Question

已知三角形ABC,延长CA,AB,BC到D,E,F,连接DE,EF,FD,使得角AED=角BFE=角CDF,若角ABC=

Answer 1

∠DFE=∠DFC+∠CFE=∠DFC+∠CDF=50°=∠BCD

∠BAC=180°-∠ABC-∠ACB=70°

∠EDF=∠EDC+∠CDF=∠EDC+∠DEA=70°