Question

如图AB=AE,AC=AD,∠BAD=∠CAE=90°.AH⊥BC于点H,HA的延长线交DE于G.求证:GD=GE.

Answer 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