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=________．
网友回答
95°
解析分析：根据三角形的外角性质求出∠EOF=∠EAF+∠AFB+∠AED，∠BCD=∠AFB+∠CDF，代入求出∠EOF=（∠EAF+∠BCD），代入求出即可．
解答：∵角AED，角AFB平分线交于O，∴∠EOF=∠OAB+∠AFB+∠OAD+∠AED，=∠EAF+∠AFB+∠AED?? ①，又∠BCD=∠AFB+∠CDF，=∠AFB+∠EAF+∠AED ②，由①②得∠EOF=∠EAF+∠AFB+∠AED=（∠EAF+∠EAF+∠AFB+∠AED）=∠EAF+∠BCD=×60°+×130°=95°．故