Clearly $\triangle BHC$ is equilateral so we are done by Ptolemy Theorem.

By very easy angle chase we get $\angle BIC = 120 = \angle BOC $. So quadrilateral $ BIOC$ is concyclic. Then $\angle BHC =60$ but H is midpoint of the arc $BC$ . We get $\triangle BHC$ is equilateral . Applying Ptolemy theorem on quadrilateral $IBHC$ $BI \cdot CH $ $+$ $CI \cdot BH$ $=$ $BC\cdot IH$ $BC$ ($BI$ $+$ $CI$) $=$ $BC \cdot IH$ $BI$ $+$ $CI$ $=$ $IH$

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$\left. \begin{array}{l} {\rm{C}}\widehat {\rm{O}}{\rm{B = C}}\widehat {\rm{I}}{\rm{B = 12}}{{\rm{0}}^o}\\ \\ {\rm{C}}{\rm I} = {\rm I}{\rm E} \end{array} \right\}\mathop {}\limits_{}^{} \Rightarrow \mathop {}\limits_{}^{} \left\{ \begin{array}{l} {\rm{CE = EI = BC}}\\ \\ {\rm{C}}{\rm H} = {\rm H}{\rm B} = {\rm{BC}} \end{array} \right.\mathop {}\limits_{}^{} \Rightarrow \mathop {}\limits_{}^{} {\rm{tr}}{\rm{.IBC = tr}}{\rm{.EHC}}\mathop {}\limits_{}^{} \Rightarrow \mathop {}\limits_{}^{} {\rm I}{\rm B} + {\rm I}{\rm{C = IE + EH = IH}}$

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Solved with L567 who found a synthetic solution and is ashamed it to post it after seeing #2 - posting for storage. Notice that $I$ lies on the circumcircle of $\triangle BOC$, then using Euler's Formula $$IH^2 = OH^2 - OI^2 = 4R^2-(R^2-2Rr) = R^2(3+4\sin \frac{\beta}{2}\sin \frac{\gamma}{2})$$On the other hand $$(BI+IC)^2 = 4R^2(\sin \frac{\beta}{2} + \sin \frac{\gamma}{2})^2$$Writing $\frac{\gamma}{2} = 60^{\circ}-\frac{\beta}{2}$ we have that $$4R^2(\sin \frac{\beta}{2} + \sin (60^{\circ}-\frac{\beta}{2}))^2 = R^2(3+4\sin \frac{\beta}{2} \frac{\sin(60^{\circ}-\frac{\beta}{2})}{2})$$after expanding using the fact that $\sin (60^{\circ}-\frac{\beta}{2})$ as $\frac{\sqrt{3}}{2}\cos \frac{\beta}{2} - \sin \frac{\beta}{2}$. $\blacksquare$