问题
证明题
如图所示,AD是△ABC的角平分线,EF是AD的垂直平分线,交BC的延长线于点F,连接AF.求证:∠BAF=∠ACF.
![](https://img.ixiawen.com/uploadfile/2017/0422/20170422082513108.png)
答案
证明:∵EF是AD的垂直平分线,
∴AF=DF,
∴∠FAD=∠ADF,
∵∠FAD=∠FAC+∠CAD,∠ADF=∠B+∠DAB,
∴AD是∠BAC的平分线,
∴∠DAB=∠CAD,
∴∠CAF=∠B,
∴∠BAC+∠FAC=∠B+∠BAC,
即∠BAF=∠ACF.
如图所示,AD是△ABC的角平分线,EF是AD的垂直平分线,交BC的延长线于点F,连接AF.求证:∠BAF=∠ACF.
证明:∵EF是AD的垂直平分线,
∴AF=DF,
∴∠FAD=∠ADF,
∵∠FAD=∠FAC+∠CAD,∠ADF=∠B+∠DAB,
∴AD是∠BAC的平分线,
∴∠DAB=∠CAD,
∴∠CAF=∠B,
∴∠BAC+∠FAC=∠B+∠BAC,
即∠BAF=∠ACF.