读了王敬如老师的《一道常见几何题的推广》(《中学生数学》2015,2(下),P14)很受启发.唯感觉推广1和推广2中的证明较为复杂,本文给出较简单的证明思路.原文大意已知:直线AB∥CD被直线EF所截.(1)如图1,∠BEF和∠EFD的角平分线交于点G.则∠G=90°;(2)推广1如图2,将(1)中∠BEF的角 Inspired by Wang Jingru’s “Promotion of a Common Geometry Problem” (Secondary School Maths, 2015, 2 (P), P14), I found that the proofs in Popularization 1 and Popularization 2 are more complex. This article gives a relatively simple (1) As shown in Fig. 1, the angle bisector of ∠BEF and ∠EFD is at point G. Then ∠G = 90 °; (2) The general idea is that the straight line AB∥CD is truncated by the straight line EF 1 As shown in Figure 2, (1) corner of ∠BEF