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Mathematics equilateral triangle in the second day of junior high school
Solution: Take a point p on MN and make NP=CN.

∫BM+CN = MN,NP=CN

∴BM=MP

And ∵△ABC is an equilateral triangle, ∴ △ ABC = ∠ ACB = 60.

And BD = DC, ∠ BDC = 120, ∴ DBC = ∠ DCB = 30.

∴∠DBM=∠DCN=90

According to the point on the bisector of the angle, the distance between the feet is equal.

∵NP=CN, ∴DN is the bisector of ∠CDP, ∴ CDN = ∠ PDN.

∵BM=MP, ∴DM is the angular bisector of ∠BDP, ∴ BDM = ∠ PDM.

∠∠MDN =∠PDN+∠PDM =∠BDC/2 = 60

This is the correct answer process, I hope to adopt ~ ~ ~ remember to add points.