Question

如图1所示,求角A+角B+角c+角D+角E+角F

Answer 1

延长DE,AF交BC于M,N

则∠A+∠B=180°-∠ANB

∠C+∠D=180°-∠CMD

∠E=180°-∠MEF

∠F=180°-∠NFE

∴∠A+∠B+∠C+∠D+∠E+∠F=720°-(∠ANB+∠CMD+∠MEF+∠NFE)=720°-360°=360°