问题补充:
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.
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.
======以下答案可供参考======
供参考答案1:
F在AD的垂直平分线上。∠FAD=∠FDA
∠FAD=∠FAB+∠BAD. ∠FDA=∠DAC+∠DCA.
AD是∠BAC的平分线.∠BAD=∠DAC
∠FAB=∠FAD-∠BAD=∠FDA-∠DAC=∠DCA=∠ACF. .