In a triangle $ABC$, let $H, I$ and $O$ be the orthocentre, incentre and circumcentre, respectively. If the points $B, H, I, C$ lie on a circle, what is the magnitude of $\angle BOC$ in degrees?
1. **Identify the given points and their properties:**
   - \( H \) is the orthocenter of \(\triangle ABC\).
   - \( I \) is the incenter of \(\triangle ABC\).
   - \( O \) is the circumcenter of \(\triangle ABC\).
   - Points \( B, H, I, C \) lie on a circle.

2. **Determine the angle \(\angle BHC\):**
   - The orthocenter \( H \) is where the altitudes of the triangle intersect.
   - The foot of the altitude from \( A \) is \( X \), from \( B \) is \( Y \), and from \( C \) is \( Z \).
   - By angle chasing, \(\angle BHX = \angle C\) and \(\angle CHX = \angle B\).
   - Therefore, \(\angle BHC = \angle B + \angle C\).

3. **Use the cyclic quadrilateral property:**
   - Since \( B, H, I, C \) lie on a circle, \(\angle BHC = \angle BIC\).
   - The incenter \( I \) is the intersection of the angle bisectors of \(\triangle ABC\).
   - Therefore, \(\angle BIC = 180^\circ - \frac{1}{2}(\angle A)\).

4. **Equate the angles:**
   - From the cyclic quadrilateral property, \(\angle BHC = \angle BIC\).
   - Thus, \(\angle B + \angle C = 180^\circ - \frac{1}{2}(\angle A)\).

5. **Solve for \(\angle A\):**
   - We know that \(\angle A + \angle B + \angle C = 180^\circ\).
   - Substitute \(\angle B + \angle C\) from the previous step:
     \[
     \angle B + \angle C = 180^\circ - \frac{1}{2}(\angle A)
     \]
     \[
     180^\circ - \angle A = 180^\circ - \frac{1}{2}(\angle A)
     \]
     \[
     \angle A = 60^\circ
     \]

6. **Determine \(\angle BOC\):**
   - Since \(\angle A = 60^\circ\), and \(\angle A\) is an inscribed angle of \(\angle BOC\), the central angle \(\angle BOC\) is twice the inscribed angle.
   - Therefore, \(\angle BOC = 2 \times 60^\circ = 120^\circ\).

The final answer is \(\boxed{120^\circ}\).