证明：
过E点作BC的垂线交BC于G,则BG=EG=(√2/2)*BE
MN^2=BM^2+BN^2=ME^2+NE^2
NE^2=EG^2+NG^2
=EG^2+(BN+EG)^2
=[(√2/2)*BE]^2+[BN+(√2/2)*BE]^2
=BE^2/2+BN^2+√2*BN*BE+BE^2/2
=BE^2+BN^2+√2*BN*BE
ME^2=BE^2+BM^2-2BE*BMcos45度
=BE^2+BM^2-√2*BM*BE
ME^2+NE^2
=BE^2+BM^2-√2*BM*BE+BE^2+BN^2+√2*BN*BE
=2BE^2+BM^2+BN^2-√2*BM*BE+√2*BN*BE
=2BE^2+BM^2+BN^2-√2*BE(BM-BN)
=BM^2+BN^2
所以
2BE^2-√2*BE(BM-BN)=0
√2(BM-BN)=2BE
BM-BN=√2*BE
证毕