已知：如图，△ABC等全于△ADE,BC的延长线交DA于F,交DE于G,∠ACB=105°,∠CAD=10°,∠B=∠D=25°

解：
∠ACB=105°
∠FCA=180°-∠ACB=75°
∠CAD=10°
∠AFC=180°-∠FCA-∠CAD=95°
∠DFB=180°-∠AFC=85°
∠DGB=∠DFB-∠D=60°

∠GCA=180°-∠ACB=180°-105°=75° 则
∠AFB=180° -∠CAD-75°=180° -10°-75°=95°
故∠DFB=180°-95°=85°

因为∠B=∠D=25°；∠DFG=∠AFB=95°
故∠DGB=180°-95°-25°=60°

∠DFB=180-[180-∠CAD-(180-∠ACB)]=85
∠DGB=180-(180-∠DFB)-∠D=60

∠DFB=∠CAD+∠ACF=10+180-105=85
∠DGB=∠DFB-∠D=85-25=60