已知BD是等腰直角三角形的腰AC上的中线,AE垂直BD于E,AE的延长线交BC于F,连结DF. 求证：∠ADB=∠CDF

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