作AF⊥BC,垂足为F.因为面ABC⊥面B1BCC1,∴AF⊥平面B1BCC1.连结B1F,则B1F是AB1在平面B1BCC1内的射影.∵BC1⊥AB1,∴BC1⊥B1F.∵四边形B1BCC1是矩形,∴∠B1BF=∠BCC1=90°,又∠FB1B=∠C1BC,∴△B1BF∽△BCC1,则 = = .又...