已知△ABC,AD//BC,连CD交AB于E,且AE:EB=1:2,过E作EF//BC,交AC于F,S△ADE=1,求S

AD//BC,△ADE∽△BEC,S△ADE/S△BEC=（AE/BE）^2=1/4,S△BEC=4.
S△BEC/S△AEC=BE/AE=2,S△AEC=S△BEC/2=2,
S△AEF/S△EFC=AF/FC,EF//BC,AF/FC=AE/BE=1/2,S△EFC/S△AEF=2,(S△EFC+S△AEF)/S△AEF=3/1,S△AEC/S△AEF=3,2/S△AEF=3,S△AEF=2/3.