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题目如图1所示,在三棱锥P-ABQ中,PB⊥平面ABQ,BA=BP=BQ,D,C,E,F分别是AQ,BQ,AP,BP的中点,AQ:2BD,PD与EQ交于点G,PC与FQ交于点H,连接GH.(Ⅰ)求证:AB//GH;(Ⅱ)求二面角D-GH-E的余弦值.1解题思路分析与解题方法(Ⅰ)思路1要证明AB∥CH,由中位线的性质EF//AB,DC//AB,只要证明EF//GH,所以关键是证
The topic is shown in Figure 1, in the triangular pyramid P-ABQ, PB⊥ plane ABQ, BA = BP = BQ, D, C, E, F are the midpoints of AQ, BQ, AP, BP, AQ: 2BD, PD and EQ to point G, PC and FQ to point H, connect GH (Ⅰ) Prove: AB // GH; (Ⅱ) Find the cosine of dihedral angle D-GH-E. And problem-solving method (I) idea 1 to prove AB // CH, by the nature of the median line EF // AB, DC // AB, as long as the certification EF / GH, so the key is the card