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證明:如圖,連接AC交BD於O,∵ABCD是平行四邊形,∴AO=OC,即O是AC的中點,連接OQ,則OQ⊂平面BDQ,且OQ是△APC的中位線,∴PC‖OQ,又PC在平面BDQ外,∴PC‖平面BDQ.
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