∴CM=BM=MA=MD，

And ∵CD⊥AB,

∴EM is the center line on the bottom of isosceles △CMD, that is, CE=ED,

And EM bisects ∠CMD, that is ∠CMA=∠CMD=2∠CME,

And ∠ CMA+∠ CME = 180, that is, 2 ∠ CME+∠ CME = 180, the solution is ∠ CME = 60.

∵CM=BM, △BCM is an equilateral triangle.

In Rt△BCE, CE=DE=BC? sin60 =3