1古籍轻断处,难度晚尤彰学过初等平面几何的人都熟知外角定理,即三角形的任一外角大于每一个不与之相邻的内角.它的传统证明可以表述为题设点D在△ABC的边BC的延长线上.题断∠ACD>∠CAB,∠ACD>∠ABC.证取边AC的中点E.连结BE并且延长它到F,使EF=BE;作射线CF(图1).因为EC=EA,∠CEF=∠AEB(对顶角相等),EF=EB,所以△CEF≌△AEB(边角边).因此 An ancient book light break at the difficulty of late especially those who have studied elementary plane geometry are familiar with the outer angle theorem, that is, any outer angle of the triangle is greater than each of the inner angles that are not adjacent to it. Its traditional proof can be expressed as a point D ABC on the extension of the edge of BC. 断 ACD> ∠ CAB, ∠ ACD> ∠ ABC. Check the edge of the middle of AC E. Link BE and extend it to F, so that EF = BE; (Figure 1). Because EC = EA, ∠CEF = ∠AEB (for vertex equal), EF = EB, so △ CEF≌ △ AEB