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对任一四面体都可以把它接补成一个平行六面体,据此,可解一类立几问题,兹举二例。例1 一元选择题:空间四点A、B、C、D,如果有AB=CD=8,AC=BD=10,AD=BC=13,则( ) (A)A、B、C、D为平行四边形的四顶点。 (B)A、B、C、D中有三点共线。 (C)A、B、C、D为不共面四点。 (D)不存在满足题设条件的四点。分析:根据题中的条件,很容易判定(A)、(B)不可能成为选择的答案,那么问题只在于A、B、C、D四点是否为不共面的四点。如果这四点不共面,我们考察以这四点为顶点的四面体,并把它补成长方体BECF-MAND(如图1)
For any tetrahedron, it can be made into a parallelepiped. According to this, one can solve a class of problems. Here are two examples. Example 1 One-variable multiple-choice questions: Space four points A, B, C, D, if there is AB=CD=8, AC=BD=10, AD=BC=13, then () (A) A, B, C, D Four vertices of a parallelogram. (B) There are three collinear points in A, B, C, and D. (C) A, B, C, and D are not coplanar. (D) There are no four points that satisfy the problem-setting conditions. Analysis: According to the conditions in the question, it is easy to determine (A), (B) cannot be the answer to the choice, then the question is whether the four points A, B, C, D are not co-planar. If these four points are not coplanar, we examine the tetrahedron with these four points as the vertices and complement it with the cuboid BECF-MAND (Figure 1).