1.如图②所示,求∠A+∠B+∠C+∠D+∠E的度数.2.如图③所示,求∠A+∠B+∠C+∠D+∠E+∠F的度数.

1、连接BC
∴∠BOC=∠DOE=180°-∠D-∠E
∵∠BOC=180°-(∠OBC+∠OCB)
∴∠D+∠E=∠OBC+∠OCB
∵∠ABC+∠ACB=180°-∠A
∴∠ABO+∠ACO=∠ABC+∠ACB-(∠OBC+∠OCB)=180°-∠A-∠D-∠E
∴∠ABO+∠ACO+∠A+∠D+∠E=180°
即∠B+∠C+∠A+∠D+∠E=180°
2、∠GHI=∠CHD=180°-(∠C+∠D)
∠GIH=∠EIF=180°-(∠E+∠F)
∠IGH=∠AGB=180°-(∠A+∠B)
∵∠CHD+∠GIH+∠IGH=180°
∴180°-(∠C+∠D)+180°-(∠E+∠F)+180°-(∠A+∠B)=180°
∴∠A+∠B+∠C+∠D+∠E+∠F=360°