解答:证明:(1)过点D作DO⊥BC,O为垂足.
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因为面DBC⊥面ABC,又面DBC∩面ABC=BC,DO?面DBC,
所以DO⊥面ABC.
又AE⊥面ABC,则AE∥DO.
又AE?面DBC,DO?面DBC,故AE∥面DBC.
(2)由(1)知DO⊥面ABC,AB?面ABC,所以DO⊥AB.
又AB⊥BC,且DO∩BC=O,DO,BC?平面DBC,则AB⊥面DBC.
因为DC?面DBC,所以AB⊥DC.
又BD⊥CD,AB∩DB=B,AB,DB?面ABD,则DC⊥面ABD.
又AD?面ABD,故可得AD⊥DC.