定理[1]在四面体A－BCD中，对棱AB和CD所成的角为θ，则如图1，E、M、N分队为BC、CA、BD的中点，则MEN为导面直线AB和CD所成的角，推论在四面体A—BCD中，对核AB与CD垂直的充要条件是AC‘＋BD’一BCZ＋ADZ（刀）应用定理的结论，我们可报方便地求出两条异面直线所成的角．因为 Theorem [1] In the tetrahedron A-BCD, the angle formed by the edges AB and CD is θ. As shown in Figure 1, the E, M, and N teams are the midpoints of BC, CA, and BD, and the MEN is the guide surface. The angle formed by the straight line AB and CD is deduced that in the tetrahedron A-BCD, the necessary and sufficient conditions for the perpendicularity of the core AB and the CD are the conclusions of the AC’+BD’-BCZ+ADZ (knife) application theorem, which we can conveniently calculate The angle formed by two different straight lines. because