Question

已知在△ABC中，∠BAC=90°，AD⊥BC，E是AC的中点，ED交AB的延长线于F，求证：AB/AC=DF/AF

Answer 1

∠BAC=90°，AD⊥BC，∠ABC+∠C=90°=∠ABC+∠BAD,∠C=∠BAD,∠DAC+∠C=90°=∠ABC+∠C,∠DAC=∠ABC,∠ADC=∠BDA=90°,RT⊿ADC∽RT⊿BDA,[AAA]AD/BD=AC/AB,AB/AC=BD/AD;∠BAC=90°，E为AC中点,所以AE=EC=DE,∠C=∠EDC=∠FDB,∠FBD=∠BAC+∠C=90°+∠C;∠FDA=∠ADB+∠FDB=90°+∠C;∠FBD=∠FDA,∠F=∠F,故∠FDB=∠FAD,⊿DBF∽⊿ADF,[AAA]DF/AF=DB/AD=AB/AC,即AB/AC=DF/AF。