Problem 1475

1. Given Conditions and Notations:
- Points $A', B', C'$ lie on sides $BC, CA, AB$ of triangle $ABC$.
- For a point $X$, we have:
$$\angle AXB = \angle A'C'B' + \angle ACB$$
$$\angle BXC = \angle B'A'C' + \angle BAC$$

2. Objective:
- Prove that the quadrilateral $XA'BC'$ is cyclic.

3. **Introduction of Point $P$:**
- Let $P$ be the intersection of the circumcircles of triangles $AB'C', A'BC', A'B'C$.

4. Angle Chasing:
- Consider the circumcircle of $\triangle AB'C'$. By the properties of cyclic quadrilaterals, we have:
$$\angle APB = \angle ACB + \angle PAC + \angle PBC$$
- Since $P$ lies on the circumcircle of $\triangle A'B'C$, we can write:
$$\angle PAC = \angle PC'B'$$
$$\angle PBC = \angle PC'A'$$
- Therefore:
$$\angle APB = \angle ACB + \angle PC'B' + \angle PC'A'$$
- Given that $\angle AXB = \angle A'C'B' + \angle ACB$, we can substitute:
$$\angle APB = \angle ACB + \angle A'C'B'$$
- This implies:
$$\angle AXB = \angle APB$$

5. **Verification for $\angle BXC$:**
- Similarly, consider the circumcircle of $\triangle A'BC'$. By the properties of cyclic quadrilaterals, we have:
$$\angle BPC = \angle BAC + \angle B'A'C'$$
- Given that $\angle BXC = \angle B'A'C' + \angle BAC$, we can substitute:
$$\angle BPC = \angle BAC + \angle B'A'C'$$
- This implies:
$$\angle BXC = \angle BPC$$

6. Conclusion:
- Since $\angle AXB = \angle APB$ and $\angle BXC = \angle BPC$, point $P$ coincides with point $X$.
- Therefore, quadrilateral $XA'BC'$ is cyclic because $X$ lies on the circumcircle of $\triangle A'BC'$.

$\blacksquare$