过C作CG⊥AC交AE延山并长线于G∵AE⊥BD于F,所以∠DBA=∠GAC(都与∠EAB互余)又∵AB=CA,∠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∴△逗春迹GCE≌△DCE(边角边森中)∴∠CGA=∠CDE∴∠ADB=∠CDE
问题呢?