Question

如图,已知等腰Rt∠AOB=90 AE=BF AE

Answer 1

(1) 等腰Rt△AOB -> AO=BO

等腰Rt△EOF -> EO=FO

∠AOB=∠EOF=90°-> ∠AOE=∠BOF

-> △AOE≌△BOF

-> AE=BF

(2)延长AE与BF交与点G,与BO交与点H

△AOE≌△BOF -> ∠OAE=∠OBF

∠AHO=∠BHG

-> ∠AOB=∠HGB=90°

-> AE⊥BF