Iranian Mathematical Olympiad

Note that
$$\frac{XB}{XC} = \frac{\sin \angle XOB}{\sin \angle XOC} = \frac{\sin \angle IOB}{\sin \angle IOC}.$$
Which is equivalent to
$$\frac{\sin\left(\frac{\angle C}{2}\right)}{\sin\left(\frac{|\angle B - \angle A|}{2}\right)} \cdot \frac{\sin\left(\frac{|\angle B - \angle A|}{2}\right)}{\sin\left(\frac{\angle C}{2}\right)}.$$
Using Ceva's theorem in triangle $\triangle BOC$ for point $I$. Now, by the Ratio lemma, we have
$$\frac{YB}{YC} = \frac{c}{b} \cdot \frac{\sin \angle YAB}{\sin \angle YAC} = \frac{c}{b} \cdot \frac{\sin \angle KAF}{\sin \angle KAE}.$$
By Ceva's theorem in $\triangle AEF$ for point $K$, we have
$$\frac{\sin \angle KAF}{\sin \angle KAE} = \frac{\sin \angle KEF}{\sin \angle KEA} \cdot \frac{\sin \angle KFA}{\sin \angle KFE} = \frac{\sin(90^\circ - \frac{\angle B}{2})}{\sin(90^\circ - \frac{\angle C}{2})} \cdot \frac{\sin(\frac{\angle B}{2})}{\sin(\frac{\angle C}{2})}.$$
So we have just to prove that
$$\begin{gather*} \frac{c}{b} \cdot \frac{\sin(90^\circ - \frac{\angle B}{2})}{\sin(90^\circ - \frac{\angle C}{2})} = \frac{\sin(\frac{\angle C}{2})}{\sin(\frac{\angle B}{2})} \\ \iff \frac{\sin \angle C}{\sin \angle B} \cdot \frac{\cos(\frac{\angle B}{2})}{\cos(\frac{\angle C}{2})} = \frac{\sin(\frac{\angle C}{2})}{\sin(\frac{\angle B}{2})}. \end{gather*}$$
Which is obviously true since $\sin x = 2 \sin(\frac{x}{2}) \cos(\frac{x}{2})$.