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I fooled around with the concept of an angle bisector, and I (thought I) found out (and some websites confirmed this, but now I'm in doubt) that the angle bisector of a vertex is the collection of points equidistant from the 2 sides of the vertex it's bisecting. However, how is this possible. Wouldn't this mean that the bisector would divide the opposite side in 2 equally long segments? But this would create a median, and obviously a median and an angle bisector are different things. Can anyone help?

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    $\begingroup$ "Equidistant from the 2 sides" ... when you measure the distance from a point to a side, you use a line-segment perpendicular to the side. But your opposite side is probably not perpendicular to the other sides. So there is no contradiction. $\endgroup$ – GEdgar Dec 1 '12 at 14:19
  • $\begingroup$ That's it! Why don't you put this in an answer, so I can give you the best answer! $\endgroup$ – JohnPhteven Dec 1 '12 at 16:35
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answer copied from comment

"Equidistant from the 2 sides" ... when you measure the distance from a point to a side, you use a line-segment perpendicular to the side. But your opposite side is probably not perpendicular to the other sides. So there is no contradiction.

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Best thing would be to draw a picture. However let me clarify: If you consider a triangle (I guess you considered one), then the angle bisector is indeed the line equidistant from the two sides whose intersection is the vertex. However, unless the sides are equal, then it will not bisect the opposite side. To get what I mean, draw a scalene triangle, bisect an angle, measure and convince yourself.

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  • $\begingroup$ I am already convinced of that fact sir, however, what concerns me, is that since it doesn't bisect the opposite side, the statement I made doesn't hold, since it is closer to one side than another. Do you understand what I mean? I must stop being equidistant from the sides at one point, thus falsifying the statement. $\endgroup$ – JohnPhteven Dec 1 '12 at 13:41
  • $\begingroup$ I think GEdgar answered your query. $\endgroup$ – Gautam Shenoy Dec 1 '12 at 20:14

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