解:(1)∵△ABC中,∠ABC的平分线与∠ACB的外角∠ACM的平分线交于点E,
∴∠ECD=
![](http://thumb.1010pic.com/pic5/latex/13.png)
∠ACD,∠EBC=
![](http://thumb.1010pic.com/pic5/latex/13.png)
∠ABC,
∵∠ACD=∠A+∠ABC,
∴∠ECD=
![](http://thumb.1010pic.com/pic5/latex/13.png)
∠ACD=
![](http://thumb.1010pic.com/pic5/latex/13.png)
(∠A+∠ABC)=
![](http://thumb.1010pic.com/pic5/latex/13.png)
∠A+∠EBC,
∴∠E=∠ECD-∠EBC=
![](http://thumb.1010pic.com/pic5/latex/13.png)
∠A+∠EBC-∠EBC=
![](http://thumb.1010pic.com/pic5/latex/13.png)
∠A,
∵∠A=70°,
∴∠E=35°;
(2)∵△ABC中,∠ABC的平分线与∠ACB的外角∠ACM的平分线交于点E,
∴∠ECD=
![](http://thumb.1010pic.com/pic5/latex/13.png)
∠ACD,∠EBC=
![](http://thumb.1010pic.com/pic5/latex/13.png)
∠ABC,
∵∠ACD=∠A+∠ABC,
∴∠ECD=
![](http://thumb.1010pic.com/pic5/latex/13.png)
∠ACD=
![](http://thumb.1010pic.com/pic5/latex/13.png)
(∠A+∠ABC)=
![](http://thumb.1010pic.com/pic5/latex/13.png)
∠A+∠EBC,
∴∠E=∠ECD-∠EBC=
![](http://thumb.1010pic.com/pic5/latex/13.png)
∠A+∠EBC-∠EBC=
![](http://thumb.1010pic.com/pic5/latex/13.png)
∠A,
∵∠A=90°,
∴∠E=45°;
(3)∵△ABC中,∠ABC的平分线与∠ACB的外角∠ACM的平分线交于点E,
∴∠ECD=
![](http://thumb.1010pic.com/pic5/latex/13.png)
∠ACD,∠EBC=
![](http://thumb.1010pic.com/pic5/latex/13.png)
∠ABC,
∵∠ACD=∠A+∠ABC,
∴∠ECD=
![](http://thumb.1010pic.com/pic5/latex/13.png)
∠ACD=
![](http://thumb.1010pic.com/pic5/latex/13.png)
(∠A+∠ABC)=
![](http://thumb.1010pic.com/pic5/latex/13.png)
∠A+∠EBC,
∴∠E=∠ECD-∠EBC=
![](http://thumb.1010pic.com/pic5/latex/13.png)
∠A+∠EBC-∠EBC=
![](http://thumb.1010pic.com/pic5/latex/13.png)
∠A,
∵∠A=130°,
∴∠E=65°.
结论:∠E=
![](http://thumb.1010pic.com/pic5/latex/13.png)
∠A.
理由:∵△ABC中,∠ABC的平分线与∠ACB的外角∠ACM的平分线交于点E,
∴∠ECD=
![](http://thumb.1010pic.com/pic5/latex/13.png)
∠ACD,∠EBC=
![](http://thumb.1010pic.com/pic5/latex/13.png)
∠ABC,
∵∠ACD=∠A+∠ABC,
∴∠ECD=
![](http://thumb.1010pic.com/pic5/latex/13.png)
∠ACD=
![](http://thumb.1010pic.com/pic5/latex/13.png)
(∠A+∠ABC)=
![](http://thumb.1010pic.com/pic5/latex/13.png)
∠A+∠EBC,
∴∠E=∠ECD-∠EBC=
![](http://thumb.1010pic.com/pic5/latex/13.png)
∠A+∠EBC-∠EBC=
![](http://thumb.1010pic.com/pic5/latex/13.png)
∠A.
分析:由△ABC中,∠ABC的平分线与∠ACB的外角∠ACM的平分线交于点E,根据角平分线的性质,可得∠ECD=
![](http://thumb.1010pic.com/pic5/latex/13.png)
∠ACD,∠EBC=
![](http://thumb.1010pic.com/pic5/latex/13.png)
∠ABC,然后利用三角形外角的性质,即可求得:∠ECD=
![](http://thumb.1010pic.com/pic5/latex/13.png)
∠ACD=
![](http://thumb.1010pic.com/pic5/latex/13.png)
∠A+∠EBC,∠E=∠ECD-∠EBC,则可求得∠E=
![](http://thumb.1010pic.com/pic5/latex/13.png)
∠A;则可将(1)∠A=70°,(2)∠A=90°,(3)∠A=130°分别代入求解即可求得答案.
点评:此题考查了三角形的外角的性质与角平分线的定义.此题难度适中,解此题的关键是注意数形结合思想与整体思想的应用.