How does one find an exterior angle bisector relative to the x-axis? Let's say we're given points $A$, $B$, and $C$, which form $\Delta ABC$. Assuming $A=(0,0)$, what is the value of the exterior angle bisector formed by $\angle A$ relative to the x-axis?

(The image is simply to make what I'm asking clearer.)
 A: The angle of inclination of a line through $(0,0)$ with slope $m$ is $\tan m$. Assume in your diagram that $B$ is the lower point (in quadrant 4) and $C$ the upper point in quadrant 1. Let $m$ be the slope of the line through the origin and $C$, and $n$ the slope of the line through the origin and $B$. We can get the slope $x$ of the bisector of angle $CAB$ by solving 
$$\frac{m-x}{1+mx}=\frac{x-n}{1+xn}.$$
This has two solutions
$$x=\frac{mn-1 \pm \sqrt{(m^2+1)(n^2+1)}}{m+n}.\tag{1}$$
I did a few cases and it seems if we use the $+$ sign here we get the slope for the bisector of angle $CAB$, provided the slopes are as in the diagram and so satisfy $m>n.$
Once you have that slope for the bisector, its angle of inclination with the $x$ axis is $\arctan x.$ Of course that's not what you're after since you want the external bisector. But the external bisector makes a 90 degree angle with the internal bisector, so that's not much more work.
Added: I think one may just choose the $-$ sign in $(1)$, calculate arctangent, and add 180 to the result, and arrive at the inclination of the external bisector in one step. (It may be best to calculate everything to make sure where the angles in question actually wind up relative to a given diagram.)
A: if you define two unit vectors in the direction of AB and -AC, you could use the fact that the angles between the bisector- let's call it X- and the two lines are equal, and so is the cosine. Using the dot product: $ \ u \cdot X = v \cdot X  $, where u and v are the unit vectors. Once you get a subspace of vectors that satisfy that condition is easy to get the angle with the horizontal axis.
