Question

∠BAC=90°，AD⊥BC、DE⊥AB、DF⊥AC，垂足分别为D、E、F。求证AB³/AC³=BE/CF

Answer 1

证明:∵∠BED=∠BAC=90度;∠B=∠B.
∴⊿BED∽⊿BAC,AB/AC=BE/DE; -------------------------------(1)

同理:⊿BAC∽⊿DEA,得AB/AC=DE/AE=DE/DF;----------------(2)
⊿BAC∽⊿DFC,得AB/AC=DF/FC.--------------------------(3)
∴(AB/AC)*(AB/AC)*(AB/AC)=(BE/DE)*(DE/DF)*(DF/CF)=BE/CF.

即AB³/AC³=BE/CF.

Answer 2

由△ABD∽△ABC易得：AB^2＝BD·BC
由△ACD∽△ABC易得：AC^2＝CD·BC
∴AB^2/AC^2＝BD/CD
∴AB^4/AC^4＝BD^2/CD^2
类似地，由△BDE∽△ABD可得：BD^2＝BE·AB
由△CDF∽△ACD可得：CD^2＝CF·AC
∴AB^4/AC^4＝(BE·AB)/(CF·AC)＝BE/CF·AB/AC
∴AB^3/AC^3＝BE/CF