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题目如图1,几何体E-ABCD是四棱锥,△ABD为正三角形,CB=CD,EC⊥BD.(Ⅰ)求证:BE=DE;(Ⅱ)若∠BCD=120°,M为线段AE的中点,求证:DM∥平面BEC.本题是2012年山东高考文科数学立体几何解答题,在知识上主要考查直线与直线、直线与平面、平面与平面的平行和垂直位置关系;在思想方法上主要考查转化与化归;在能力上主要考查空间想象能力、推理论证能力.
The subject is shown in Figure 1, geometry E-ABCD is a quadrangular pyramid, △ ABD is a triangle, CB = CD, EC ⊥ BD. (Ⅰ) Proof: BE = DE; The middle point, verify: DM // plane BEC. This question is 2012 Shandong entrance liberal arts mathematics three-dimensional geometric solution questions, the main knowledge in the test of straight line and straight line, straight line and plane, plane and plane parallel and vertical position relationship; On the main test of conversion and return; ability mainly on the spatial imagination, reasoning and argumentation ability.