如图,点B,F,C,E在一条直线上,FB=CE,AB//ED,AC//FD.求证AB=DE,AC=DF图片如下:11题啊file://C:\Documents and Settings\Administrator\Local Settings\Temporary Internet Files\Content.IE5\IFGXM18V\W020080505326796997009[1].jpg
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![如图,点B,F,C,E在一条直线上,FB=CE,AB//ED,AC//FD.求证AB=DE,AC=DF图片如下:11题啊file://C:\Documents and Settings\Administrator\Local Settings\Temporary Internet Files\Content.IE5\IFGXM18V\W020080505326796997009[1].jpg](/uploads/image/z/11551255-7-5.jpg?t=%E5%A6%82%E5%9B%BE%2C%E7%82%B9B%2CF%2CC%2CE%E5%9C%A8%E4%B8%80%E6%9D%A1%E7%9B%B4%E7%BA%BF%E4%B8%8A%2CFB%3DCE%2CAB%2F%2FED%2CAC%2F%2FFD.%E6%B1%82%E8%AF%81AB%3DDE%2CAC%3DDF%E5%9B%BE%E7%89%87%E5%A6%82%E4%B8%8B%EF%BC%9A11%E9%A2%98%E5%95%8Afile%3A%2F%2FC%3A%5CDocuments+and+Settings%5CAdministrator%5CLocal+Settings%5CTemporary+Internet+Files%5CContent.IE5%5CIFGXM18V%5CW020080505326796997009%5B1%5D.jpg)
如图,点B,F,C,E在一条直线上,FB=CE,AB//ED,AC//FD.求证AB=DE,AC=DF图片如下:11题啊file://C:\Documents and Settings\Administrator\Local Settings\Temporary Internet Files\Content.IE5\IFGXM18V\W020080505326796997009[1].jpg
如图,点B,F,C,E在一条直线上,FB=CE,AB//ED,AC//FD.求证AB=DE,AC=DF
图片如下:11题啊
file://C:\Documents and Settings\Administrator\Local Settings\Temporary Internet Files\Content.IE5\IFGXM18V\W020080505326796997009[1].jpg
如图,点B,F,C,E在一条直线上,FB=CE,AB//ED,AC//FD.求证AB=DE,AC=DF图片如下:11题啊file://C:\Documents and Settings\Administrator\Local Settings\Temporary Internet Files\Content.IE5\IFGXM18V\W020080505326796997009[1].jpg
1.先说题外话:你给的图片地址在你自己C盘上,你又没有上传图片,所以别人是看不到的-_-!
2.证明如下:
因为:FB=CE,
所以:BC=EF;
因为:AB//ED,
所以:角ABC=角DEF;
又因为:AC//FD,
所以:角ACB=角DFE;
根据角边角对应相等的三角形全等,
所以AB=DE,AC=DF