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9.证明:(1)连接BG并延长,交AD于M,连接MC,在△ABD中,O为BD中点,G在AO上,AG=2GOBG 2G为△ABD的重心∴GM IBE_2.BG⊥BEI GM EC GEMCGEg平面ACD,ACc平面ACD,∴GE∥平面ACD;(2)在△ABD中,O为BD中点,BD=2,AB=AD=√,∴AO⊥BD∴AO=√AB2-BO2=1,在△BCD中,BC=CD=BD=2,O为BD中点,连接OC,则OC=√3又CA=2,∴OA2 OC2=CA2,∴AO⊥OC由AO⊥OC,AO⊥BD,OC∩BD=O,OC,BDc平面BCD得AO⊥平面BCD,又AOc平面ABD,∴平面ABD⊥平面BCD