三角形好
来源:学生作业帮 编辑:搜狗做题网作业帮 分类:数学作业 时间:2024/07/02 18:38:58
![](http://img.wesiedu.com/upload/6/2c/62c77e6d29ba0c2049099f462d123f24.jpg)
![三角形好](/uploads/image/z/20313627-51-7.jpg?t=%E4%B8%89%E8%A7%92%E5%BD%A2%E5%A5%BD)
解题思路: 利用三角形的外角性质来解。
解题过程:
证明:连接AO并延长到P。 ∵∠AED。∠AFB的角平分线交与O ∴∠POE=∠OAE+∠AEO=∠OAE+1/2∠AED。同理∠POF=∠OAF+1/2∠AFB
∴ ∠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+1/2∠EAF+1/2∠AFB+1/2∠AE D
=1/2∠EAF+1/2∠BCD
=1/2(∠EAF+∠BCD)
∴角EOF=1/2(角EAF+角BCD)
最终答案:略
解题过程:
证明:连接AO并延长到P。 ∵∠AED。∠AFB的角平分线交与O ∴∠POE=∠OAE+∠AEO=∠OAE+1/2∠AED。同理∠POF=∠OAF+1/2∠AFB
∴ ∠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+1/2∠EAF+1/2∠AFB+1/2∠AE D
=1/2∠EAF+1/2∠BCD
=1/2(∠EAF+∠BCD)
∴角EOF=1/2(角EAF+角BCD)
最终答案:略