《几何》第二册中有一个题目:已知:如图1,点C为线段AB上一点,△ACM、△CBN是等边三角形.求证:AN=BM.此题内涵丰富,可以开发出很多“产品”:设AN与CM相交于点D,BM与CN相交丁点E,则有1.∠ANC=∠MBC=∠ABM; There is a title in the second volume of Geometry: known: as shown in Figure 1, point C is a point on the line AB, △ACM, △CBN is an equilateral triangle. Proof: AN=BM. This question has rich connotation and can be developed Many “products”: Let AN and CM intersect at point D, and BM and CN meet at point E, then 1. ANC=∠MBC=∠ABM;