Question

在△ABC中,∠BAC=90°,AD⊥BC,E为AC中点,ED延长线交AB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.