BD²＋EC²＝DE²證明：方法1：如圖1，將△ABD繞點A逆時針旋轉90°，得到△ACF，連結EF則△ACF≌△ABD，∴CF＝BD，AF＝AD，∠CAF＝∠BAD，∠ACF＝∠B＝45°∴∠ECF＝90°，∴CF²＋EC²＝EF²，BD²＋EC²＝EF²∵∠BAC＝90°，∠DAE＝45°，∴∠BAD＋∠CAE＝45°又∵∠CAF＝∠BAD，∠EAF＝∠CAF+∠CAE∴∠EAF＝45°，∴∠EAF＝∠DAE又∵AF＝AD，AE＝AE∴△EAF≌△DAE（SAS），∴EF＝DE∴BD²＋EC²＝DE²方法2：如圖2，以A為頂點引射線AF，使∠FAD＝∠BAD，且AF＝AB，連結DF、EF則有△ADF≌△ADB，∴DF＝BD，∠DAF＝∠BAD，∠ADF＝∠ADB∵在△ABC中，∠BAC＝90°，AB＝AC，∴AF＝AC又∵∠DAE＝45°，∴∠BAD＋∠CAE＝45°，∴∠CAE＝45°－∠BAD而∠FAE＝∠DAE－∠DAF＝45°－∠DAF＝45°－∠BAD，∴∠CAE＝∠FAE∴△CAE≌△FAE，∴EC＝EF，∠AEC＝∠AEF又∵∠EDF＋∠DEF＝(∠ADF－∠ADC)＋(∠AEF－∠AEB)＝(∠ADB－∠ADC)＋(∠AEC－∠AEB)∠ADB＝∠DAE＋∠AEB＝45°＋∠AEB，∠AEC＝∠DAE＋∠ADC＝45°＋∠ADC∴∠EDF＋∠DEF＝90°，∴∠DFE＝90°∴DF²＋EF²＝DE²∴BD²＋EC²＝DE²