设BB1=a,则AB=√2a.
记AB,BB1,B1C1,B1A1中点分别为K,L,M,N.
KL平行且等于AB1/2,
KL^2=AB1^2/4=3a^2/4,
LM平行且等于C1B/2,
LM^2=3a^2/4,
MN平行且等于A1C1/2,
MN^2=a^2/2,
KN平行且等于BB1,
KN^2=a^2,
KM^2=MN^2+KN^2=3a^2/2,
KM^2=KL^2+LM^2,所以,KL⊥LM,
AB1与C1B所成角的大小为90度.