AD是三角形ABC的中线,E是AD上一点,且AE:ED=1:3,BE的延长线交AC于F,求AF/FC的值
人气:311 ℃ 时间:2020-02-06 07:17:24
解答
作DG∥BF,交AC于点G
∵D是BC的中点
∴G是FC的中点
即GF=GC
∵AE/ED=1/3
∴AF/FG=1/3
∴AF/FC=1/6
推荐
- AD是三角形ABC的中线,E是AD上一点,且AE:ED=1:3,BE的延长线交AC于F,求AF/FC的值
- 如图,已知AD是△ABC的中线,E是AD上的点,且AE=2DE,连接BE并延长交AC于F. (1)求证:AF=FC; (2)求BF/EF的值.
- ·在三角形ABC中AD为中线,BE交AD于F,交AC于E且AF=FD求证AE=三分之一AC
- AD是三角形ABC的中线,点E在线段AD上,延长BE交AC于F,若AE=2ED,求AF:FC
- 在三角形ABC中,AD是中线,BE交AD于点F,且AE=AF,求证AC=BF
- I did what everyone does who has no idea what to do with themselves when they got out of college and went on to graduate
- Tom has over ten books.(改同义句) Tom has ____ ____ ten books.
- 成语‘狡兔三窟’的来历,请告诉.
猜你喜欢