设AM与EF相交于O,根据题意,EA=EM,设EA=EM=X,则BE=2-X,
在RTΔBME中,EM^2=BE^2+BM^2
X^2=(2-X)^2+1,X=5/4,
即AE=5/4.
∵AM=√(AB^2+BM^2)=√5,
∴AO=1/2AM=√5/2,OE=√(AE^2-AO^2)=√5/4,
∠AEO为公共角,∴RTΔEAO∽RTΔFA,
∴AE/EF=OE/AE,
∴EF=AE^2/OE=5√5/4.