(1) ∵A1C1//AC ∴∠DA1C1便为AC与A1D所成角
连接C1D,∵A1C1,A1D,C1D都是正方形对角线
∴A1C1=A1D=C1D,即∠DA1C1=60°
即AC与A1D所成角为60°
(2) 连接AB中点E和C1D1中点M
则AD1//EM,且EM交A1C1于O
则∠COE即为A1C与AD1所成角
连接CE,CM,A1E,A1M,可知
CE=CM=A1E=A1M,∴A1ECM是菱形
∴EM和A1C互相垂直平分,即∠COE=90°
即A1C与AD1所成角为90°
(3) 连接BD,B1D1,可知EF//BD//B1D1
而B1D1⊥A1C1,∴EF⊥A1C1
即A1C1与EF所称角为90°