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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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Do you know the coordinates of the three points? –  Henry Sep 10 '12 at 0:03
    
Yes, I know the coordinates –  George Sep 10 '12 at 0:05
    
Hold on: are you able to do operations such as rotation and translation with computer? Now simplify your problem: move the lines to origin and rotate them so that one line match one axis. Then you have a simple angle. –  hhh Sep 10 '12 at 0:23
    
That would be possible, but I need to find you the angle in geographic orientation. –  George Sep 10 '12 at 0:26
    
You do know, don’t you, that the external and the internal bisector are the same line? –  Lubin Sep 10 '12 at 2:02

1 Answer 1

up vote 0 down vote accepted

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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CA, CD, DB, AB are lengths? –  George Sep 10 '12 at 0:27
    
Yes. Unless they are written as vectors, they are lengths. –  N. S. Sep 10 '12 at 0:31
    
So, let me see if I understood this correctly: I have a triangle, ABC, with the following lenghts: AB 3, BC 4, AC 5. If I change the values in your formula, this will give me: CD/DB = 5/3. I need further help getting past that :P –  George Sep 10 '12 at 0:45
    
Yes, now you reduced the problem to the following: find the point $D$ between $C,B$ which divides $CD/DB=5/3$. This is a pretty standard linear algebra problem, do you know how to work with vectors? If not it is easy to find a sligtly more complicated solution without vectors. –  N. S. Sep 10 '12 at 0:56
    
@Lubin N.S. wrote the answer as if $A$ were the vertex of the angle being bisected, rather than $B$, but otherwise it's correct. Why do you think there has to be square roots? –  Ted Sep 10 '12 at 1:07

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