As shown in the figure, ABC is an equilateral triangle, BD is the middle line, extend BC to e, so that CE = CD

As shown in the figure, ABC is an equilateral triangle, BD is the middle line, extend BC to e, so that CE = CD

It is proved that: ∵ ABC is an equilateral triangle, BD is the middle line, ∵ ABC = ∠ ACB = 60 °, DBC = 30 ° (isosceles triangle three lines in one), and ∵ CE = CD, ∵ CDE = ∠ CED. And ∵ BCD = ∠ CDE + ∠ CED, ∵ CDE = ∠ CED = 12 ∠ BCD = 30 °. ∵ DBC = ∠ Dec. ∵ DB = de (equiangular to equilateral)