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Consider this image:

triangle

will the angle bisector of angle AOB always pass through the midpoint of AB, regardless of the lengths of AO and BO?

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    Consider $O=(0,0)$, $B=(1,0)$ and $A=(2,1)$. Then the angle $\angle AOB$ is $\tan^{-1}(1/2)$. The midpoint $M$ of $AB$ is $M=(3/2,1/2)$ which makes an angle of $\angle MOB=\tan(\frac{1/2}{3/2})=\tan(1/3)$ with the $x$-axis. But [wolfram alpha](http://www.wolframalpha.com/input/?i=%281%2F2%29Arctan%5B1%2F2%5D-Arctan%5B1%2F3%5D) shows that $$\frac{\angle AOB}{2}~=~\frac{1}{2}\tan^{-1}\left(\frac{1}{2}\right)~\ne~\tan^{-1}\left(\frac{1}{3}\right)~=~\angle MOB.$$ They are off by about 0.09. I picked these numbers so they would be easy to follow.2012-11-16
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    Just consider some extreme cases, like $O=(0,0)$, $A=(1,0)$, and $B = (0,1000000)$.2012-11-16
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    NO ${}{}{}{}{}$2012-11-16
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    Carefully **draw** an angle, and its bisector. The positive $x$ and $y$ axes make a nice angle. Now draw the line that goes through $(0,1)$ and $(10,0)$.2012-11-16

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