由题意,|AB|=sqrt(2),|BC|=2
AE=AB+BE=AB+BC/2
BF=BC+CF=BC-FC,又：|DF|=|DC|sqrt(2)/2=1,故：|FC|=|DC|(sqrt(2)-1)/sqrt(2)
=(1-sqrt(2)/2)|DC|,故：BF=BC-FC=BC-(1-sqrt(2)/2)AB
故：AE dot BF=(AB+BC/2) dot (BC-(1-sqrt(2)/2)AB)=AB dot BC+|BC|^2/2
-(1-sqrt(2)/2)|AB|^2-(1-sqrt(2)/4)AB dot BC,因：AB dot BC=0
故：AE dot BF=|BC|^2/2-(1-sqrt(2)/2)|AB|^2=2-2(1-sqrt(2)/2)=sqrt(2)