:已知三角形ABC中,角ABC的角平分线与角ACB的外角平分线交于点D,DE平行BC交AC于点F.求证:BE-CF=EF.
问题描述:
:已知三角形ABC中,角ABC的角平分线与角ACB的外角平分线交于点D,DE平行BC交AC于点F.求证:BE-CF=EF.
答
延长BC到G.
已知,DE‖BC,
可得:∠EDB = ∠DBC = ∠EBD ,∠FDC = ∠DCG = ∠FCD ,
则有:DE = BE ,DF = CF ;
所以,BE-CF = DE-DF = EF .