Question

如图角abc+角def=55度,求角A+角C+角D+角F的度数

Answer 1

将EF与AB的交点设为P,EF与BC的交点设为Q

∵∠A+∠F＝∠BPQ,∠CQE＝∠ABC+∠BPQ

∴∠CQE＝∠A+∠B+∠F

∵∠C+∠D+∠DEF+∠CQE＝360

∴∠A+∠ABC+∠C+∠D+∠DEF+∠F＝360

∴∠A+∠C+∠D +∠F＝360-（∠ABC+∠DEF）＝360-55＝305°

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