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I have a simple question, and I've looked over the internet and can't find a numerical way of doing it.

I only found how to do this using a ruler and a compass, which I can't use, since I'm doing a computer program.

I need to calculate the bisection angle for line segments ABC. I need to calculate the inner bisection angle and the outer bisection angle.

EDIT: dividing the angle by two won't work.

Let me explain it a bit better: this linestring, which contains the vertices A, B and C is used to generate a polygonal buffer. I use the linestring to find out the angle (medium angle between AB and BC) that will be used to construct a line, used to split the polygon mentioned earlier.

Example:

LINE: 0 0, 10enter image description here

Thanks

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    You do know, don’t you, that the external and the internal bisector are the same line?2012-09-10

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The bisector theorem should help you in this case.If $AD$ is the bisector, with $D \in (BC)$, then

$\frac{CD}{DB}=\frac{CA}{AB} \,.$

From the coordinates of $A,B,C$, you should be able to find the coordinates of $D$.

Note that the same happens for the external bisector, this time $D'$ is a point on $BC$, outside $(BC)$. Then

$\frac{CD}{DB}=\frac{CA}{AB} \,.$

P.S. If you know a little linear algebra,finding $D$ is easy:

$\frac{CD}{DB}=\frac{CA}{AB} \Rightarrow \frac{CD}{BC}=\frac{CA}{AB+AC} \,.$

Thus, since $D$ is between $B$ and $C$, as vectors

$\vec{CD}=\frac{CA}{AB+AC} \vec{CB} \,.$

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    Well, I definitely got egg on my face on this one. I completely forgot that one formula for $\tan(x/2)$ is $\sin x/(1+\cos x)$. So there’s no philosophical problem at all. Sorry, but I can’t remove the downvote.2012-09-10