Problem 1539

1. Identify the given points and ratios:
- Point $X$ on side $AB$ such that $\frac{AX}{AB} = \frac{4}{5}$.
- Point $Y$ on segment $CX$ such that $\frac{CY}{YX} = 2$.
- Point $Z$ on the ray $CA$ such that $\angle CXZ = 180^\circ - \angle ABC$.

2. **Determine the coordinates of point $X$:**
- Let $A = (0, 0)$ and $B = (b, 0)$.
- Since $\frac{AX}{AB} = \frac{4}{5}$, the coordinates of $X$ are $X = \left(\frac{4b}{5}, 0\right)$.

3. **Determine the coordinates of point $Y$:**
- Let $C = (c_x, c_y)$.
- The line segment $CX$ can be parameterized as $C + t(X - C)$ for $t \in [0, 1]$.
- Since $\frac{CY}{YX} = 2$, we have $CY = 2YX$, implying $Y$ divides $CX$ in the ratio $2:1$.
- Using the section formula, the coordinates of $Y$ are:
$$Y = \left( \frac{2 \cdot \frac{4b}{5} + c_x}{3}, \frac{2 \cdot 0 + c_y}{3} \right) = \left( \frac{8b + 5c_x}{15}, \frac{c_y}{3} \right)$$

4. **Determine the coordinates of point $Z$:**
- Point $Z$ lies on the ray $CA$ such that $\angle CXZ = 180^\circ - \angle ABC$.
- This implies that $Z$ is on the extension of $CA$ past $A$.

5. **Calculate the angle $\angle XYZ$:**
- Given $\angle XYZ = 45^\circ$, we need to use the properties of the triangle and the given conditions to find the relationship between the angles of the triangle.

6. Prove similarity of triangles:
- To prove that all triangles in $\Sigma$ are similar, we need to show that the angles of the triangles are the same.
- Since $\angle XYZ = 45^\circ$ and the other conditions are met, we can use the properties of similar triangles and angle chasing to show that the triangles are similar.

7. Find the measure of the smallest angle:
- Let $\angle A = \alpha$, $\angle B = \beta$, and $\angle C = \gamma$.
- Using the given conditions and the fact that the sum of the angles in a triangle is $180^\circ$, we can find the measure of the smallest angle.

8. Conclusion:
- By proving the similarity of the triangles and using the given conditions, we can determine the measure of the smallest angle.

The final answer is $\boxed{15^\circ}$