Question

如图,AB为⊙O的直径,点C为⊙O上一点,AD⊥面ABC,AE⊥BD于E,AD⊥CD于F.求证：BD⊥平面AEF

Answer 1

因为AB是直径,所以BC⊥AC；又AD⊥面ABC,所以BC⊥AD,于是知BC⊥面ACD,可知BC⊥AF.又AF⊥CD,且AF⊥BC,所以AF⊥面BCD,即知AF⊥BD.又BD⊥AE,所以BD⊥平面AEF.