Four. (20 points) Let the line $y=\sqrt{3} x+b$ intersect the parabola $y^{2}=2 p x(p>0)$ at points $A$ and $B$. The circle passing through $A$ and $B$ intersects the parabola $y^{2}=2 p x(p>0)$ at two other distinct points $C$ and $D$. Find the size of the angle between the lines $AB$ and $CD$.


Let points $A, B, C, D$ lie on a circle with the equation
$$
x^{2}+y^{2}+d x+e y+f=0 \text {, }
$$

Then the coordinates of $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), C\left(x_{3}, y_{3}\right), D\left(x_{4}, y_{4}\right)$ are the solutions to the system of equations
$$
\left\{\begin{array}{l}
x^{2}+y^{2}+d x+e y+f=0, \\
y^{2}=2 p x
\end{array}\right.
$$

Eliminating $x$ yields
$$
y^{4}+\left(4 p^{2}+2 p d\right) y^{2}+4 p^{2} e y+4 p^{2} f=0 .
$$

Since $y_{1}, y_{2}, y_{3}, y_{4}$ are the four roots of equation (1), by Vieta's formulas, we have
$$
y_{1}+y_{2}+y_{3}+y_{4}=0 .
$$

And $k_{A B}=\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\frac{y_{1}-y_{2}}{\frac{y_{1}^{2}}{2 p}-\frac{y_{2}^{2}}{2 p}}=\frac{2 p}{y_{1}+y_{2}}$.
Similarly, $k_{C D}=\frac{2 p}{y_{3}+y_{4}}$.
Thus, $k_{A B}+k_{C D}=\frac{2 p}{y_{1}+y_{2}}+\frac{2 p}{y_{3}+y_{4}}$
$$
=\frac{2 p\left(y_{1}+y_{2}+y_{3}+y_{4}\right)}{\left(y_{1}+y_{2}\right)\left(y_{3}+y_{4}\right)}=0 \text {. }
$$

Therefore, the slopes of lines $A B$ and $C D$ are complementary.
Given that the slope of line $A B$ is $\sqrt{3}$, then $\alpha=60^{\circ}, \beta=120^{\circ}$.
Hence, the angle $\theta$ between lines $A B$ and $C D$ is $\theta=\beta-\alpha=60^{\circ}$.