证明 设BF=x,CE=y,EF=z,正ΔABC的边长为2.则有 AF=2-x,AE=2-y,BD=CD=1.正ΔABC周长为6.因为∠EDF=60°,∠FBD=∠ECD=60°，所以△BDF∽△CED即有BF/BD=CD/CE ，故得:xy=1.在△BDF,△CED,△AEF,△DEF中，由余弦定理得:DF^2=x^2-x+1;CE^2=y^2-y+1;z^2=(2-x)^2+(2-y)^2-(2-x)*(2-y)，即 z^2=x^2+y^2-2x-2y+3 (1)z^2=x^2-x+1+y^2-y+1-√[(x^2-x+1)*(y^2-y+1)];注意到:(x^2-x+1)*(y^2-y+1)=x^2+y^2-2x-2y+3=z^2,所以得:z^2=x^2-x+1+y^2-y+1-z (2)(1)-(2)得:z+1=x+y <==> z+(2-x)+(2-y)=3.故ΔAEF的周长是ΔABC周长的一半.