# Does the angle bisector always pass through the midpoint of any line segment between the two sides of the angle?

Consider this image:

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 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. – anon Nov 16 '12 at 19:39
Just consider some extreme cases, like $O=(0,0)$, $A=(1,0)$, and $B = (0,1000000)$. – Sean Eberhard Nov 16 '12 at 19:40
NO ${}{}{}{}{}$ – Will Jagy Nov 16 '12 at 19:44
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 $(10,0)$. – André Nicolas Nov 16 '12 at 20:47

To be specific, if the angle bisector hits $AB$ in $P$ then we have the wonderful theorem $$PA:BP = OA:OB.$$ Thus $P$ is is the middle of $A$ and $B$ if and only if $OA=OB$ (i.e. the triangle is isosceles).