Some Amazing Triangle

Construct the A-height and call the foot of the altitude $D$ .

Notice that $\angle BHC = \angle BIC = 120^{\circ}$ . This implies that $BIHC$ is cyclic. Because opposite angles in a cyclic quadrilateral are supplements and $\angle IBC = \frac{\angle B}{2}$ , we know that $\angle IHC = 180^{\circ} - \frac{\angle B}{2}$ .

Since $\angle DAC = 90^{\circ} - \angle C$ and $\angle ACH = 90^{\circ} - A$ , we know that $\angle DHC = 180 - \angle A - \angle C = \angle B$ .

Since $\angle DHC = \angle B$ $\angle IHC = 180^{\circ} - \frac{\angle B}{2}$ , we know that $\angle IHD = \angle IHC - \angle DHC = 180^{\circ} - \frac{3}{2}\angle B \implies \angle AHI = \boxed{\frac{3}{2}\angle B}.$