It is given an acute triangle  $ABC$ , $AB \neq AC$  where the feet of altitude from $A$ its $H$.  In the extensions of the sides $AB$ and $AC$  (in the direction of $B$ and $C$) we take the points $P$ and $Q$ respectively such that $HP=HQ$ and the points $B,C,P,Q$ are concyclic.
Find the ratio $\tfrac{HP}{HA}$.
1. **Identify the given elements and conditions:**
   - Triangle \(ABC\) is acute with \(AB \neq AC\).
   - \(H\) is the foot of the altitude from \(A\) to \(BC\).
   - Points \(P\) and \(Q\) are on the extensions of \(AB\) and \(AC\) respectively such that \(HP = HQ\).
   - Points \(B, C, P, Q\) are concyclic.

2. **Consider the circle with center \(H\) and radius \(HA\):**
   - Let this circle intersect \(AB\) at \(P'\) and \(AC\) at \(Q'\).
   - Since \(HP' = HQ' = HA\), points \(P'\) and \(Q'\) lie on the circle centered at \(H\) with radius \(HA\).

3. **Analyze the angles formed:**
   - \(\angle P'HQ' = 2\angle A\) because \(P'\) and \(Q'\) are on the circle centered at \(H\) with radius \(HA\).
   - \(\angle HP'B = 90^\circ - \angle B\) because \(H\) is the foot of the altitude from \(A\) to \(BC\).

4. **Determine the cyclic nature of quadrilateral \(BP'Q'C\):**
   - \(\angle BP'Q' = \angle BCA\) because \(\angle HP'B = 90^\circ - \angle B\) and \(\angle HP'Q' = 2\angle A\).
   - Since \(\angle BP'Q' = \angle BCA\), quadrilateral \(BP'Q'C\) is cyclic.

5. **Conclude the uniqueness and ratio:**
   - By the uniqueness of points \(P\) and \(Q\) (since \(AB \neq AC\)), we have \(P = P'\) and \(Q = Q'\).
   - Therefore, \(HP = HA\).

6. **Calculate the ratio \(\frac{HP}{HA}\):**
   - Since \(HP = HA\), the ratio \(\frac{HP}{HA} = 1\).

\(\blacksquare\)

The final answer is \(\boxed{1}\)