一九八五年全国中学生数学竞赛有一道立体几何题: 如图1,在正方体ABCD-A_1B_1C_1D_1中,E是BC的中点,F在AA_1上,且A_1F:FA=1:2,求平面B_1EF与底面A_1B_1C_1D_1所成的二面角。该题标准答案提供的解法不仅难,而且还很繁。本文用另法来解它。我们先引入如下结论:如图2,若正方体内一空间三角形xyz,在三个面上的投影面积分别为S_1,S_2,S_3,则这个空间三角形 In the 1985 National Mathematics Competition for Middle School Students, there was a three-dimensional geometric problem: As shown in Figure 1, in the box ABCD-A_1B_1C_1D_1, E is the midpoint of BC, F is on AA_1, and A_1F:FA=1:2, and the plane B_1EF The dihedral angle to the bottom surface A_1B_1C_1D_1. The solution provided by the standard answer to the question is not only difficult but also very complicated. This article uses another method to solve it. We first introduce the following conclusion: As shown in Fig. 2, if the projective area of ​​a spatial triangle xyz in a square body on the three surfaces is S_1, S_2, S_3, then this spatial triangle