摘要:在△ABC的三边上分别取点P1.P2.P3.P4.P5.--.使得P1.P4.P7.--在AC上.P2.P5.P8.--在AB上.P3.P6.P9.--在BC上.且AP1=AP2.BP2=BP3.CP3=CP4.AP4=AP5.BP5=BP6.--.则P2与P2000的距离为 .
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(Ⅰ)求证:C1F∥平面B1GE;
(Ⅱ)求证:PF⊥平面B1EF.
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(1)证明:DE∥平面BCF;
(2)证明:CF⊥平面ABF;
(3)当AD=
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