如图,△ABC中,∠BAC=Rt∠,AD⊥BC于D,BF平分∠ABC,交AD于E,求证DE:AE=AF:CF

问题描述:

如图,△ABC中,∠BAC=Rt∠,AD⊥BC于D,BF平分∠ABC,交AD于E,求证DE:AE=AF:CF

已知,BF平分∠ABC,可得:DE∶AE = BD∶AB ,AF∶CF = AB∶BC ;
而且,由射影定理可得:AB^2 = BD·BC ,即有:BD∶AB = AB∶BC ;
所以,DE∶AE = AF∶CF .