(V.Protasov, 9--10) The Euler line of a non-isosceles triangle is parallel to the bisector of one of its angles. Determine this
angle (There was an error in published condition of this problem).
1. Let \( O \), \( H \), and \( N \) be the circumcenter, orthocenter, and 9-point center of \( \triangle ABC \), respectively.
2. Assume that \( OH \) (the Euler line) is parallel to the internal bisector of \( \angle BAC \).
3. This implies that \( HO \) bisects \( \angle BHC \). This is because \( HB \), \( HC \), and the external bisector of \( \angle BAC \) form an isosceles triangle with apex \( H \).
4. Let \( U \) and \( V \) denote the midpoints of \( HC \) and \( HB \), respectively. The quadrilateral \( HUNV \) is cyclic, where \( N \) is the midpoint of the arc \( UV \) of its circumcircle.
5. By angle chasing, we need to show that \( \angle UNV + \angle BHC = 180^\circ \).

   Since \( HUNV \) is cyclic, we have:
   \[
   \angle UNV = 180^\circ - \angle UHV
   \]
   Also, since \( U \) and \( V \) are midpoints, \( \angle UHV = \angle BHC \).

6. Therefore:
   \[
   \angle UNV = 180^\circ - \angle BHC
   \]
   Adding \( \angle BHC \) to both sides:
   \[
   \angle UNV + \angle BHC = 180^\circ - \angle BHC + \angle BHC = 180^\circ
   \]

7. This implies:
   \[
   2\angle BHC = 180^\circ \implies \angle BHC = 60^\circ
   \]

8. Since \( \angle BHC = 180^\circ - \angle BAC \), we have:
   \[
   180^\circ - \angle BAC = 60^\circ \implies \angle BAC = 120^\circ
   \]

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