Question

如图，△ABC≌△ADE，

如图，△ABC≌△ADE，BC的延长线交DA于点F、DE于点G，∠AED=105°，∠CAD=10°，∠B=25°，求∠DFB和∠DGB的度数

Answer 1

△ABC≌△ADE
所以∠B=∠D=25 ∠AED=∠ACB=105
∠ACF=180-∠ACB=180-105=75
∠AFC=∠ACB-∠CAD=105-10=95
∠DFB=∠AFG=180-95=85
∠CAB=∠DAE=180-105-25=50
∠DGB=360-∠AED-∠DAE-∠AFG=360-105-50-85=120

Answer 2

因为△ABC≌△ADE所以，∠D=∠B=25°，因为∠AED=105°所以，∠EAD=50°所以∠DGF=50°
∠DFB=∠DGF+∠DAC=60°

Answer 3

∵△ABC≌△ADE
∴∠D=∠B=25°
∠ACB=∠AED=105°
∴∠EAD=∠CAB=180°-∠AED-∠D=180°-105°-25°=50°
∵∠ACB=∠CAD+∠AFC
∴∠AFC=∠ACB-∠CAD=105°-10°=95°
∴∠DFB=180°-∠AFC=180°-95°=85°
∵∠DFB=∠D+∠DGF=∠D+∠DGB
∴∠DGB=∠DFB-∠D=85°-25°=60°