di-zhang-fdu/OlympiadExtended huggingface dataset:
Subfield: Solid Geometry


Question: $如图，平行六面体ABCD-A_{1}B_{1}C_{1}D_{1}的底面ABCD是菱形，\angle C_{1}CB=\angle C_{1}CD=\angle BCD=60^\circ .$<img_2837>
$(1)求证：C_{1}C\perp BD；$

Solution:
$证明：如图，连接A_{1}C_{1}，AC，设AC和BD交于点O，连接C_{1}O.$<img_3053>$\because 四边形ABCD是菱形，\therefore AC\perp BD，BC=CD.又\because \angle C_{1}CB=\angle C_{1}CD，C_{1}C是公共边，\therefore \triangle C_{1}BC\cong \triangle C_{1}DC，\therefore C_{1}B=C_{1}D.\because DO=OB，\therefore C_{1}O\perp BD.又\because AC\cap C_{1}O=O，\therefore BD\perp 平面ACC_{1}A_{1}.又\because C_{1}C\subset 平面ACC_{1}A_{1}，\therefore C_{1}C\perp BD.$
Subfield: Solid Geometry


Question: $如图，平行六面体ABCD-A_{1}B_{1}C_{1}D_{1}的底面ABCD是菱形，\angle C_{1}CB=\angle C_{1}CD=\angle BCD=60^\circ .$![Image](https://cdntk.53zaixian.com/tq/1702600667896090626/img/compression/M_c0c1b90220befdd479872cf04d03e40f.jpg)
$(1)求证：C_{1}C\perp BD；$

Solution:
$证明：如图，连接A_{1}C_{1}，AC，设AC和BD交于点O，连接C_{1}O.$![Image](https://cdntk.53zaixian.com/tq/1702600668151943170/img/compression/M_eb1c0062d47b7e8740664d72b512f370.jpg)$\because 四边形ABCD是菱形，\therefore AC\perp BD，BC=CD.又\because \angle C_{1}CB=\angle C_{1}CD，C_{1}C是公共边，\therefore \triangle C_{1}BC\cong \triangle C_{1}DC，\therefore C_{1}B=C_{1}D.\because DO=OB，\therefore C_{1}O\perp BD.又\because AC\cap C_{1}O=O，\therefore BD\perp 平面ACC_{1}A_{1}.又\because C_{1}C\subset 平面ACC_{1}A_{1}，\therefore C_{1}C\perp BD.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，平行六面体ABCD-A_{1}B_{1}C_{1}D_{1}的底面ABCD是菱形，\angle C_{1}CB=\angle C_{1}CD=\angle BCD=60^\circ .$![Image](https://cdntk.53zaixian.com/tq/1702600667896090626/img/compression/M_c0c1b90220befdd479872cf04d03e40f.jpg)
$(2)当\frac{CD}{CC_1}的值为多少时，可使A_{1}C\perp 平面C_{1}BD?$

Solution:
$当\frac{CD}{CC_1}=1时，可使A_{1}C\perp 平面C_{1}BD.理由如下：由(1)知BD\perp 平面ACC_{1}A_{1}.\because A_{1}C\subset 平面ACC_{1}A_{1}，\therefore BD\perp A_{1}C.当\frac{CD}{CC_1}=1时，平行六面体的六个面是全等的菱形.同理,可证BC_{1}\perp A_{1}C.又\because BD\cap BC_{1}=B，\therefore A_{1}C\perp 平面C_{1}BD.故当\frac{CD}{CC_1}=1时，A_{1}C\perp 平面C_{1}BD.$

Final answer: $当\frac{CD}{CC_1}=1时，可使A_{1}C\perp 平面C_{1}BD.$
Subfield: Solid Geometry


Question: $如图，平行六面体ABCD-A_{1}B_{1}C_{1}D_{1}的底面ABCD是菱形，\angle C_{1}CB=\angle C_{1}CD=\angle BCD=60^\circ .$<img_2837>
$(2)当\frac{CD}{CC_1}的值为多少时，可使A_{1}C\perp 平面C_{1}BD?$

Solution:
$当\frac{CD}{CC_1}=1时，可使A_{1}C\perp 平面C_{1}BD.理由如下：由(1)知BD\perp 平面ACC_{1}A_{1}.\because A_{1}C\subset 平面ACC_{1}A_{1}，\therefore BD\perp A_{1}C.当\frac{CD}{CC_1}=1时，平行六面体的六个面是全等的菱形.同理,可证BC_{1}\perp A_{1}C.又\because BD\cap BC_{1}=B，\therefore A_{1}C\perp 平面C_{1}BD.故当\frac{CD}{CC_1}=1时，A_{1}C\perp 平面C_{1}BD.$



Final answer: 1
Subfield: Solid Geometry


Question: $如图，几何体P-ABC是底面边长为1的正三棱锥，D,E,F分别为棱PA,PB,PC上的点，截面DEF\parallel 底面ABC，且棱台DEF-ABC与棱锥P-ABC的棱长和相等.(棱长和是指多面体中所有棱的长度之和)$![Image](https://cdntk.53zaixian.com/tq/1702600708257628164/img/compression/M_d696af5e4911787ff2dc85dc1f424632.jpg)
$(1)证明：几何体P-ABC为正四面体；$

Solution:
$证明：\because 棱台DEF-ABC与棱锥P-ABC的棱长和相等，\therefore DE+EF+FD=PD+PE+PF.\because 平面DEF\parallel 平面ABC,三棱锥P-ABC为正三棱锥,\therefore DE=EF=FD=PD=PE=PF,\therefore \angle DPE=\angle EPF=\angle FPD=60^\circ ,\therefore PA=PB=PC=AB=AC=BC,\therefore 几何体P-ABC为正四面体.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，几何体P-ABC是底面边长为1的正三棱锥，D,E,F分别为棱PA,PB,PC上的点，截面DEF\parallel 底面ABC，且棱台DEF-ABC与棱锥P-ABC的棱长和相等.(棱长和是指多面体中所有棱的长度之和)$<img_2413>
$(1)证明：几何体P-ABC为正四面体；$

Solution:
$证明：\because 棱台DEF-ABC与棱锥P-ABC的棱长和相等，\therefore DE+EF+FD=PD+PE+PF.\because 平面DEF\parallel 平面ABC,三棱锥P-ABC为正三棱锥,\therefore DE=EF=FD=PD=PE=PF,\therefore \angle DPE=\angle EPF=\angle FPD=60^\circ ,\therefore PA=PB=PC=AB=AC=BC,\therefore 几何体P-ABC为正四面体.$
Subfield: Solid Geometry


Question: $如图，几何体P-ABC是底面边长为1的正三棱锥，D,E,F分别为棱PA,PB,PC上的点，截面DEF\parallel 底面ABC，且棱台DEF-ABC与棱锥P-ABC的棱长和相等.(棱长和是指多面体中所有棱的长度之和)$<img_2413>
$(2)若PD=\frac{1}{2}PA，求二面角D-BC-A的大小；(结果用反三角函数值表示)$

Solution:
$如图，取BC的中点M，连接PM,DM,AM.$<img_2894>$\because PB=PC,AB=AC，\therefore PM\perp BC,AM\perp BC.\because AM,PM\subset 平面PAM,AM\cap PM=M,\therefore BC\perp 平面PAM,\because DM\subset 平面PAM,\therefore BC\perp DM,\therefore \angle DMA为二面角D-BC-A的平面角,由(1)知，三棱锥P-ABC各棱长均为1,\therefore PM=AM=\frac{\sqrt{3}}{2},\because PD=\frac{1}{2}PA,\therefore DM\perp PA,\therefore \sin  \angle DMA=\frac{AD}{AM}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{\sqrt{3}}{3},\therefore \angle DMA=arc\sin  \frac{\sqrt{3}}{3},即二面角D-BC-A的大小为arc\sin  \frac{\sqrt{3}}{3}.$

Final answer: $\arcsin  \frac{\sqrt{3}}{3}$
Subfield: Solid Geometry


Question: $如图，几何体P-ABC是底面边长为1的正三棱锥，D,E,F分别为棱PA,PB,PC上的点，截面DEF\parallel 底面ABC，且棱台DEF-ABC与棱锥P-ABC的棱长和相等.(棱长和是指多面体中所有棱的长度之和)$![Image](https://cdntk.53zaixian.com/tq/1702600708257628164/img/compression/M_d696af5e4911787ff2dc85dc1f424632.jpg)
$(2)若PD=\frac{1}{2}PA，求二面角D-BC-A的大小；(结果用反三角函数值表示)$

Solution:
$如图，取BC的中点M，连接PM,DM,AM.$![Image](https://cdntk.53zaixian.com/tq/1702600708513480706/img/compression/M_78391451929aadd467a9c0be34677b79.jpg)$\because PB=PC,AB=AC，\therefore PM\perp BC,AM\perp BC.\because AM,PM\subset 平面PAM,AM\cap PM=M,\therefore BC\perp 平面PAM,\because DM\subset 平面PAM,\therefore BC\perp DM,\therefore \angle DMA为二面角D-BC-A的平面角,由(1)知，三棱锥P-ABC各棱长均为1,\therefore PM=AM=\frac{\sqrt{3}}{2},\because PD=\frac{1}{2}PA,\therefore DM\perp PA,\therefore \sin  \angle DMA=\frac{AD}{AM}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{\sqrt{3}}{3},\therefore \angle DMA=arc\sin  \frac{\sqrt{3}}{3},即二面角D-BC-A的大小为arc\sin  \frac{\sqrt{3}}{3}.$

Final answer: $\arcsin  \frac{\sqrt{3}}{3}$
Subfield: Solid Geometry


Question: $如图，在三棱柱ABC-DEF中，侧面ADFC是边长为2的正方形，M是棱EF的中点.$![Image](https://cdntk.53zaixian.com/tq/1702591452474458115/img/compression/M_23899c7d663b610d048cd32618b4fedd.jpg)
$(1)证明：AF\parallel 平面BDM；$

Solution:
$证明：如图，连接AE，与BD交于点N，连接MN.$![Image](https://cdntk.53zaixian.com/tq/1702591452885499906/img/compression/M_aaec757d835b974de55880b670c1290a.jpg)$因为侧面ABED是平行四边形，所以N是AE的中点. (2分)因为M是棱EF的中点，所以MN是\triangle AEF的中位线，所以MN\parallel AF. (4分)因为AF\not\subset 平面BDM，MN\subset 平面BDM，所以AF\parallel 平面BDM. (6分)$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在三棱柱ABC-DEF中，侧面ADFC是边长为2的正方形，M是棱EF的中点.$<img_2876>
$(1)证明：AF\parallel 平面BDM；$

Solution:
$证明：如图，连接AE，与BD交于点N，连接MN.$<img_3004>$因为侧面ABED是平行四边形，所以N是AE的中点. (2分)因为M是棱EF的中点，所以MN是\triangle AEF的中位线，所以MN\parallel AF. (4分)因为AF\not\subset 平面BDM，MN\subset 平面BDM，所以AF\parallel 平面BDM. (6分)$
Subfield: Solid Geometry


Question: $如图，在三棱柱ABC-DEF中，侧面ADFC是边长为2的正方形，M是棱EF的中点.$<img_2876>
$(2)若三棱锥B-DEF的体积为4，求点B到平面ADFC的距离.$

Solution:
$因为三棱锥B-DEF的体积为4，所以三棱柱ABC-DEF的体积为12， 则四棱锥B-ADFC的体积为12-4=8.(8分)因为侧面ADFC是边长为2的正方形，所以侧面ADFC的面积为2\times 2=4. (10分)设点B到平面ADFC的距离为h，则\frac{1}{3}\times 4h=8，解得h=6. 故点B到平面ADFC的距离为6.(12分)$

Final answer: $6$
Subfield: Solid Geometry


Question: $如图，在三棱柱ABC-DEF中，侧面ADFC是边长为2的正方形，M是棱EF的中点.$![Image](https://cdntk.53zaixian.com/tq/1702591452474458115/img/compression/M_23899c7d663b610d048cd32618b4fedd.jpg)
$(2)若三棱锥B-DEF的体积为4，求点B到平面ADFC的距离.$

Solution:
$因为三棱锥B-DEF的体积为4，所以三棱柱ABC-DEF的体积为12， 则四棱锥B-ADFC的体积为12-4=8.(8分)因为侧面ADFC是边长为2的正方形，所以侧面ADFC的面积为2\times 2=4. (10分)设点B到平面ADFC的距离为h，则\frac{1}{3}\times 4h=8，解得h=6. 故点B到平面ADFC的距离为6.(12分)$

Final answer: $6$
Subfield: Solid Geometry


Question: $如图，在棱长为1的正四面体OABC中，M是棱BC的中点，点N在线段OM上，点P在线段AN上，且MN=\frac{1}{2}ON，AP=\frac{3}{4}AN.$<img_2128>
$(1)用向量\overrightarrow{OA}，\overrightarrow{OB}，\overrightarrow{OC}表示\overrightarrow{AN}；$

Solution:
$\overrightarrow{AN}=\overrightarrow{AO}+\overrightarrow{ON}=－\overrightarrow{OA}+\frac{2}{3}\overrightarrow{OM}=－\overrightarrow{OA}+\frac{1}{3}\overrightarrow{OB}+\frac{1}{3}\overrightarrow{OC}.$



Final answer: $\overrightarrow{AN}=－\overrightarrow{OA}+\frac{1}{3}\overrightarrow{OB}+\frac{1}{3}\overrightarrow{OC}$
Subfield: Solid Geometry


Question: $如图，在棱长为1的正四面体OABC中，M是棱BC的中点，点N在线段OM上，点P在线段AN上，且MN=\frac{1}{2}ON，AP=\frac{3}{4}AN.$![Image](https://cdntk.53zaixian.com/tq/1658643153781022722/img/compression/M_b5ff8145ecfa80c3d50a60ed60228756.jpg)
$(1)用向量\overrightarrow{OA}，\overrightarrow{OB}，\overrightarrow{OC}表示\overrightarrow{AN}；$

Solution:
$\overrightarrow{AN}=\overrightarrow{AO}+\overrightarrow{ON}=－\overrightarrow{OA}+\frac{2}{3}\overrightarrow{OM}=－\overrightarrow{OA}+\frac{1}{3}\overrightarrow{OB}+\frac{1}{3}\overrightarrow{OC}.$

Final answer: $\overrightarrow{AN}=－\overrightarrow{OA}+\frac{1}{3}\overrightarrow{OB}+\frac{1}{3}\overrightarrow{OC}.$
Subfield: Solid Geometry


Question: $如图，在棱长为1的正四面体OABC中，M是棱BC的中点，点N在线段OM上，点P在线段AN上，且MN=\frac{1}{2}ON，AP=\frac{3}{4}AN.$<img_2128>
$(2)求|\overrightarrow{OP}|.$

Solution:
$\overrightarrow{OP}=\overrightarrow{OA}+\overrightarrow{AP}=\overrightarrow{OA}+\frac{3}{4}\overrightarrow{AN}=\overrightarrow{OA}+\frac{3}{4}\left(\right.-\overrightarrow{OA}+\frac{1}{3}\overrightarrow{OB}+\frac{1}{3}\overrightarrow{OC}\left)\right.=\frac{1}{4}\overrightarrow{OA}+\frac{1}{4}\overrightarrow{OB}+\frac{1}{4}\overrightarrow{OC}，\therefore |\overrightarrow{OP}|^{2}=\frac{1}{16}(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC})^{2}=\frac{1}{16}(\overrightarrow{OA}^2+\overrightarrow{OB}^2+\overrightarrow{OC}^2+2\overrightarrow{OA}\cdot \overrightarrow{OB}+2\overrightarrow{OA}\cdot \overrightarrow{OC}+2\overrightarrow{OB}\cdot \overrightarrow{OC})=\frac{1}{16}\left(\right.1+1+1+2\times 1\times 1\times \frac{1}{2}\times 3\left)\right.=\frac{3}{8}，\therefore |\overrightarrow{OP}|=\frac{\sqrt{6}}{4}.$

Final answer: $\frac{\sqrt{6}}{4}$
Subfield: Solid Geometry


Question: $如图，在棱长为1的正四面体OABC中，M是棱BC的中点，点N在线段OM上，点P在线段AN上，且MN=\frac{1}{2}ON，AP=\frac{3}{4}AN.$![Image](https://cdntk.53zaixian.com/tq/1658643153781022722/img/compression/M_b5ff8145ecfa80c3d50a60ed60228756.jpg)
$(2)求|\overrightarrow{OP}|.$

Solution:
$\overrightarrow{OP}=\overrightarrow{OA}+\overrightarrow{AP}=\overrightarrow{OA}+\frac{3}{4}\overrightarrow{AN}=\overrightarrow{OA}+\frac{3}{4}\left(\right.-\overrightarrow{OA}+\frac{1}{3}\overrightarrow{OB}+\frac{1}{3}\overrightarrow{OC}\left)\right.=\frac{1}{4}\overrightarrow{OA}+\frac{1}{4}\overrightarrow{OB}+\frac{1}{4}\overrightarrow{OC}，\therefore |\overrightarrow{OP}|^{2}=\frac{1}{16}(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC})^{2}=\frac{1}{16}(\overrightarrow{OA}^2+\overrightarrow{OB}^2+\overrightarrow{OC}^2+2\overrightarrow{OA}\cdot \overrightarrow{OB}+2\overrightarrow{OA}\cdot \overrightarrow{OC}+2\overrightarrow{OB}\cdot \overrightarrow{OC})=\frac{1}{16}\left(\right.1+1+1+2\times 1\times 1\times \frac{1}{2}\times 3\left)\right.=\frac{3}{8}，\therefore |\overrightarrow{OP}|=\frac{\sqrt{6}}{4}.$

Final answer: $\frac{\sqrt{6}}{4}$
Subfield: Solid Geometry


Question: $如图所示,在平行六面体ABCD-A_{1}B_{1}C_{1}D_{1}中,E，F分别在棱B_{1}B和D_{1}D上,且BE=\frac{1}{3}B_{1}B，DF=\frac{2}{3}DD_{1}.$<img_2822>
$(1)求证:A，E，C_{1}，F四点共面;$

Solution:
$证明:连接AC_{1}，AC，则\overrightarrow{AC_1}=\overrightarrow{AC}+\overrightarrow{CC_1}=\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{AA_1}=\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AA_1}+\overrightarrow{AD}+\frac{2}{3}\overrightarrow{AA_1}=(\overrightarrow{AB}+\overrightarrow{BE})+(\overrightarrow{AD}+\overrightarrow{DF})=\overrightarrow{AE}+\overrightarrow{AF}, \therefore 由共面向量定理可知A,E,C_{1},F四点共面.$
Subfield: Solid Geometry


Question: $如图所示,在平行六面体ABCD-A_{1}B_{1}C_{1}D_{1}中,E，F分别在棱B_{1}B和D_{1}D上,且BE=\frac{1}{3}B_{1}B，DF=\frac{2}{3}DD_{1}.$![Image](https://cdntk.53zaixian.com/tq/1658643127211999234/img/compression/M_84bb335cb8c688f3ecb9443443b606ea.jpg)
$(1)求证:A，E，C_{1}，F四点共面;$

Solution:
$证明:连接AC_{1}，AC，则\overrightarrow{AC_1}=\overrightarrow{AC}+\overrightarrow{CC_1}=\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{AA_1}=\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AA_1}+\overrightarrow{AD}+\frac{2}{3}\overrightarrow{AA_1}=(\overrightarrow{AB}+\overrightarrow{BE})+(\overrightarrow{AD}+\overrightarrow{DF})=\overrightarrow{AE}+\overrightarrow{AF}, \therefore 由共面向量定理可知A,E,C_{1},F四点共面.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图所示,在平行六面体ABCD-A_{1}B_{1}C_{1}D_{1}中,E，F分别在棱B_{1}B和D_{1}D上,且BE=\frac{1}{3}B_{1}B，DF=\frac{2}{3}DD_{1}.$<img_2822>
$(2)若\overrightarrow{EF}=x\overrightarrow{AB}+y\overrightarrow{AD}+z\overrightarrow{AA_1},求x+y+z的值.$

Solution:
$\overrightarrow{EF}=\overrightarrow{AF}-\overrightarrow{AE}=\overrightarrow{AD}+\overrightarrow{DF}-(\overrightarrow{AB}+\overrightarrow{BE}) =\overrightarrow{AD}+\frac{2}{3}\overrightarrow{DD_1}-\overrightarrow{AB}-\frac{1}{3}\overrightarrow{BB_1} =-\overrightarrow{AB}+\overrightarrow{AD}+\frac{1}{3}\overrightarrow{AA_1},又\overrightarrow{EF}=x\overrightarrow{AB}+y\overrightarrow{AD}+z\overrightarrow{AA_1},\therefore x=-1,y=1,z=\frac{1}{3},\therefore x+y+z=\frac{1}{3}.$

Final answer: $\frac{1}{3}$
Subfield: Solid Geometry


Question: $如图所示,在平行六面体ABCD-A_{1}B_{1}C_{1}D_{1}中,E，F分别在棱B_{1}B和D_{1}D上,且BE=\frac{1}{3}B_{1}B，DF=\frac{2}{3}DD_{1}.$![Image](https://cdntk.53zaixian.com/tq/1658643127211999234/img/compression/M_84bb335cb8c688f3ecb9443443b606ea.jpg)
$(2)若\overrightarrow{EF}=x\overrightarrow{AB}+y\overrightarrow{AD}+z\overrightarrow{AA_1},求x+y+z的值.$

Solution:
$\overrightarrow{EF}=\overrightarrow{AF}-\overrightarrow{AE}=\overrightarrow{AD}+\overrightarrow{DF}-(\overrightarrow{AB}+\overrightarrow{BE}) =\overrightarrow{AD}+\frac{2}{3}\overrightarrow{DD_1}-\overrightarrow{AB}-\frac{1}{3}\overrightarrow{BB_1} =-\overrightarrow{AB}+\overrightarrow{AD}+\frac{1}{3}\overrightarrow{AA_1},又\overrightarrow{EF}=x\overrightarrow{AB}+y\overrightarrow{AD}+z\overrightarrow{AA_1},\therefore x=-1,y=1,z=\frac{1}{3},\therefore x+y+z=\frac{1}{3}.$

Final answer: $\frac{1}{3}$
Subfield: Solid Geometry


Question: $如图所示,在三棱锥A-BCD中,DA,DB,DC两两垂直,且DB=DC=DA=2,E为BC的中点.$![Image](https://cdntk.53zaixian.com/tq/1658643135858282498/img/compression/M_7fe3470d097013e0ffe6c4dc64ec2714.jpg)
$(1)证明:AE\perp BC;$

Solution:
$证明:连接DE，则\overrightarrow{AE}=\overrightarrow{DE}-\overrightarrow{DA}=\frac{1}{2}(\overrightarrow{DB}+\overrightarrow{DC})-\overrightarrow{DA},\overrightarrow{BC}=\overrightarrow{DC}-\overrightarrow{DB},所以\overrightarrow{AE}\cdot \overrightarrow{BC}=\left(\frac{1}{2}\overrightarrow{DB}+\frac{1}{2}\overrightarrow{DC}-\overrightarrow{DA}\right)\cdot (\overrightarrow{DC}-\overrightarrow{DB})=\frac{1}{2}\overrightarrow{DB}\cdot \overrightarrow{DC}-\frac{1}{2}\overrightarrow{DB}\cdot \overrightarrow{DB}+\frac{1}{2}\overrightarrow{DC}\cdot \overrightarrow{DC}-\frac{1}{2}\overrightarrow{DC}\cdot \overrightarrow{DB}-\overrightarrow{DA}\cdot \overrightarrow{DC}+\overrightarrow{DA}\cdot \overrightarrow{DB}=0-2+2-0-0+0=0,所以AE\perp BC.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图所示,在三棱锥A-BCD中,DA,DB,DC两两垂直,且DB=DC=DA=2,E为BC的中点.$<img_2242>
$(1)证明:AE\perp BC;$

Solution:
$证明:连接DE，则\overrightarrow{AE}=\overrightarrow{DE}-\overrightarrow{DA}=\frac{1}{2}(\overrightarrow{DB}+\overrightarrow{DC})-\overrightarrow{DA},\overrightarrow{BC}=\overrightarrow{DC}-\overrightarrow{DB},所以\overrightarrow{AE}\cdot \overrightarrow{BC}=\left(\frac{1}{2}\overrightarrow{DB}+\frac{1}{2}\overrightarrow{DC}-\overrightarrow{DA}\right)\cdot (\overrightarrow{DC}-\overrightarrow{DB})=\frac{1}{2}\overrightarrow{DB}\cdot \overrightarrow{DC}-\frac{1}{2}\overrightarrow{DB}\cdot \overrightarrow{DB}+\frac{1}{2}\overrightarrow{DC}\cdot \overrightarrow{DC}-\frac{1}{2}\overrightarrow{DC}\cdot \overrightarrow{DB}-\overrightarrow{DA}\cdot \overrightarrow{DC}+\overrightarrow{DA}\cdot \overrightarrow{DB}=0-2+2-0-0+0=0,所以AE\perp BC.$
Subfield: Solid Geometry


Question: $如图所示,在三棱锥A-BCD中,DA,DB,DC两两垂直,且DB=DC=DA=2,E为BC的中点.$<img_2242>
$(2)求直线AE与DC所成角的余弦值.$

Solution:
$\overrightarrow{AE}\cdot \overrightarrow{DC}=\left(\frac{1}{2}\overrightarrow{DB}+\frac{1}{2}\overrightarrow{DC}-\overrightarrow{DA}\right)\cdot \overrightarrow{DC}=\frac{1}{2}\overrightarrow{DB}\cdot \overrightarrow{DC}+\frac{1}{2}\overrightarrow{DC}\cdot \overrightarrow{DC}-\overrightarrow{DA}\cdot \overrightarrow{DC}=0+2-0=2,|\overrightarrow{AE}|=\sqrt{(\sqrt{2}{)}^{2}+{2}^{2}}=\sqrt{6},所以\\cos  \langle \overrightarrow{AE},\overrightarrow{DC}\rangle =\frac{\overrightarrow{AE} \cdot \overrightarrow{DC}}{|\overrightarrow{AE}||\overrightarrow{DC}|}=\frac{2}{\sqrt{6} \times 2}=\frac{\sqrt{6}}{6},即直线AE与DC所成角的余弦值为\frac{\sqrt{6}}{6}.$

Final answer: $\frac{\sqrt{6}}{6}$
Subfield: Solid Geometry


Question: $如图所示,在三棱锥A-BCD中,DA,DB,DC两两垂直,且DB=DC=DA=2,E为BC的中点.$![Image](https://cdntk.53zaixian.com/tq/1658643135858282498/img/compression/M_7fe3470d097013e0ffe6c4dc64ec2714.jpg)
$(2)求直线AE与DC所成角的余弦值.$

Solution:
$\overrightarrow{AE}\cdot \overrightarrow{DC}=\left(\frac{1}{2}\overrightarrow{DB}+\frac{1}{2}\overrightarrow{DC}-\overrightarrow{DA}\right)\cdot \overrightarrow{DC}=\frac{1}{2}\overrightarrow{DB}\cdot \overrightarrow{DC}+\frac{1}{2}\overrightarrow{DC}\cdot \overrightarrow{DC}-\overrightarrow{DA}\cdot \overrightarrow{DC}=0+2-0=2,|\overrightarrow{AE}|=\sqrt{(\sqrt{2}{)}^{2}+{2}^{2}}=\sqrt{6},所以\\cos  \langle \overrightarrow{AE},\overrightarrow{DC}\rangle =\frac{\overrightarrow{AE} \cdot \overrightarrow{DC}}{|\overrightarrow{AE}||\overrightarrow{DC}|}=\frac{2}{\sqrt{6} \times 2}=\frac{\sqrt{6}}{6},即直线AE与DC所成角的余弦值为\frac{\sqrt{6}}{6}.$

Final answer: $\frac{\sqrt{6}}{6}$
Subfield: Solid Geometry


Question: $如图，在四棱锥P－ABCD中，底面ABCD为直角梯形，AD\parallel BC，\angle BAD=90^\circ ，PA\perp 底面ABCD，且PA=AD=AB=2BC，M为PC的中点.$![Image](https://cdntk.53zaixian.com/tq/1658643137523421187/img/compression/M_322a744740c30f8a1dcb6aac18d4ac28.jpg)
$(1)求证：PB\perp DM；$

Solution:
$证明：结合题图，可得\overrightarrow{PB}=\overrightarrow{AB}－\overrightarrow{AP}，\overrightarrow{DM}=\frac{1}{2}(\overrightarrow{DP}+\overrightarrow{DC})=\frac{1}{2}\left(\right.\overrightarrow{AP}－\overrightarrow{AD}+\overrightarrow{AB}－\frac{1}{2}\overrightarrow{AD}\left)\right.=\frac{1}{2}\overrightarrow{AP}+\frac{1}{2}\overrightarrow{AB}－\frac{3}{4}\overrightarrow{AD}，则\overrightarrow{PB}\cdot \overrightarrow{DM}=(\overrightarrow{AB}－\overrightarrow{AP})\cdot \left(\right.\frac{1}{2}\overrightarrow{AP}+\frac{1}{2}\overrightarrow{AB}－\frac{3}{4}\overrightarrow{AD}\left)\right.=\frac{1}{2}|\overrightarrow{AB}|^{2}－\frac{1}{2}|\overrightarrow{AP}|^{2}=0,\therefore \overrightarrow{PB}\perp \overrightarrow{DM}，即PB\perp DM.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在四棱锥P－ABCD中，底面ABCD为直角梯形，AD\parallel BC，\angle BAD=90^\circ ，PA\perp 底面ABCD，且PA=AD=AB=2BC，M为PC的中点.$<img_2114>
$(1)求证：PB\perp DM；$

Solution:
$证明：结合题图，可得\overrightarrow{PB}=\overrightarrow{AB}－\overrightarrow{AP}，\overrightarrow{DM}=\frac{1}{2}(\overrightarrow{DP}+\overrightarrow{DC})=\frac{1}{2}\left(\right.\overrightarrow{AP}－\overrightarrow{AD}+\overrightarrow{AB}－\frac{1}{2}\overrightarrow{AD}\left)\right.=\frac{1}{2}\overrightarrow{AP}+\frac{1}{2}\overrightarrow{AB}－\frac{3}{4}\overrightarrow{AD}，则\overrightarrow{PB}\cdot \overrightarrow{DM}=(\overrightarrow{AB}－\overrightarrow{AP})\cdot \left(\right.\frac{1}{2}\overrightarrow{AP}+\frac{1}{2}\overrightarrow{AB}－\frac{3}{4}\overrightarrow{AD}\left)\right.=\frac{1}{2}|\overrightarrow{AB}|^{2}－\frac{1}{2}|\overrightarrow{AP}|^{2}=0,\therefore \overrightarrow{PB}\perp \overrightarrow{DM}，即PB\perp DM.$
Subfield: Solid Geometry


Question: $如图，在四棱锥P－ABCD中，底面ABCD为直角梯形，AD\parallel BC，\angle BAD=90^\circ ，PA\perp 底面ABCD，且PA=AD=AB=2BC，M为PC的中点.$<img_2114>
$(2)求AC与PD所成角的余弦值.$

Solution:
$设PA=AD=AB=2BC=2a(a>0)，由于\overrightarrow{PD}=\overrightarrow{AD}－\overrightarrow{AP}，\overrightarrow{AC}=\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AD}，因此\overrightarrow{PD}\cdot \overrightarrow{AC}=(\overrightarrow{AD}－\overrightarrow{AP})\cdot \left(\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AD}\right)=2a^{2}，又|\overrightarrow{PD}|=\sqrt{|\overrightarrow{PA}|^2+|\overrightarrow{AD}|^2}=2\sqrt{2}a，|\overrightarrow{AC}|=\sqrt{|\overrightarrow{AB}|^2+|\overrightarrow{BC}|^2}=\sqrt{5}a，\therefore \\cos  \langle \overrightarrow{AC},\overrightarrow{PD}\rangle =\frac{\overrightarrow{AC} \cdot \overrightarrow{PD}}{|\overrightarrow{AC}||\overrightarrow{PD}|}=\frac{2a^2}{\sqrt{5}a \times 2\sqrt{2}a}=\frac{\sqrt{10}}{10}.\therefore AC与PD所成角的余弦值为\frac{\sqrt{10}}{10}.$

Final answer: $\frac{\sqrt{10}}{10}$
Subfield: Solid Geometry


Question: $如图，在四棱锥P－ABCD中，底面ABCD为直角梯形，AD\parallel BC，\angle BAD=90^\circ ，PA\perp 底面ABCD，且PA=AD=AB=2BC，M为PC的中点.$![Image](https://cdntk.53zaixian.com/tq/1658643137523421187/img/compression/M_322a744740c30f8a1dcb6aac18d4ac28.jpg)
$(2)求AC与PD所成角的余弦值.$

Solution:
$设PA=AD=AB=2BC=2a(a>0)，由于\overrightarrow{PD}=\overrightarrow{AD}－\overrightarrow{AP}，\overrightarrow{AC}=\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AD}，因此\overrightarrow{PD}\cdot \overrightarrow{AC}=(\overrightarrow{AD}－\overrightarrow{AP})\cdot \left(\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AD}\right)=2a^{2}，又|\overrightarrow{PD}|=\sqrt{|\overrightarrow{PA}|^2+|\overrightarrow{AD}|^2}=2\sqrt{2}a，|\overrightarrow{AC}|=\sqrt{|\overrightarrow{AB}|^2+|\overrightarrow{BC}|^2}=\sqrt{5}a，\therefore \\cos  \langle \overrightarrow{AC},\overrightarrow{PD}\rangle =\frac{\overrightarrow{AC} \cdot \overrightarrow{PD}}{|\overrightarrow{AC}||\overrightarrow{PD}|}=\frac{2a^2}{\sqrt{5}a \times 2\sqrt{2}a}=\frac{\sqrt{10}}{10}.\therefore AC与PD所成角的余弦值为\frac{\sqrt{10}}{10}.$

Final answer: $\frac{\sqrt{10}}{10}$
Subfield: Solid Geometry


Question: $如图，在四棱锥P-ABCD中，PA\perp 底面ABCD，AD\perp DC，AB\parallel DC，AD=DC=AP=2，AB=1，点E为棱PC的中点.$![Image](https://cdntk.53zaixian.com/tq/1658643137963823107/img/compression/M_142f1c4dd7802804242f604c62d29864.jpg)
$(1)证明：BE\perp PD；$

Solution:
$证明：\because E为PC的中点，\therefore \overrightarrow{BE}=\frac{1}{2}(\overrightarrow{BP}+\overrightarrow{BC})=\frac{1}{2}(\overrightarrow{AP}-\overrightarrow{AB}+\overrightarrow{AC}-\overrightarrow{AB})=\frac{1}{2}(\overrightarrow{AP}+\overrightarrow{AC}-\overrightarrow{DC})=\frac{1}{2}(\overrightarrow{AP}+\overrightarrow{AD})，又\overrightarrow{DP}=\overrightarrow{AP}-\overrightarrow{AD},\therefore \overrightarrow{BE}\cdot \overrightarrow{DP}=\frac{1}{2}(\overrightarrow{AP}^{2}-\overrightarrow{AD}^{2})=0,\therefore BE\perp PD.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在四棱锥P-ABCD中，PA\perp 底面ABCD，AD\perp DC，AB\parallel DC，AD=DC=AP=2，AB=1，点E为棱PC的中点.$<img_2376>
$(1)证明：BE\perp PD；$

Solution:
$证明：\because E为PC的中点，\therefore \overrightarrow{BE}=\frac{1}{2}(\overrightarrow{BP}+\overrightarrow{BC})=\frac{1}{2}(\overrightarrow{AP}-\overrightarrow{AB}+\overrightarrow{AC}-\overrightarrow{AB})=\frac{1}{2}(\overrightarrow{AP}+\overrightarrow{AC}-\overrightarrow{DC})=\frac{1}{2}(\overrightarrow{AP}+\overrightarrow{AD})，又\overrightarrow{DP}=\overrightarrow{AP}-\overrightarrow{AD},\therefore \overrightarrow{BE}\cdot \overrightarrow{DP}=\frac{1}{2}(\overrightarrow{AP}^{2}-\overrightarrow{AD}^{2})=0,\therefore BE\perp PD.$
Subfield: Solid Geometry


Question: $如图，在四棱锥P-ABCD中，PA\perp 底面ABCD，AD\perp DC，AB\parallel DC，AD=DC=AP=2，AB=1，点E为棱PC的中点.$<img_2376>
$(2)若F为棱PC上一点，满足BF\perp AC，求线段PF的长.$

Solution:
$\because F为PC上一点，\therefore 设\overrightarrow{PF}=\lambda \overrightarrow{PC}(0\leq \lambda \leq 1),\therefore \overrightarrow{BF}=\overrightarrow{BP}+\overrightarrow{PF}=\overrightarrow{AP}-\overrightarrow{AB}+\lambda \overrightarrow{PC}=\overrightarrow{AP}-\overrightarrow{AB}+\lambda (\overrightarrow{AD}+2\overrightarrow{AB}-\overrightarrow{AP})=(1-\lambda )\overrightarrow{AP}-(1-2\lambda )\overrightarrow{AB}+\lambda \overrightarrow{AD}.\because BF\perp AC,\therefore \overrightarrow{BF}\cdot \overrightarrow{AC}=[(1-\lambda )\overrightarrow{AP}-(1-2\lambda )\overrightarrow{AB}+\lambda \overrightarrow{AD}]\cdot (\overrightarrow{AD}+2\overrightarrow{AB})=-2(1-2\lambda )+4\lambda =0,解得\lambda =\frac{1}{4}.易知|\overrightarrow{PC}|=\sqrt{{(\overrightarrow{AD}+2\overrightarrow{AB}-\overrightarrow{AP})}^2}=\sqrt{4+4+4}=2\sqrt{3},\therefore PF=\frac{1}{4}PC=\frac{\sqrt{3}}{2}，即线段PF的长为\frac{\sqrt{3}}{2}.$

Final answer: $\frac{\sqrt{3}}{2}$