Q:

Find a specific vector equation of a line that divides a angle in half.

I've been studying a little geometry on my own, and I just recently stumbled on this problem, that I'm unable to answer:
Given the points A=(2,-1), B=(5,4) and C=(-7,8), find a vector equation of a line that divides the angle BAC in half(Same angle for each size).
Thank you
-Dom

A:

Using user georg's approach:
$$
\begin{align}
u &= \frac{B - A}{|B-A|} = (0.51450, 0.85749) \\
v &= \frac{C - A}{|C-A|} = (-0.70711, 0.70711) \\
z &= \frac{u+v}{|u+v|} = (-0.12218, 0.99251)
\end{align}
$$
where $z$ is the unit vector on the desired line
$$
g(t) = t z \quad (t \in \mathbb{R})
$$
Checking the angles gives:
$$
\begin{align}
u \cdot v &= \cos \alpha \iff \alpha = 75.964^\circ \\
u \cdot z &= \cos \beta \iff \beta = 37.982^\circ
\end{align}
$$
The interesting bit is to prove, that $\beta = \alpha /2$.
In the parallelogram of $u+v$ (draw in the other diagonal $v-u$ as well) one can read from the triangle with the sides $u$, $v-u$, $v$
$$
180^\circ = \alpha + 2 \gamma 
$$
and from the triangle with the sides $(u+v)/2$, $(v-u)/2$ and $v$
$$
90^\circ = \beta + \gamma
$$
which gives
$$
\beta = \frac{\alpha}{2}
$$

