Let $k$ be a given circle and $A$ is a fixed point outside $k$. $BC$ is a diameter of $k$. Find the locus of the orthocentre of $\triangle ABC$ when $BC$ varies.

[i]Proposed by T. Vitanov, E. Kolev[/i]
1. **Define the problem and setup:**
   - Let \( k \) be a given circle with center \( O \) and radius \( R \).
   - Let \( A \) be a fixed point outside \( k \).
   - Let \( BC \) be a diameter of \( k \).
   - We need to find the locus of the orthocenter \( H \) of \( \triangle ABC \) as \( BC \) varies.

2. **Identify key points and properties:**
   - Since \( BC \) is a diameter of \( k \), \( \angle BOC = 180^\circ \).
   - The orthocenter \( H \) of \( \triangle ABC \) is the intersection of the altitudes of the triangle.
   - Let \( D \) and \( E \) be the feet of the perpendiculars from \( B \) and \( C \) to \( AC \) and \( AB \), respectively.
   - Let \( F \) and \( G \) be the points where the tangents from \( A \) touch the circle \( k \).

3. **Use trigonometric identities and properties of the circle:**
   - The power of a point theorem states that for a point \( P \) outside a circle, the power of \( P \) with respect to the circle is given by \( PA^2 - r^2 \), where \( PA \) is the distance from \( P \) to the center of the circle and \( r \) is the radius of the circle.
   - For the orthocenter \( H \), we have:
     \[
     BH \cdot HD = 4R^2 \cos \angle A \cos \angle B \cos \angle C
     \]
   - The power of \( A \) with respect to the circle \( (BCDE) \) is:
     \[
     AE \cdot AC = 4R^2 \sin \angle C \sin \angle B \cos \angle A = AF^2
     \]

4. **Calculate the power of \( H \) with respect to the circle \( (A, AF) \):**
   - The power of \( H \) with respect to the circle \( (A, AF) \) is:
     \[
     AF^2 - AH^2
     \]
   - Given \( AH = 2R \cos \angle A \), we have:
     \[
     AF^2 - AH^2 = BH \cdot HD
     \]

5. **Determine the locus of \( H \):**
   - Since \( H \) lies on the radical axis of the circles \( (BCDE) \) and \( (A, AF) \), and the radical axis is a fixed line \( GF \), the locus of \( H \) is this fixed line.

Conclusion:
The locus of the orthocenter \( H \) of \( \triangle ABC \) as \( BC \) varies is the line \( GF \).

The final answer is the line \( \boxed{ GF } \).