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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

2 Answers 2

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).

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What you're asking is the following: in a given triangle, is the perpendicular bisector of some angle the median to the side in front of this angle?

The answer is clearly no, as the above condition characterises isosceles triangles...

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