题目
题型:不详难度:来源:
答案
(2)要证明线面平行,关键是要通过线线平行的证明,结合判定定理来得到,通过做适当的辅助线,结合三角形的中位线平移,得到EF∥DQ.
解析
试题分析:证明:(1)因为PD⊥面ABCD,
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051803-76464.png)
所以PD⊥AB. 2分
在平面ABCD中,D作DM//AB,则由AB=12得
DM=12.又BC=10,AD=
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051803-26055.png)
于是在△CDM中,CD=13,DM=12,CM=5,则
由
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051803-95745.png)
又DM//AB,BC//AD,所以AD⊥AB.
又PD∩AD=D,所以AB⊥面PAD. 6分
(2)[证法一] 取AB的中点N,连结EN、FN.
因为点E是棱PB的中点,所以在△ABP中,EN//
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051803-26055.png)
又PAÌ面PAD,所以EN//面PAD. 8分
因为点F分别是边CD的中点,所以在梯形ABCD中,FN//AD.
又ADÌ面PAD,所以FN//面PAD. 10分
又EN∩FN=N,PA∩DA=A,所以面EFN//面PAD. 12分
又EFÌ面EFN,则EF//面PAD. 14分
[证法二] 延长CD,BA交于点G.
连接PG,EG,EG与PA交于点Q.
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051803-15554.png)
由题设AD∥BC,且AD=
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051803-26055.png)
=AG,即点A为BG的中点.
又因为点E为棱PB的中点,所以EA为△BPG的中位线,即EA∥PG,且EA:PG=1:2,故有EA:PG=EQ:QG=1:2. 10分
又F是边CD的中点,并由CD=DG,则有FD:DG
=1:2. 12分
在△GFE中,由于EQ:QG=1:2,FD:DG=1:2,所以EF∥DQ.
又EFË面PAD,而DQÌ面PAD,所以EF∥面PAD. 14分
点评:解决该试题的关键是熟练的结合线面平行和垂直的判定定理,找到线线的平行和垂直关系,属于基础题。
核心考点
试题【如图,在四棱锥P-ABCD中,PD⊥面ABCD,AD∥BC,CD=13,AB=12,BC=10,AD =12 BC. 点E、F分别是棱PB、边CD的中点.(1)】;主要考察你对线线角等知识点的理解。[详细]
举一反三
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051748-34490.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051748-34490.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051748-25230.png)
(1)求异面直线AE与A1 F所成角的大小;
(2)求平面AEF与平面ABC所成角的余弦值.
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051731-48179.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051731-74444.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051731-70641.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051731-82594.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051732-13917.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051732-26048.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051732-48238.jpg)
(Ⅰ)求证
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051732-79693.png)
(Ⅱ)求证
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051733-20229.png)
(Ⅲ)若
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051733-92367.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051733-63569.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051712-48158.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051713-38611.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051713-97059.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051713-25239.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051713-40808.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051714-28582.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051713-40808.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051714-30805.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051714-47625.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051714-92914.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051714-30805.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051714-49135.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051715-67291.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051715-63869.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051714-30805.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051715-42244.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051715-49580.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051716-74022.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051716-69886.png)
(Ⅰ) 求证:平面
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051716-47183.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051716-32532.png)
(Ⅱ)求平面
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051716-98940.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051717-35482.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051706-94285.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051706-95186.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051706-54524.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051706-67308.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051706-39791.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051707-51883.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051707-25589.jpg)
(Ⅰ)证明:平面
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051707-83773.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051707-66437.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051708-20247.png)
(Ⅱ)若
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051708-33310.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051708-90322.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051706-94285.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022051700-51039.png)
(1)求直线MN与平面ABCD所成角的正弦值;
(2)求异面直线ME与BN所成角的余弦值。
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