Question

如图,已知三角形ABC中,角B=2角C.求证:AB^2-AC^2=AC·BC

Answer 1

你的题目有问题,B=2C,B>C,AC>AB,AB^2-AC^2<0?

我把它改为：AC^2-BC^2=AB*BC,

延长CB至D,使BD=AB,连结AD,

△ABD是等腰△,〈D=〈DAB,〈ABC=〈D+〈DAB=2〈D,

而已知,〈ABC=2〈C,

故〈C=〈D,

△ADC是等腰△,AD=AC,

〈D=〈D,〈DAB=〈C,

△ADB∽△CDA,

AD/CD=BD/AD,

AD^2=CD*BD,

AC^2=AB*(BC+BD)=AB*(BC+AB)=AB^2+AB*BC,

∴AC^2-AB^2=AB*BC.

你可把