Points $X$ and $Y$ are the midpoints of arcs $AB$ and $BC$ of the circumscribed circle of triangle $ABC$. Point $T$ lies on side $AC$. It turned out that the bisectors of the angles $ATB$ and $BTC$ pass through points $X$ and $Y$ respectively. What angle $B$ can be in  triangle $ABC$?
1. **Claim:** \( \angle B = 90^\circ \).

2. **Proof:**
   - Given points \(X\) and \(Y\) are the midpoints of arcs \(AB\) and \(BC\) of the circumscribed circle of triangle \(ABC\).
   - Point \(T\) lies on side \(AC\).
   - The bisectors of angles \(ATB\) and \(BTC\) pass through points \(X\) and \(Y\) respectively.

3. **Step-by-Step Proof:**
   - Since \(X\) is the midpoint of arc \(AB\) (not containing \(C\)), \(X\) lies on the angle bisector of \(\angle AOB\) where \(O\) is the center of the circumscribed circle.
   - Similarly, \(Y\) is the midpoint of arc \(BC\) (not containing \(A\)), so \(Y\) lies on the angle bisector of \(\angle BOC\).

4. **Angle Bisector Property:**
   - The angle bisector of \(\angle ATB\) passing through \(X\) implies that \(X\) is equidistant from \(A\) and \(B\).
   - The angle bisector of \(\angle BTC\) passing through \(Y\) implies that \(Y\) is equidistant from \(B\) and \(C\).

5. **Perpendicularity:**
   - Since \(X\) and \(Y\) are midpoints of arcs, they are also the points where the perpendicular bisectors of \(AB\) and \(BC\) intersect the circle.
   - Therefore, \(TX \perp AB\) and \(TY \perp BC\).

6. **Right Angle at \(T\):**
   - Since \(TX \perp AB\) and \(TY \perp BC\), and \(X\) and \(Y\) are on the angle bisectors, it follows that \(\angle XTY = 90^\circ\).

7. **Conclusion:**
   - Since \(TX \perp AB\) and \(TY \perp BC\), and \(\angle XTY = 90^\circ\), it implies that \(AB \perp BC\).
   - Therefore, \(\angle B = 90^\circ\).

\(\blacksquare\)

The final answer is \( \boxed{ 90^\circ } \).