First, it is proved that △ BCF △ BAE is known that BF=BE, and within △BEF, BE=BF, ∠ EBF = 60, so △BEF is an equilateral triangle, that is, EF=BE=BF, ∠ CBF+∠ Abe =.

After thinking about it, I found that it seems to involve the simplification of trigonometric functions. Now that I am a junior, those complicated formulas have long been forgotten, and I can't help, I can only say something.

If b is the perpendicular of EF, it intersects with point W, and then it is decomposed into two right-angled triangles. The sum of the angles corresponding to these two sides (EW and FW) is always 60, and AE and CF sides are actually the same. In the two right-angled triangles, the sum of the diagonal angles is also 60!

Alas, I am old with a sigh!