Question

已知三角形ABC中AD是角BAC的平分线,求证AD的平方=AB乘AC减BC乘CD

Answer 1

证明：∵cos∠BAD=(AB^2+AD^2-BD^2)/(2AB*AD)

cos∠CAD=(AC^2+AD^2-CD^2)/(2AC*AD)

∠BAD=∠CAD

∴AB/BD=AC/DC

∴(AB^2+AD^2-BD^2)/(2AB*AD)=(AC^2+AD^2-CD^2)/(2AC*AD)

AC*AB^2+AD^2*AC-BD^2*AC=AC^2*AB+AD^2*AB-CD^2*AB

AC*AB(AB-AC)+CD^2*AB-BD^2*AC=AD^2(AB-AC)

AC*AB(AB-AC)+CD*AC*BD-BD*AB*DC=AD^2(AB-AC)

AC*AB(AB-AC)+CD*BD(AC-AB)=AD^2(AB-AC)

∴AD^2=AC*AB-CD*BD