In the triangle $ABC$ with sides $AC = b$ and $AB = c$, the  extension of the bisector of angle $A$ intersects it's circumcircle at point  with $W$. Circle $\omega$ with center at $W$ and radius $WA$ intersects lines $AC$ and $AB$ at points $D$ and $F$, respectively. Calculate the lengths of segments $CD$ and $BF$.

(Evgeny Svistunov)
[img]https://cdn.artofproblemsolving.com/attachments/7/e/3b340afc4b94649992eb2dccda50ca8f3f7d1d.png[/img]
1. **Identify Key Points and Properties:**
   - Given triangle \(ABC\) with sides \(AC = b\) and \(AB = c\).
   - The angle bisector of \(\angle A\) intersects the circumcircle of \(\triangle ABC\) at point \(W\).
   - Circle \(\omega\) with center \(W\) and radius \(WA\) intersects lines \(AC\) and \(AB\) at points \(D\) and \(F\), respectively.

2. **Use Angle Bisector Theorem:**
   - The angle bisector theorem states that the angle bisector of \(\angle A\) divides the opposite side \(BC\) into segments proportional to the adjacent sides:
     \[
     \frac{BD}{DC} = \frac{AB}{AC} = \frac{c}{b}
     \]

3. **Properties of Circle \(\omega\):**
   - Since \(W\) is on the circumcircle of \(\triangle ABC\) and \(WA\) is the radius of circle \(\omega\), \(W\) is equidistant from points \(D\) and \(F\).
   - This implies that \(WD = WA = WF\).

4. **Symmetry and Isosceles Triangle:**
   - The triangles \(BWE\) and \(CWE\) are isosceles with \(WB = WE\) and \(WC = WE\) because \(W\) lies on the circumcircle of \(\triangle ABC\).
   - Since \(W\) is the midpoint of the arc \(BC\) not containing \(A\), \(W\) is equidistant from \(B\) and \(C\).

5. **Calculate Lengths of Segments \(CD\) and \(BF\):**
   - Since \(D\) and \(F\) lie on the circle \(\omega\) centered at \(W\) with radius \(WA\), and \(W\) is the midpoint of the arc \(BC\), the segments \(CD\) and \(BF\) are equal to the segments \(AC\) and \(AB\) respectively.
   - Therefore, \(CD = AC - AD\) and \(BF = AB - AF\).

6. **Using the Angle Bisector Theorem:**
   - Let \(AD = x\) and \(AF = y\).
   - By the angle bisector theorem:
     \[
     \frac{BD}{DC} = \frac{c}{b} \implies BD = \frac{c}{c+b} \cdot BC \quad \text{and} \quad DC = \frac{b}{c+b} \cdot BC
     \]
   - Since \(D\) and \(F\) are on the circle \(\omega\), \(AD = AF = WA\).

7. **Final Calculation:**
   - Since \(D\) and \(F\) are on the circle \(\omega\) and \(W\) is the midpoint of the arc \(BC\), the segments \(CD\) and \(BF\) are:
     \[
     CD = AC - AD = b - x \quad \text{and} \quad BF = AB - AF = c - y
     \]
   - Given \(AD = AF = WA\), we have:
     \[
     CD = b - WA \quad \text{and} \quad BF = c - WA
     \]

The final answer is \( \boxed{ CD = b - WA } \) and \(BF = c - WA\).