帮下忙第三题
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帮下忙第三题
![](http://img.wesiedu.com/upload/d/17/d17bb7aa7c6f38644c80a2f69ad3f216.jpg)
![](http://img.wesiedu.com/upload/d/17/d17bb7aa7c6f38644c80a2f69ad3f216.jpg)
![帮下忙第三题](/uploads/image/z/18243215-71-5.jpg?t=%E5%B8%AE%E4%B8%8B%E5%BF%99%E7%AC%AC%E4%B8%89%E9%A2%98%26nbsp%3B)
成立
作CF平行于AD,交AB于点F,则∠ACF=∠CAD ∠BCF=∠D=90
∵∠B=180-90-∠BAD 则∠B+∠BAD=90 ∵∠BAE+∠EAC+∠CAD=∠BAD
∠EAD=∠EAC+∠CAD
∴∠ACB-∠B=90+∠ACF-90+∠BAD=∠ACF+∠BAD
∵∠BAE+∠EAC+∠CAD=∠BAD
∴∠BAE+∠EAC+∠CAD+∠ACF=∠ACB-∠B
∵平分线∴∠BAE=∠EAC ∴2∠EAD=∠ACB-∠B
作CF平行于AD,交AB于点F,则∠ACF=∠CAD ∠BCF=∠D=90
∵∠B=180-90-∠BAD 则∠B+∠BAD=90 ∵∠BAE+∠EAC+∠CAD=∠BAD
∠EAD=∠EAC+∠CAD
∴∠ACB-∠B=90+∠ACF-90+∠BAD=∠ACF+∠BAD
∵∠BAE+∠EAC+∠CAD=∠BAD
∴∠BAE+∠EAC+∠CAD+∠ACF=∠ACB-∠B
∵平分线∴∠BAE=∠EAC ∴2∠EAD=∠ACB-∠B