AC^2+BC^2=

AM^2+BM^2+CM^2=2AM+2BM+2CM-3

AM^2+BM^2+CM^2-2AM+2BM+2CM+3=0

(AM^2+2AM+1)+(BM^2+2BM+1)+(CM^2+2CM+1)=0

（AM-1）^2+（BM-1）^2+（CM-1）^2=0

AM-1=0
BM-1=0
CM-1=0

推出

AM=1
BM=1
CM=1

AC^2+BC^2=AB^2=(AM+BM)^2=4

AC^2+BC^2=4

解: AM^2+BM^2+CM^2=2AM+2BM+2CM-3

AM^2 - 2AM + 1 + BM^2 - 2BM + 1 + CM^2 - 2CM +1 = 0

(AM - 1)^2 + (BM -1 )^2 + (CM - 1)^2 = 0

AM = 1
BM = 1
CM = 1

AB= AM+BM =2

对图分析可证明是直角三角形,AB是斜边,所以AC^2 + BC^2= AB^2 = 4