过C作CG⊥AC交AE延长线于G∵AE⊥BD于F，所以∠DBA=∠GAC(都与∠EAB互余）∵∠ABC=∠ACB=45∴AB=CA 又∵∠DAB=∠GCA=90°∴△DAB≌△GCA(角边角)∴∠ADB=∠CGA,AD=CG又∵AD=DC,所以CD=CG又∵∠GCE=∠DCE=45°，CE=CE∴△GCE≌△DCE（边角边）∴∠CGA=∠CDE∴∠ADB=∠CDE
