(2008? Yantai) As shown in the figure, the side length of rhombic ABCD is 2, BD=2, E and F are two moving points on the sides of AD and CD respectively, AE+CF = 2.

( 1)AE+CF=2=CD=DF+CF

∴AE=DF

AB=BD

∠A=∠BDF=60

∴△BDE is equal to △ △BCF.

(2) BE=BF from (1)

And ∠ EBF = ∠ EBD+∠ DBF = ∠ EBD+∠ Abe = ∠ Abd = 60.

△ BEF is an equilateral triangle.

③3√3/4 & lt; = S & lt=√3