Question

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

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=______．

Answer 1

∵角AED，角AFB平分线交于O， ∴∠EOF=∠OAB+ 1 2 ∠AFB+∠OAD+ 1 2 ∠AED， =∠EAF+ 1 2 ∠AFB+ 1 2 ∠AED   ①， 又∠BCD=∠AFB+∠CDF， =∠AFB+∠EAF+∠AED ②， 由①②得∠EOF=∠EAF+ 1 2 ∠AFB+ 1 2 ∠AED = 1 2 （∠EAF+∠EAF+∠AFB+∠AED） = 1 2 ∠EAF+ 1 2 ∠BCD = 1 2 ×60°+ 1 2 ×130° =95°． 故答案为：95°．