Question

如图,在△ABC中,已知∠A=90°时,AD⊥BC于D,E为直角边AC的中点,过D,E作直线交AB的延长线于F

Answer 1

证明:∠ADC=∠BAC=90°;∠C=∠C.则⊿ACD∽⊿BCA,AB/AC=AD/CD;

作AM∥BC,交DF于M.

则AM/CD=AE/CE=1,AM=CD,故AB/AC=AD/AM;(1)

DE=AC/2=AE,则:∠ACE=∠DAE=∠B=∠FAM;

又∠F=∠F,则⊿AFM∽⊿DFA,DF/AF=AD/AM;(2)

∴AB/AC=DF/AF,AB*AF=AC*DF.