In a $\Delta ABC$ with incenter $I$, prove that the circumcenter of $\Delta AIB$ lies on $BI$ In a $\Delta ABC$ with incenter $I$, prove that the circumcenter of $\Delta AIB$ lies on $BI$. 
Consider the following figure: 
$AD, BE, CF$ are internal angle bisectors concurrent at $I$, the incenter. Let the perpendicular bisector of $AI$ meet $BF$ at $G$ and $AI$ at $K$ (not marked in fig). 
Clearly, $GA = GI$, since $\Delta AGK \cong \Delta IGK$.
Now all that remains to be proven is that $GA = GB$, or $\angle GAB = \angle GBA$. If we can prove that $AGBC$ is cylic, then
$$\angle ABG = \angle ACG = \angle GCB = \angle GAB$$
as desired. However, despite numerous efforts I was not able to prove the required conjecture. 
I was, however, able to obtain an alternate solution by drawing the excenter $I_c$ of $\Delta ABC$, and then showing that $II_c$ was a diameter. 
I would also like to prove it using this approach. Any help would be much appreciated. 
 A: 
Let $I_c$ is the circumcenter of $\triangle ABI$,(= your $G$)
$AI_c=BI_c=II_c, \angle I_cAB=\angle I_cBA=\alpha,\angle I_cAI=\angle I_cIA=\alpha+\dfrac{A}{2},\angle I_cBI=\angle I_cIB=\alpha+\dfrac{B}{2},\angle I_cIB+\angle I_cIA=\pi-\dfrac{B}{2}-\dfrac{A}{2}=\dfrac{\pi}{2}+\dfrac{C}{2}=2\alpha+\dfrac{B}{2}+\dfrac{A}{2}=2\alpha+\dfrac{\pi}{2}-\dfrac{C}{2},2\alpha=C$ 
$\angle I_cAC=B+\alpha=A+\dfrac{C}{2},\angle I_cBF=\pi-\angle I_cBC=\pi-(B+\dfrac{C}{2})=A+\dfrac{C}{2}=\angle I_cAC$
$I_cE$ and $I_cF$ is the distances,
it is trivial $\triangle I_cAE \cong \triangle I_cBF \implies I_cE=I_cF $
so $I_cC$ is the bisector of $\angle C$
A: Recall that the incenter, $I$ lies on the angle bisectors of $\angle A$, $\angle B$, and $\angle C$; define
$$\alpha = \frac{1}{2}A \qquad \beta = \frac{1}{2}B \qquad \gamma = \frac{1}{2}C$$

Now, extend $\overline{CI}$ to $D$ on the circumcircle of $\triangle IAB$.
Applying the Exterior Angle Theorem to $\triangle AIC$, we have $\angle AID = \alpha + \gamma$. Since this angle and $\angle ABD$ both subtend the same $\stackrel{\frown}{AD}$, the Inscribed Angle Theorem gives $\angle ABD = \alpha + \gamma$ as well. Therefore
$$\angle IBD = \alpha + \beta + \gamma = \frac{1}{2}( A + B + C ) = 90^\circ$$
By the Inscribed Angle Theorem, then, $\overline{ID}$ is a diameter of the circle, so that it ---and therefore also $\overline{CD}$--- contains its center.
