解:(1)过C作CF∥AB,∵AB∥ED,∴CF∥ED,∴∠FCD=∠D,∵AB∥ED,∴∠B=∠BCF,∴∠BCF+∠FCD=∠B+∠D,∴∠C=∠B+∠D;(2)过C作CF∥AB,则∠BCF=∠B,∵∠C=∠BCF+∠DCF=∠B+∠DCF,又∵∠C=∠B+∠D,∴∠DCF=∠D,∴CF∥ED,∴AB∥ED.