This isn't a matter of real versus complex geometry -- the equations you give aren't enough to encode Morley's theorem over the reals. Let $z_1 z_2 z_3$ be our original triangle; I will use indices that are cyclic modulo $3$. Let $\ell^0_i$ be the trisector of the angle at $z_i$ which is closer to $z_{i-1}$ and let $m^0_i$ be the trisector which is closer to $z_i$. Let $\ell^1_i$ and $\ell^2_i$ be the rotations of $\ell^0_i$ by $120^{\circ}$ and $240^{\circ}$, and define $m^1_i$ and $m^2_i$ likewise.

<p>For $r=1$, $2$ and $3$, let $(\delta_r, \epsilon_r)$ equal to one of $(0,0)$, $(1,2)$ or $(2,1)$. Let $c_r$ be the intersection of $m^{\epsilon_r}_r$ and $\ell^{\delta_{r+1}}_{r+1}$. Then $\angle z_r z_{r+1} c_r = \angle c_r z_{r+1} c_{r+1} = \angle c_{r+1} z_{r+1} z_{r+2}$ so your equations hold for $(x_r, y_r) = z_r$ and $(a_r, b_r) = c_r$. </p>

Of the $81$ possible choices for $(\delta_r, \epsilon_r)$, I believe that the triangle $c_1 c_2 c_3$ is equilateral only for $54$ of them. I am basing this on [Connes' proof][1] of Morley's theorem, where a key hypothesis is that 
$$\sum_{r=1}^3 \angle c_{r-1} z_{r} c_{r} = \pm 60^{\circ} \ \mbox{not} \ 180^{\circ}.$$
Connes' rotation $g_r$ is a rotation around $z_r$ by angle $2 \angle c_{r-1} z_{r} c_{r}$, so what he says is that $g_1 g_2 g_3$ should be a nontrivial rotation.

It's easy enough to encode this condition algebraically. (If you have trouble with it I'll write more, but I bet you won't.) Once you do that, I think it should be easy to convert Connes proof into the sort of proof you are looking for. You'll also need to impose that certain things are nonzero, because Connes divides at times.


  [1]: http://www.alainconnes.org/docs/morley.pdf