(1)解:∵CE⊥AB,∠ECB=30°.∴∠B=60°;BC=2BE=2.又BC=CD.∴CD=BC=2.(2)证明:在AE上取点F,使EF=EB,连接CF.∵CE垂直平分BF.∴CF=BC=CD,则∠2=∠B;∠3=∠1.∵∠B+∠ADC=180°.∴∠2+∠D=180°.又∠2+∠AFC=180°.∴∠ADC=∠AFC.又∠1=∠3.∴∠5=∠4,AF=AD.∵AF=AD,CF=CD,AC=AC.∴⊿AFC≌⊿ADC(SSS),∠DAC=∠FAC.故:AC平分∠BAD.(3)解:∵∠BAD=60°.∴∠BAD+∠ADC=180°,则AB平行于CD; ∠BAD=∠B=60°.则四边形ABCD为等腰梯形.∴AD=BC=2,AB=CD+2BE=4.
故:四边形ABCD周长为AB+BC+CD+AD=10.