As shown in the figure, C is a point on the line AB, △ ACD and △ BCE are equilateral triangles. AE intersects CD at point m, BD intersects CE at point n, AE intersects AE at point O. The results are as follows: (1) AOB = 120 °; (2) cm = cn; (3) Mn ‖ ab

As shown in the figure, C is a point on the line AB, △ ACD and △ BCE are equilateral triangles. AE intersects CD at point m, BD intersects CE at point n, AE intersects AE at point O. The results are as follows: (1) AOB = 120 °; (2) cm = cn; (3) Mn ‖ ab

It is proved that: (1) both ∵ ACD and ∵ BCE are equilateral triangles, ∵ AC = CD, CE = CB, ∵ ACD = ∵ BCE = 60 °, ∵ ACD + ∵ DCE = ∵ BCE + ∵ DCE, ? ace = ? DCB. In ? ace and ? DCB, AC = CD ? ace = ≌ DCE = CB, ≌ ace ≌ DCB (SAS), ? cam = ≂ CDN