已知在△ABC中,∠BAC＝90°,AB=AC,D为AC的中点,AE⊥BD于E,延长AE交BC于F,求证：∠ADB=∠C

作AG平分∠BAC交BD于G
∵∠BAC=90°
∴∠CAG= ∠BAG=45°
∵∠BAC=90° AC=AB
∴∠C=∠ABC=45°
∴∠C=∠BAG
∵AE⊥BD
∴∠ABE+∠BAE=90°
∵∠CAF+∠BAE=90°
∴∠CAF=∠ABE
∵ AC=AB
∴△ACF ≌△BAG
∴CF=AG
∵∠C=∠DAG =45° CD=AD
∴△CDF ≌△ADG
∴∠CDF=∠ADB