如图,AB=AD,AC=AE,

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如图,AB=AD,AC=AE,

①在BC上截取BG'=AG
∵∠BAD=∠CAE=∠AHB=∠AHC=90°
∴∠BAH+∠ABC=∠BAH+∠DAG=∠CAH+∠BCA=∠CAH+∠EAG=90°
∴∠CBA=∠DAG,∠BCA=∠EAG
又∵AB=AD,AG=BG'
∴△ABG'≌△ADG(SAS)
∴DG=AG',∠DGA=∠BG'A
∴∠EGA=∠CG'A
又∵∠BCA=∠EAG,AC=AE
∴△ACG'≌△AEG(AAS)
∴GE=AG'=GD（方法一） ②证明：过D、E作DN⊥AH,EM⊥AH,交HA延长线于N、M∵∠BAD=90°,∴∠DAN+∠BAH=90°又AH⊥BC∴∠ABH+∠BAH=90°∴∠DAN=∠ABH(等角的余角相等）DN⊥AH,AH⊥BC,∴∠DNA=∠AHB=90°AB=AD∴△ABH≌△DAN∴DN=AH同理可证△AHC≌△EMA∴EM=AH∴DN=EM在△DNG和△EMG中DN=EM∠DNG=∠EMG=90°∠DGN=∠EGM∴△DNG≌△EMG∴DG=GE方法二

（方法二图）请采纳,谢谢

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