如图所示,求角A+角B+角C+角D+角E+角F的度数

来源：学生作业帮编辑：灵鹊做题网作业帮分类：数学作业时间：2024/05/02 15:37:15

连接BE,令DE,BC交于G
∵∠DGC=∠BGE
∴180°-∠DGC=180°-∠BGE
即∠D+∠C=∠CBE+∠DEB
∠A+∠B+∠C+∠D+∠E+∠F=∠A+(∠ABC+∠CBE)+(∠DEB+∠FED)+∠F
=∠A+∠ABE+∠FEB+∠F
=180°*(4-2)
=360°
再问： o(≧v≦)o
再问：（￣▽￣）b
再答：满意请采纳

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