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°．