如图，延长四边形ABCD对边AD，BC交于F；DC，AB交于E．如果∠AED，∠AFB平分线交于O，∠A=60°，∠BCD=130°，则∠EOF=______．

题型：不详难度：来源：

如图，延长四边形ABCD对边AD，BC交于F；DC，AB交于E．如果∠AED，∠AFB平分线交于O，∠A=60°，∠BCD=130°，则∠EOF=______．

答案

∵角AED，角AFB平分线交于O，
∴∠EOF=∠OAB+

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∠AFB+∠OAD+

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∠AED，
=∠EAF+

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∠AFB+

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∠AED   ①，
又∠BCD=∠AFB+∠CDF，
=∠AFB+∠EAF+∠AED ②，
由①②得∠EOF=∠EAF+

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∠AFB+

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∠AED
=

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（∠EAF+∠EAF+∠AFB+∠AED）
=

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∠EAF+

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∠BCD
=

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×60°+

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×130°
=95°．
故答案为：95°．