13:40提问如图.CE是△ABC的角平分线,过点E作BC的平行线,交AC于点D,交外角∠ACG的平分线与点F,求证DE=DF.http://hi.baidu.com/%CE%D2%B0%AE%CA%FD%D1%A7oox/album/item/b96b19f3d6506a827931aab4.html
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/03 02:35:04
![13:40提问如图.CE是△ABC的角平分线,过点E作BC的平行线,交AC于点D,交外角∠ACG的平分线与点F,求证DE=DF.http://hi.baidu.com/%CE%D2%B0%AE%CA%FD%D1%A7oox/album/item/b96b19f3d6506a827931aab4.html](/uploads/image/z/11448670-22-0.jpg?t=13%3A40%E6%8F%90%E9%97%AE%E5%A6%82%E5%9B%BE.CE%E6%98%AF%E2%96%B3ABC%E7%9A%84%E8%A7%92%E5%B9%B3%E5%88%86%E7%BA%BF%2C%E8%BF%87%E7%82%B9E%E4%BD%9CBC%E7%9A%84%E5%B9%B3%E8%A1%8C%E7%BA%BF%2C%E4%BA%A4AC%E4%BA%8E%E7%82%B9D%2C%E4%BA%A4%E5%A4%96%E8%A7%92%E2%88%A0ACG%E7%9A%84%E5%B9%B3%E5%88%86%E7%BA%BF%E4%B8%8E%E7%82%B9F%2C%E6%B1%82%E8%AF%81DE%3DDF.http%3A%2F%2Fhi.baidu.com%2F%25CE%25D2%25B0%25AE%25CA%25FD%25D1%25A7oox%2Falbum%2Fitem%2Fb96b19f3d6506a827931aab4.html)
13:40提问如图.CE是△ABC的角平分线,过点E作BC的平行线,交AC于点D,交外角∠ACG的平分线与点F,求证DE=DF.http://hi.baidu.com/%CE%D2%B0%AE%CA%FD%D1%A7oox/album/item/b96b19f3d6506a827931aab4.html
13:40提问
如图.CE是△ABC的角平分线,过点E作BC的平行线,交AC于点D,交外角∠ACG的平分线与点F,求证DE=DF.
http://hi.baidu.com/%CE%D2%B0%AE%CA%FD%D1%A7oox/album/item/b96b19f3d6506a827931aab4.html
13:40提问如图.CE是△ABC的角平分线,过点E作BC的平行线,交AC于点D,交外角∠ACG的平分线与点F,求证DE=DF.http://hi.baidu.com/%CE%D2%B0%AE%CA%FD%D1%A7oox/album/item/b96b19f3d6506a827931aab4.html
∵DE‖BC
∴∠DEC=∠ECB ∠F=∠FCG
又∵CE平分∠ACB ∴∠ACE=∠ECB
∴∠ECB=∠DEC ∴DE=DC
同理得DF=DC
∴DE=DF
EF ‖BC
∠FEC=∠BCE=∠ECA
DE=CD
∠DFC=∠FCG=∠ACF
DF=CD
DE=DF
做辅助线。连结AF .证明出是个矩形,就得出 DE=DF
∠DCE=∠ECB=∠CED
∴DC=DE
∠DCF=∠F=∠FCG
∴DC=DF
∴DF=DE
EF//BC
角DEC=角ECB
CE是角ACB的角平分线
角ECB=角DCE
角DEC=角DCE
DE=DC
同理可证DC=DF
所以DE=DF