连接BE,DE与BC交于O∵∠C+∠D+∠COD=180∠CBE+∠DEB+∠BOE=180∠BOE=∠COD∴∠C+∠D=∠CBE+∠BDE∴∠A+∠ABC+C+∠D+∠E+∠F=∠A+∠ABE+∠BEF+∠F=360°
连接B E,BC、DE交点O,三角形内角和180,DOC=BOE,C+D=CBE+DEB,四边形ABEF内角和360