解:过d点作de垂直ab1,延长交bb1于f∵a1c1=b1c1,a1d=b1d∴c1d⊥a1b1∴c1d⊥面a1abb1∴c1d⊥ab1∵de⊥ab1∴ab1⊥面c1df即f即为所求在矩形a1abb1中,b1b=a1a=A,ab=a1b1=√2A,b1d=a1d=√2A/2,∠b1df=∠b1aa1b1f/b1d=a1b1/a1ab1f=A,即f点于b点重合
神马东东......直三棱柱 还是锥? 还没听过直三棱锥 = -
图呢
不懂