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对于某些立几问题,作适当的辅助几何体,把问题归结到一个联系已知量与待求量的几何体中讨论将有助于使问题得到解决。例在长方体ABCD-A＇B＇C＇D＇中,已知B＇C＇=a,BB＇=b,B＇A＇=c,求异面直线BD＇与B＇C所成的角以及它们之问的距离。解如图,作辅助长方体CC_1D_1D-C＇C_1＇D_1＇D＇,使C＇C_1＇=a。连结D_1D＇_9,则D＇D_1(?)B＇C,∠BD＇D_1即为BD＇与B＇C所成的角。C到平面BD_1D＇的距离即为B＇C到平面BD_1D＇的距离即为B＇C与BD＇,之间的距离,设为n 在△BD_1D＇中。BD_1=((2a)~2+c~2)~(1/2)
For certain vertical problems, the appropriate auxiliary geometry is used, and the problem is attributed to a discussion of the geometry that links the known quantity to the required quantity. This will help solve the problem. For example, in the cuboid ABCD-A’B’C’D’, it is known that B’C’=a, BB’=b, and B’A’=c, and the angle formed by the different straight lines BD’ and B’C is obtained. And their distance to ask. As shown in the figure, an auxiliary rectangular body CC_1D_1D-C’C_1’D_1’D’ is used to make C’C_1’=a. Link D_1D’_9, then D’D_1(?)B’C, ∠BD’D_1 is the angle formed by BD’ and B’C. The distance from C to the plane BD_1D′ is the distance between B′C and the plane BD_1D′, that is, the distance between B′C and BD′, and n is in ΔBD_1D′. BD_1=((2a)~2+c~2)~(1/2)