等边三角形的许多性质在解题中有广泛的应用,在竞赛中深受命题者的青睐.现举三例,供读者参考.例1如图1,在△ABC中,∠ACB=90°,∠CAD=30°,AC=BC=AD.求证:CD=BD.分析要证CD=BD,应有∠DCB=∠DBC,但是很难证得.根据∠ACD=75°,∠DCB=15°,它们的差是60°.由此可以联想构造等边三角形.证明如图1,作等边△CDE,连接AE, Many properties of equilateral triangles in the problem has a wide range of applications in the competition by the proponents of the favor. Now give three cases for readers reference. Example 1 Figure 1, △ ABC, ∠ ACB = 90 ° , ∠CAD = 30 °, AC = BC = AD. Proof: CD = BD. Analysis to prove that CD = BD, ∠DCB = ∠DBC, but it is difficult to prove.According to ∠ACD = 75 °, ∠DCB = 15 °, their difference is 60 ° .This can be associated with the construction of equilateral triangles.Proved as shown in Figure 1, for equilateral △ CDE, connecting AE,