设AB＝a,AD＝b,AA1＝c.则a²＝b²＝c²,ab＝ac＝bc＝0（数积）
⑴ DO·BC1＝（-b+a+（b+c）/2）·（b+c）＝（a-b/2+c/2）·（b+c）＝
＝-b²/2+c²/2＝0. ∴DO⊥BC1.
⑵ A1C·BB1＝（-c+a+b）·c＝-c²≠0, A1C与BB1不垂直!
题目打错,应该是A1C1垂直BB1,留给楼主证明吧!
⑶ A1C·AB1＝（-c+a+b）·（a+c）＝0 ,A1C⊥AB1
A1C·AD1＝（-c+a+b）·（b+c）＝0 , A1C·AD1,∴A1C⊥平面AB1D1.