Let B,C be fixed points and A be roving point. Let H, G be orthecentre and centroid of triagle ABC. Known midpoint of HG lies on BC, find locus of A
1. **Set up the coordinate system:**
   - Place the midpoint of \(BC\) at the origin \(O(0,0)\).
   - Let \(B(-a,0)\) and \(C(a,0)\).
   - Let \(A(x,y)\) be the roving point.

2. **Find the centroid \(G\) of \(\triangle ABC\):**
   - The centroid \(G\) is given by:
     \[
     G\left(\frac{x- a + a}{3}, \frac{y + 0 + 0}{3}\right) = \left(\frac{x}{3}, \frac{y}{3}\right)
     \]

3. **Let the midpoint of \(GH\) be \(M\) on \(BC\):**
   - Since \(M\) lies on \(BC\), \(M\) has coordinates \((m,0)\).

4. **Express the orthocenter \(H\) in terms of \(M\) and \(G\):**
   - The midpoint of \(GH\) is \(M\), so:
     \[
     M = \left(\frac{\frac{x}{3} + H_x}{2}, \frac{\frac{y}{3} + H_y}{2}\right)
     \]
   - Given \(M = (m,0)\), we have:
     \[
     m = \frac{\frac{x}{3} + H_x}{2} \quad \text{and} \quad 0 = \frac{\frac{y}{3} + H_y}{2}
     \]
   - Solving these, we get:
     \[
     H_x = 2m - \frac{x}{3} \quad \text{and} \quad H_y = -\frac{y}{3}
     \]

5. **Use the condition \(AH \perp BC\):**
   - Since \(AH \perp BC\), the x-coordinate of \(A\) must satisfy:
     \[
     x = 2m - \frac{x}{3}
     \]
   - Solving for \(x\):
     \[
     x + \frac{x}{3} = 2m \implies \frac{4x}{3} = 2m \implies x = \frac{3m}{2}
     \]

6. **Use the condition \(BH \perp AC\):**
   - The vector \(\vec{BH}\) is:
     \[
     \vec{BH} = \left(2m - \frac{x}{3} + a, -\frac{y}{3}\right)
     \]
   - The vector \(\vec{AC}\) is:
     \[
     \vec{AC} = (x - a, y)
     \]
   - Since \(\vec{BH} \cdot \vec{AC} = 0\):
     \[
     \left(2m - \frac{x}{3} + a\right)(x - a) + \left(-\frac{y}{3}\right)y = 0
     \]
   - Substituting \(x = \frac{3m}{2}\):
     \[
     \left(2m - \frac{\frac{3m}{2}}{3} + a\right)\left(\frac{3m}{2} - a\right) - \frac{y^2}{3} = 0
     \]
     \[
     \left(2m - \frac{m}{2} + a\right)\left(\frac{3m}{2} - a\right) - \frac{y^2}{3} = 0
     \]
     \[
     \left(\frac{3m}{2} + a\right)\left(\frac{3m}{2} - a\right) - \frac{y^2}{3} = 0
     \]
     \[
     \left(\frac{9m^2}{4} - a^2\right) - \frac{y^2}{3} = 0
     \]
     \[
     \frac{9m^2}{4} - a^2 = \frac{y^2}{3}
     \]
     \[
     9m^2 - 4a^2 = \frac{4y^2}{3}
     \]
     \[
     27m^2 - 12a^2 = 4y^2
     \]
     \[
     27m^2 - 4y^2 = 12a^2
     \]
     \[
     \frac{27m^2}{12} - \frac{4y^2}{12} = a^2
     \]
     \[
     \frac{9m^2}{4} - \frac{y^2}{3} = a^2
     \]
     \[
     x^2 - \frac{y^2}{3} = a^2
     \]

Thus, the locus of \(A\) is a hyperbola given by the equation:
\[
x^2 - \frac{y^2}{3} = a^2
\]

The final answer is \(\boxed{x^2 - \frac{y^2}{3} = a^2}\)