First of all, ∫△ABC is an equilateral triangle, and the point ∴C is on the negative semi-axis of the X axis, and C (- 1, 0).

∫ straight line y=kx+2k? ∴ Its intersection with the X axis is (-2,0).

∫A(0，√3)、B( 1，0)？ ∴ The analytical formula of straight line AB is y=-√3(x- 1) (undetermined coefficient method).

If S△DEC=S△AEF, then S△DBF=S△ABC=√3.

The coordinates of point F are ((√3-2k)/(k+√3), (k(√3-2k))/(k+√3)+2k).

That is, there is DB×F (vertical) ×? =√3

This gives k=2√3/7.