如图,四边形ABCD为正方形,平面PQC⊥平面DQC,PD∥QA,QA=AB=1/2PD (1)证明:平面ABCD⊥AQ
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![如图,四边形ABCD为正方形,平面PQC⊥平面DQC,PD∥QA,QA=AB=1/2PD (1)证明:平面ABCD⊥AQ](/uploads/image/z/1798477-61-7.jpg?t=%E5%A6%82%E5%9B%BE%2C%E5%9B%9B%E8%BE%B9%E5%BD%A2ABCD%E4%B8%BA%E6%AD%A3%E6%96%B9%E5%BD%A2%2C%E5%B9%B3%E9%9D%A2PQC%E2%8A%A5%E5%B9%B3%E9%9D%A2DQC%2CPD%E2%88%A5QA%2CQA%3DAB%3D1%EF%BC%8F2PD+%281%29%E8%AF%81%E6%98%8E%EF%BC%9A%E5%B9%B3%E9%9D%A2ABCD%E2%8A%A5AQ)
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