问题补充:
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:
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
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=________.