# How to get the angle between the bisector (with 3 points given) and the abscisse.

I have three random points in a 2D space, $$A$$, $$B$$ and $$C$$, how can I compute the angle between the abscisse and the bisector of $$ABC$$:

So I am looking for the angle $$α$$ (in red).

I'm using a computer and I can easily compute the azimuth $$AZIMUTH_{BA}$$ and $$AZIMUTH_{BC}$$

For this specific example I can do:

$$\frac{AZIMUTH_{BA}-AZIMUTH_{BA}}{2} + \frac{5\pi}{2} - AZIMUTH_{BA}$$

To get the angle $$α$$.

But how can I get a generalized formula to always get the right $$α$$ angle between $$0$$ and $$2\pi$$ ?

• Well, if you have the coordinates of of all the points, you could calculate the vector $$\vec{v}_{\text{bisector}} = \frac{\vec{BA} + \vec{BC}}{2}$$ And then you could just use dot product to calculate the angle $\alpha$. – Matti P. Nov 13 at 10:40
• The dot product between $v_{bisector}$ and the abscisse ? – obchardon Nov 13 at 10:47
• Yeah, you could take $\vec{v}_{\text{bisector}} \cdot (- \hat{i})$. – Matti P. Nov 13 at 10:54
• @MattiP. That only produces the bisector when $\lvert\overrightarrow{BA}\rvert=\lvert\overrightarrow{BC}\rvert$. If the vectors are of different lengths, you need to normalize them first. Try, for example, $A=(10,0)$, $B=(0,0)$ and $C=(0,1)$. – amd Nov 13 at 23:28