Question 48
In an acute triangle $ABC$ with $\angle B \gt \angle C$, let $I$ be its incenter, $R$ its circumradius, and $D$ the foot of the perpendicular from $A$ to $BC$. Point $K$ lies on line $AD$ such that $AK = 2R$ and $D$ lies between $A$ and $K$. Prove that
$$
\angle KID = \frac{\angle B - \angle C}{2}.
$$
$$
\angle KID = \frac{\angle B - \angle C}{2}.
$$