Solution: Connect CE to BC midpoint F, connect EF, and then EF∨AC.

∴∠A=∠BEF,EF= 1/2AC……①

If CD⊥AB is known from the title, then ∠ ADC = ∠ BDC = 90.

That is, △ACD and △BCD are right triangles.

And \f is the midpoint of the hypotenuse BC of the right angle △BCD.

∴DE= 1/2BC=BF=CF

∴∠ 1=∠2,∠B=∠3……②

∠BEF is the external angle of △DEF, that is, ∠BEF=∠3+∠4.

Namely: ∠A=∠3+∠4=∠B+∠4 (obtained from ① and ②) ...

∫∠A = 2∠B

∴∠B=∠4……④

∴∠3=∠4……⑤

∴DE=EF= 1/2AC

I hope it will be useful to you ~ ~ ~