题目内容：

如图,在三角形ABC∠B=25°,∠C=55°,AE平分∠BAC交BC于E,F是AE上的一点,FD垂直于BC于D.求∠DFE.

优质解答

∵三角形内角和180°
∴∠BAC = 180°-∠B- ∠C = 180°-25°-55° = 100°
∵AE是角平分线
∴∠BAE = ∠CAE = ∠BAC/2 = 50°
∵三角形一个外角等于另外两个内角之和
∴∠AEC = ∠B+∠BAE = 25°+50° = 75°
∵△DFE内角和180°
∴∠DFE = 180° -90°-∠AEC = 90°-75° = 15°