(I) Proof: Suppose AD=1, then DQ=2, DP=2,

∵PD∥QA, ∴∠PDQ=∠AQD=45°,

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In △DPQ, according to the cosine theorem, PQ=2,

∴DQ2+PQ2=DP2, ∴PQ⊥DQ,

and ∵PD⊥plane ABCD , ∴PD⊥DC,

∵CD⊥DA, DA∩PD=D, ∴CD⊥Plane ADPQ

∵PQ?Plane ADPQ, ∴CD⊥PQ,

Also ∵CD∩DQ=D, ∴PQ⊥Plane DCQ,

∵PQ?Plane PQC, ∴Plane DCQ⊥Plane PQC.

(II) Establish a spatial rectangular coordinate system as shown in the figure, assuming AD=1, AB=m.

Then C (0, 0, m), P (0, 2, 0), B (1, 0, m), Q (1, 1, 0),

CB=(1,0,0), BP=(-1,2,-m),

Assume n=(x,y,z) is the normal vector of plane PBC, then n? Liked and Disliked What is your evaluation of this answer? Comments are closed // High quality or satisfactory or special type or recommended answer time window.iPerformance && window.iPerformance.mark('c_best', +new Date); Recommended lawyer service: If your problem is not solved, please provide details Describe your problem and get free professional consultation through Baidu Lulin. Other similar problems 2012-05-26 (2010·Linyi) As shown in Figure 1, the rectangle ABCD is known, point C is the midpoint of side DE, and AB=2AD. 1272016-12-01 (2014? Dongcheng District Second Model) As shown in the figure, in the square pyramid E-ABCD, plane EAD⊥plane ABCD, DC∥AB, BC⊥CD, EA⊥ED, and AB=4, BC=CD= 352012-01-22 As shown in the figure, the quadrilateral ABCD is a square, PD⊥ plane ABCD, PD∥QA, QA=AB=1/2PD (1) Prove: Plane PQ1232012-10-07 As shown in the figure, the quadrilateral ABCD is a square, QA⊥ Plane ABCD, PD∥QA, QA=AB=742016-08-20 (2013? Zibo No. 1 Model) In the geometry as shown in the figure, quadrilateral ABCD is a rhombus, ADNM is a rectangle, plane ADNM⊥plane ABCD, P is DN 32013-04-28 As shown in the figure, the quadrilateral ABCD is a square, PD⊥ plane ABCD, PD∥QA, QA=AB=1/2PD72013-01-25 Quadrilateral ABCD is a square, PD⊥ plane ABCD, PD∥QA, QA =AB=1/2PD62012-12-18 It is known that as shown in the figure: the quadrilateral ABCD is a rectangle, the PA⊥ plane ABCD, M and N are the midpoints of AB and PC respectively.

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