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            • 1.
              如图,在正三棱柱\(ABC-A_{1}B_{1}C_{1}\)中,点\(D\)在棱\(BC\)上,\(AD⊥C_{1}D\),点\(E\),\(F\)分别是\(BB_{1}\),\(A_{1}B_{1}\)的中点.
              \((1)\)求证:\(D\)为\(BC\)的中点;
              \((2)\)求证:\(EF/\!/\)平面\(ADC_{1}\).
            • 2.
              如图,在直四棱柱\(ABCD-A_{1}B_{1}C_{1}D_{1}\)中,底面\(ABCD\)为平行四边形,\(C_{1}B=C_{1}\)D.求证:
              \((1)B_{1}D_{1}/\!/\)平面\(C_{1}BD\);
              \((2)\)平面\(C_{1}BD⊥\)平面\(AA_{1}C_{1}\)C.
            • 3.
              如图,四边形\(ABCD\)中,\(AB⊥AD\),\(AD/\!/BC\),\(AD=6\),\(BC=2AB=4\),\(E\),\(F\)分别在\(BC\),\(AD\)上,\(EF/\!/AB\),现将四边形\(ABCD\)沿\(EF\)折起,使平面\(ABEF⊥\)平面\(EFDC\).
              \((1)\)若\(BE=1\),是否在折叠后的线段\(AD\)上存在一点\(P\),且\( \overrightarrow{AP}=λ \overrightarrow{PD}\),使得\(CP/\!/\)平面\(ABEF\)?若存在,求出\(λ\)的值,若不存在,说明理由;
              \((2)\)求三棱锥\(A-CDF\)的体积的最大值,并求出此时二面角\(E-AC-F\)的余弦值.
            • 4.
              如图,直四棱柱\(ABCD-A_{1}B_{1}C_{1}D_{1}\)中,\(AB/\!/CD\),\(AD⊥AB\),\(AB=2\),\(AD= \sqrt {2}\),\(AA_{1}=3\),\(E\)为\(CD\)上一点,\(DE=1\),\(EC=3\)
              \((1)\)证明:\(BE⊥\)平面\(BB_{1}C_{1}C\);
              \((2)\)求点\(B_{1}\)到平面\(EA_{1}C_{1}\) 的距离.
            • 5.
              三棱锥被平行于底面\(ABC\)的平面所截得的几何体如图所示,截面为\(A_{1}B_{1}C_{1}\),\(∠BAC=90^{\circ}\),\(A_{1}A⊥\)平面\(ABC\),\(A_{1}A= \sqrt {3}\),\(AB= \sqrt {2}\),\(AC=2\),\(A_{1}C_{1}=1\),\( \dfrac {BD}{DC}= \dfrac {1}{2}\).
              \((\)Ⅰ\()\)证明:\(BC⊥\)平面\(A_{1}AD\)
              \((\)Ⅱ\()\)求二面角\(A-CC_{1}-B\)的余弦值.
            • 6.
              如图所示,在四棱锥\(P-ABCD\)中,底面\(ABCD\)是正方形,侧棱\(PD⊥\)底面\(ABCD\),\(PD=DC=2\),\(E\)是\(PC\)的中点,过\(E\)点作\(EF⊥PB\)交\(PB\)于点\(F\).
              \((1)\)证明:\(PA/\!/\)平面\(EDB\);
              \((2)\)证明:\(PB⊥\)平面\(EFD\);
              \((3)\)求三棱锥\(E-BCD\)的体积.
            • 7.
              如图,在直三棱柱\(ABC-A_{1}B_{1}C_{1}\)中,已知\(AC⊥BC\),\(BC=CC_{1}\),设\(AB_{1}\)的中点为\(D\),\(BC_{1}∩B_{1}C=E.\)求证:
              \((\)Ⅰ\()DE/\!/\)平面\(AA_{1}C_{1}C\);
              \((\)Ⅱ\()BC_{1}⊥AB_{1}\).
            • 8.
              如图\(1\),在高为\(2\)的梯形\(ABCD\)中,\(AB/\!/CD\),\(AB=2\),\(CD=5\),过\(A\)、\(B\)分别作\(AE⊥CD\),\(BF⊥CD\),垂足分别为\(E\)、\(F.\)已知\(DE=1\),将梯形\(ABCD\)沿\(AE\)、\(BF\)同侧折起,使得\(AF⊥BD\),\(DE/\!/CF\),得空间几何体\(ADE-BCF\),如图\(2\).

              \((\)Ⅰ\()\)证明:\(BE/\!/\)面\(ACD\);
              \((\)Ⅱ\()\)求三棱锥\(B-ACD\)的体积.
            • 9.
              如图,正三棱柱的所有棱长都为\(2\),\(D\)为\(CC_{1}\)中点\(.\)用空间向量进行以下证明和计算:
              \((1)\)求证:\(AB_{1}⊥\)面\(A_{1}BD\);
              \((2)\)求二面角\(A-A_{1}D-B\)的正弦值;
              \((3)\)求点\(C\)到面\(A_{1}BD\)的距离.
            • 10.
              如图,已知\(AB⊥\)平面\(ACD\),\(DE/\!/AB\),\(\triangle ACD\)是正三角形,\(AD=DE=2AB\),且\(F\)是\(CD\)的中点.
              \((1)\)求证:\(AF/\!/\)平面\(BCE\);
              \((2)\)求证:平面\(BCE⊥\)平面\(CDE\);
              \((3)\)求平面\(BCE\)与平面\(ACD\)所成锐二面角的大小.
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