∵△ABC≌△ADE
∴∠CAB=∠EAD ∠ACB=∠DEA
∵∠EAB=120° ∠CAD=10°
∴∠CAB=∠EAD=55°
∴∠ACB=∠AED=110°
∵四边形ABFE内角和360°
∴∠BFD=75°①
∵∠AGB=∠DGF ∠B=∠D
∴△ACB∽△FCD
∴∠GFD=∠GAB=∠GAC+∠BAC=10°+55°=65°
∴∠DGB=90°②
角BAD=角CAB=(120-10)/2=55度, 角B=角D=25度, 角CAD=10度
角DGB=角GAB+角B=10+55+25=90度,
角DFB=90-25=65度
∠EAD=∠CAB=(120°-10°)÷2=55°,∠E=∠ACB=180°-25°-55°=100°
∠FCA=180°-100°=80°
四边形ACFE的内角和=360°=∠E100°+∠EAC65°+∠ACF80°+180°-∠DFB=360°
∠DFB=115°
∠DGB=∠DFB+∠D=115°+25°=140°
稍等,我在纸上把答案写给你