如图所示,已知三角形ABC全等于三角形ADE,BC的延长线交DA于点F,交DE于点G,角ACB=角AED=105°,角C

解∠ACB=∠AED=105°
又∠B=25°
即∠CAB=180°-∠ACB-∠B=50°
即∠FAB=∠CAB+∠CAD=75°
即∠DFB=∠FAB+∠B=100°
又∠D=∠B=25°
即∠DGB=180°-∠DFG-∠D=75°