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.


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 $2R\sin\dfrac{\angle C}{2}$
