In an acute triangle $ABC$ , the bisector  of angle $\angle A$ intersects the circumscribed circle of the triangle $ABC$  at the point $W$. From  point $W$ , a parallel is drawn to the side $AB$, which intersects this circle at the point $F \ne W$. Describe the construction of the triangle $ABC$, if given are  the segments $FA$  , $FW$ and $\angle FAC$.

(Andrey Mostovy)
1. **Claim**: \(FW = AC\)

   To prove this, we need to show that the segments \(FW\) and \(AC\) subtend equal angles at the points on the circle. 

   - Since \(W\) is the intersection of the angle bisector of \(\angle A\) with the circumcircle, \(\angle BAW = \angle CAW\).
   - Given that \(WF \parallel AB\), we have \(\angle WFA = \angle BAW\).
   - Therefore, \(\angle FAW = \angle FAB - \angle BAW = \angle ABW - \angle WBC = \angle B\).

   Since \(ABWF\) forms an isosceles trapezium (because \(WF \parallel AB\) and \(W\) is on the circumcircle), we have \(\angle B = \angle FAW\). This implies that \(FW\) and \(AC\) subtend equal angles at \(A\) and \(C\) respectively, thus \(FW = AC\).

2. **Constructing Triangle \(FAC\)**:
   - We are given \(FA\), \(FW\), and \(\angle FAC\).
   - Using the Side-Angle-Side (SAS) criterion, we can construct triangle \(FAC\).

3. **Drawing the Circumcircle of \(\triangle FAC\)**:
   - Once \(\triangle FAC\) is constructed, we draw its circumcircle.

4. **Locating Point \(W\)**:
   - Since \(CW = BW = AF\), we can locate point \(W\) on the circumcircle of \(\triangle FAC\) such that \(CW = AF\).

5. **Finding Point \(B\)**:
   - With \(W\) located, we can determine point \(B\) as the intersection of the circumcircle of \(\triangle FAC\) and the line through \(W\) parallel to \(AC\).

Thus, the construction of triangle \(ABC\) is complete.

The final answer is the construction of triangle \( \boxed{ ABC } \) as described.