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The first chapter of mathematics in grade three will go to school tomorrow.
Solution: CM is the center line of hypotenuse AB of Rt△ABC,

∴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