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A math problem in the second day of junior high school (equilateral triangle)
Because △ABC and △CDE are equilateral triangles, angle ACB= angle ECD = 60, so ∠ACE=∠BCD, because CD=CE and CB=AC, so △BDC is equal to triangle AEC.

So ∠CAE=∠CBD.

Because BAE+CAE = BAE+CBD = CBE+Abe = 60, AEB = 180-Abe-Abe.

Therefore, ∠ AEB-∠ EBD = 60.

I'm exhausted. Typing is so tired! ! ! Be sure to give points.