The measure of the angle $\angle A$ of the acute triangle $ABC$ is $60^o$, and $HI = HB$, where $I$ and $H$ are the incenter and the orthocenter of the triangle $ABC$. Find the measure of the angle  $\angle  B$.
1. Given that $\angle A = 60^\circ$ in the acute triangle $ABC$, and $HI = HB$, where $I$ is the incenter and $H$ is the orthocenter of the triangle $ABC$.

2. Since $\angle A = 60^\circ$, triangle $ABC$ is not only acute but also has a special property: $B, H, I, C$ are concyclic. This means that points $B, H, I, C$ lie on a common circle.

3. We need to find the measure of $\angle B$. To do this, we use the fact that $B, H, I, C$ are concyclic. This implies that the sum of opposite angles in the cyclic quadrilateral is $180^\circ$.

4. Consider the angle $\angle BHI$. Since $B, H, I, C$ are concyclic, we have:
   \[
   \angle BHI = 180^\circ - \angle \frac{C}{2}
   \]
   But we also know that:
   \[
   \angle BHI = 120^\circ + \angle \frac{B}{2}
   \]

5. We also know that $\angle HBI = \angle HIB = \angle \frac{B}{2} - 30^\circ$. Therefore:
   \[
   \angle BHI = 240^\circ - \angle B
   \]

6. Equating the two expressions for $\angle BHI$, we get:
   \[
   120^\circ + \angle \frac{B}{2} = 240^\circ - \angle B
   \]

7. Solving for $\angle B$, we first isolate $\angle B$:
   \[
   120^\circ + \frac{B}{2} = 240^\circ - B
   \]

8. Combine like terms:
   \[
   \frac{3B}{2} = 120^\circ
   \]

9. Multiply both sides by $\frac{2}{3}$ to solve for $\angle B$:
   \[
   \angle B = 80^\circ
   \]

The final answer is $\boxed{80^\circ}$