circumradii of $\triangle{IAB}$ Let $I$ be the incenter of $\triangle{ABC}$. Let $R$ be the radius of the circle that circumscribes $\triangle{IAB}$. Find a formula for $R$ in term of other elements $a, b, c, A, B, C, r, R$ of $\triangle{ABC}$. I need this formula in order to prove a geometric inequality.
 A: Remember the extended law of sines:
$$R = \frac{a}{2\sin A} = \frac{b}{2\sin B} = \frac{c}{2\sin C}$$
The above gives the circumradius ($R$) for a triangle whose one angle is $A$ and the opposite side is $a$ and so on. Well, the labeling is a bit different here.

For our triangle $IAB$, the extended law of sines becomes: $$R = \frac{a}{2\sin\angle AIB}$$
So, we need to calculate $\angle AIB$, and we are done. This can be done by realizing that $IB$ is the angle bisector of $\angle A$ and $IA$ angle bisector of $\angle A$. So, $\angle AIB$ is:
$$180^{\circ} - (\angle IAB + \angle IBA) = 180^{\circ} - (\frac{\angle A}{2} + \frac{\angle B}{2}) = 180^{\circ} - (\frac{180^{\circ} - \angle C}{2}) = 90^{\circ} + \frac{C}{2} $$
Note that $\sin (90^{\circ} + \theta) = \cos\theta$. So we substitute back and our formula becomes:
$$R=\frac{AB}{2\cos\frac{C}{2}}$$
If you need proof of anything that I have used ask me. If you want to express $\cos\frac{C}{2}$ in terms of $\cos C$, you can do that by using the identity [and considering the positive root]:
$$\cos^2\frac{\theta}{2}=\frac{1+\cos\theta}{2}$$
A: 
suppose that $AI$ cut the circumcircle of $\triangle ABC$ at $D$.
$\angle DBI=\angle IBC+\angle DBC=\angle IBA+\angle BAD=\angle BID$
thsu, $\triangle DBI$ is isosceles and likewise, $\triangle DCI$ is isosceles.
that is, $DB=DI=DC$ and then $D$ is the circumcenter of $\triangle IBC$.
It's called the theorem of Mention, who is French mathematician.
hence, the circumradius of $\triangle IBC$ is $DB$.
by sine law, $\dfrac{BD}{2R}=\sin\dfrac{\angle A}{2}$ and then $BD=2R\sin\dfrac{\angle A}{2}$
Likewise, we know that the circumradius of $\triangle IAB$ is $2R\sin\dfrac{\angle C}{2}$
