证明:(1)在△ABC中,AC=BC∴∠CAB=∠CBA.在△ECD中,CE=CD∴∠CED=∠CDE.∵∠CBA=∠CDE,∴∠ACB=∠ECD.∴∠ACB-∠ACD=∠ECD-∠ACD.∴∠ACE=∠BCD.又CE=CD,AC=BC,∴△ACE≌△BCD.∴AE=BD.(2)若AC⊥BC,∵∠ACB=∠ECD,∴∠ECD=90°,∠CED=∠CDE=45°.∴DE= 2 CD.又∵AD+BD=AD+EA=ED,∴AD+BD= 2 CD.