Question

如图,已知△ABC全等于△ADE,BC延长线交AD于点F,交ED于点G,且∠CAD=25°,∠B=∠D=30°,∠EAB

Answer 1

已知△ABC全等于△ADE ==> ∠CAB=∠DAE

∠EAB=125°,∠CAD=25°==> ∠CAB=∠DAE=(∠EAB-∠CAD)/2=50°

∠FCA=∠B+∠CAB=50°+30°=80°

∠DFB=∠CAD+∠FCA=25°+80°=105°

∠DGB=∠DFB-∠D=105°-30°=75°