三角形ABC,AB边有点E,AC边有点F,EF平行BC,三角形AEF与三角形BCE面积相等,三角ABC面积为1,求三角形CEF面

∵AE/EB=S△ACE/S△BCES△BCE=S△AEFAC/AF=S△ACE/S△AEF∴AE/EB=AC/AF又∵EF‖BC∴AE/EB=AF/FC∴AC/AF=AF/FC(AF+FC)/AF=AF/FC1+FC/AF=AF/FC设AF/FC=x则1+1/x=x解得x=(√5+1)/2∴AE/EB=AF/FC=(√5+1)/2S△BCE/S△ABC...