Solution:

The simplest way to do this is using complex numbers. Let $\omega = (-1 + i \sqrt{3}) / 2$, a cube root of unity, and we have $1 + \omega = -\omega^{2}$. Then a triangle $XYZ$ is equilateral (with vertices labeled in counterclockwise order) iff segment $ZX$ is the rotation image of $YZ$ through angle $2\pi/3$; representing the points by their values in the complex plane, this says $X - Z = \omega(Z - Y)$ or, equivalently, $X + \omega Y + \omega^{2} Z = 0$. In particular, when $Y = 0$, we get $X = -\omega^{2} Z$.

Now, position the given points so that $O$ is at the origin; then the given yields $B = -\omega^{2} A$, $D = -\omega^{2} C$, $F = -\omega^{2} E$. Also, clearly $K = (B + C)/2$, $L = (D + E)/2$, $M = (F + A)/2$. Thus
$$K + \omega L + \omega^{2} M = \left(\omega^{2} A + B + C + \omega D + \omega E + \omega^{2} F\right)/2 = \left(\omega^{2} A - \omega^{2} A + C - C + \omega E - \omega E\right)/2 = 0$$
giving what we need.