Corner angle bisectors

what is the point perpindcular to?

if it''s to the edges, you''re just finding the opposing point of a square. but i''m sure that''s not what you''re asking. maybe a diagram would help?

         /
        /
       /
      /\ perp. dist.
     /  \   
    /    * (this is the point I want to find)
   /   / |  
  /  /   | perp. dist.
 / /     |
/ /      |
*--------------------------------
 

I hope this ascii drawing comes out ok. Both perpendicular distances from corner edges
must be equal. The point is somewhere on the bisector vector.

Every point along the ray of bisection would be of equal distance from the two rays making up the angle.

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"Sneftel is correct, if rather vulgar." --Flarelocke

you know the angle between the two vectors

so sin(a/2) = (perp_dist / d)

d = perp_dist / sin(angle/2)

quote:

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Every point along the ray of bisection would be of equal distance from the two rays making up the angle.

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That's true if you draw a perpendicular line through the bisector, but I need lines
perpendicular to the edges, not the bisector.

quote:

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d = perp_dist / sin(angle/2)

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I guess this must be true for both edges, since the perp. dist. is the same, and the angle is the same.

Thanks!

[edited by - afaik on November 19, 2003 4:32:27 PM]

quote:

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Original post by Afaik

quote:

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Every point along the ray of bisection would be of equal distance from the two rays making up the angle.

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That's true if you draw a perpendicular line through the bisector, but I need lines
perpendicular to the edges, not the bisector.

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No, it's true for all points. Think about it: the edges each have a perpendicular line through them, to the point on the bisector you've chosen. That, along with the bisector, makes a right triangle. ANY point you pick on the bisector will make congruent right triangles.

EDIT: or did you mean, you want it equidistant from all three edges of the triangle?

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"Sneftel is correct, if rather vulgar." --Flarelocke

[edited by - sneftel on November 20, 2003 5:34:15 AM]

I understand what you''re saying.

Don''t worry about it. I got it working!

quote:

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Original post by Afaik
I understand what you''re saying.

Don''t worry about it. I got it working!

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How exactly did you get "it" working, since any point on that line would be valid...