(Ⅰ)證明:取BC的中點E,連結(jié)DE,則ABED為正方形.
過P作PO⊥平面ABCD,垂足為O.
連結(jié)OA,OB,OD,OE.
由
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和
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都是等邊三角形知PA=PB=PD,
所以O(shè)A=OB=OD,即點O為正方形ABED對角線的交點,
故
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,
從而
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. 3分
因為O是BD的中點,E是BC的中點,所以O(shè)E//CD.因此
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. 5分
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(Ⅱ)解法一:
由(Ⅰ)知
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,
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,
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.
故
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平面PBD.
又
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平面PBD,所以
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.
取PD的中點F,PC的中點G,連結(jié)FG,
則FG//CD,F(xiàn)G//PD.
連結(jié)AF,由
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為等邊三角形可得AF⊥PD.
所以
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為二面角A-PD-C的平面角. 8分
連結(jié)AG,EG,則EG//PB.
又PB⊥AE,所以EG⊥AE.
設(shè)AB=2,則
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,
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,
故
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.
在
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中,
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,
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,
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,
所以
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.
因此二面角A-PD-C的大小為
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. 12分
解法二:
由(Ⅰ)知,OE,OB,OP兩兩垂直.
以O(shè)為坐標原點,
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的方向為x軸的正方向建立如圖所示的空間直角坐標系O-xyz.
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設(shè)
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,則
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,
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,
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,
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.
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,
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.
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,
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.
設(shè)平面PCD的法向量為
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,則
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,
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,
可得
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,
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.
取
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,得
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,故
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. 8分
設(shè)平面PAD的法向量為
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,則
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,
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,
可得
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.
取m=1,得
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,故
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.
于是
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.
由于
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等于二面角A-PD-C的平面角,
所以二面角A-PD-C的大小為
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. 12分
(1)解題的關(guān)鍵是輔助線的添加,取BC的中點E是入手點,然后借助三垂線定理進行證明;(2)利用三垂線定理法或者空間向量法求解二面角. 求二面角:關(guān)鍵是作出或找出其平面角,常用做法是利用三垂線定理定角法,先找到一個半平面的垂線,然后過垂足作二面角棱的垂線,再連接第三邊,即可得到平面角。若考慮用向量來求:要求出二個面的法向量,然后轉(zhuǎn)化為
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,要注意兩個法向量的夾角與二面角可能相等也可能互補,要從圖上判斷一下二面角是銳二面角還是鈍二面角,然后根據(jù)余弦值確定相等或互補即可。
【考點定位】本題考查線線垂直的證明和二面角的求解,考查學生的空間想象能力和計算能力。