1如图,过平行四边形ABCD的顶点A作一条直线,分别交BD,CD及BC的延长线于GFE,若CE=1/2 BC,则DF/FC = ,AG/GE=?2 如图点D E 分别在AB BC上 DE//AC,过点D的直线DF交AC于G,交BC得延长线与F.且BE:EC:CF=3:2:1,则AG/ cg =
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![1如图,过平行四边形ABCD的顶点A作一条直线,分别交BD,CD及BC的延长线于GFE,若CE=1/2 BC,则DF/FC = ,AG/GE=?2 如图点D E 分别在AB BC上 DE//AC,过点D的直线DF交AC于G,交BC得延长线与F.且BE:EC:CF=3:2:1,则AG/ cg =](/uploads/image/z/10297973-29-3.jpg?t=1%E5%A6%82%E5%9B%BE%2C%E8%BF%87%E5%B9%B3%E8%A1%8C%E5%9B%9B%E8%BE%B9%E5%BD%A2ABCD%E7%9A%84%E9%A1%B6%E7%82%B9A%E4%BD%9C%E4%B8%80%E6%9D%A1%E7%9B%B4%E7%BA%BF%2C%E5%88%86%E5%88%AB%E4%BA%A4BD%2CCD%E5%8F%8ABC%E7%9A%84%E5%BB%B6%E9%95%BF%E7%BA%BF%E4%BA%8EGFE%2C%E8%8B%A5CE%3D1%2F2+BC%2C%E5%88%99DF%2FFC+%3D+%2CAG%2FGE%3D%3F2+%E5%A6%82%E5%9B%BE%E7%82%B9D+E+%E5%88%86%E5%88%AB%E5%9C%A8AB+BC%E4%B8%8A+DE%2F%2FAC%2C%E8%BF%87%E7%82%B9D%E7%9A%84%E7%9B%B4%E7%BA%BFDF%E4%BA%A4AC%E4%BA%8EG%2C%E4%BA%A4BC%E5%BE%97%E5%BB%B6%E9%95%BF%E7%BA%BF%E4%B8%8EF.%E4%B8%94BE%3AEC%3ACF%3D3%EF%BC%9A2%3A1%2C%E5%88%99AG%2F+cg+%3D)
1如图,过平行四边形ABCD的顶点A作一条直线,分别交BD,CD及BC的延长线于GFE,若CE=1/2 BC,则DF/FC = ,AG/GE=?2 如图点D E 分别在AB BC上 DE//AC,过点D的直线DF交AC于G,交BC得延长线与F.且BE:EC:CF=3:2:1,则AG/ cg =
1如图,过平行四边形ABCD的顶点A作一条直线,分别交BD,CD及BC的延长线于GFE,若CE=1/2 BC,则DF/FC = ,AG/GE=?
2 如图点D E 分别在AB BC上 DE//AC,过点D的直线DF交AC于G,交BC得延长线与F.且BE:EC:CF=3:2:1,则AG/ cg =
1如图,过平行四边形ABCD的顶点A作一条直线,分别交BD,CD及BC的延长线于GFE,若CE=1/2 BC,则DF/FC = ,AG/GE=?2 如图点D E 分别在AB BC上 DE//AC,过点D的直线DF交AC于G,交BC得延长线与F.且BE:EC:CF=3:2:1,则AG/ cg =
1.CEF与BEA相似,则AB/CF=BE/CE=3则AB=3CF,CD=3CF.所以F是CD三等分点,DF/FC =2
连接AC,G是ACD重心,则AG=2GF,CEF与BEA相似,AE/EF=BE/CE=3,则AE=3EF=AG+GF=3GF,则GF=EF,AG=2GF=GF+EF=GE,则AG/GE=1
2.CFG与EFD相似,则CG/DE=CF/EF=1/3,则DE=3CG,BDE与BAC相似,则DE/AC=BE/BC=3/5,则3AC=5DE,3(3CG)=(AG+CG),可解得AG/CG=4