已知AD、AE分别为三角形ABC的内、外角平分线,M为DE中点,求证AB^2/AC^2=BM/CM

Question

已知AD、AE分别为三角形ABC的内、外角平分线,M为DE中点,求证AB^2/AC^2=BM/CM

Answer 1

∵AD、AE平分△ABC的内外角
∴∠DAE＝90°
∴AF＝DF＝EF
∴∠DAF＝∠ADF
而∠DAF＝∠DAC＋∠CAF,∠ADF＝∠B＋∠BAD
∴∠DAC＋∠CAF＝∠B＋∠BAD
∴∠CAF＝∠B
∴△CAF∽△ABF
∴BF/AF＝AB/AC,AF/CF＝AB/AC
两式相乘即得：AB^2/AC^2＝BF/CF