∴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