Problem 1081

Answer: Through the fixed point $F\left(\frac{1}{4}, 0\right)$, i.e., the focus of $P$
Difficulty: Challenging
Assessment: Parabola, angle calculation, circumcircle and circumcenter
Analysis: Let the coordinates of $A, B$ be $\left(y_{1}^{2}, y_{1}\right),\left(y_{2}^{2}, y_{2}\right)$, then $A C: y_{\nu} y=\frac{y_{1}^{2}+x}{2}, B C: y_{2} y=\frac{y_{2}^{2}+x}{2}$ $\Rightarrow C\left(y_{1} y_{2}, \frac{y_{1}+y_{2}}{2}\right), k_{A C}=\frac{1}{2 y_{1}}, k_{B C}=\frac{1}{2 y_{2}},(5$ points $)$
Also, $k_{F C}=\frac{\frac{y_{1}+y_{2}}{2}}{y_{1} y_{2}-\frac{1}{4}}, k_{F B}=\frac{y_{2}}{y_{2}^{2}-\frac{1}{4}}$,
then $\tan \measuredangle F B C=\frac{k_{B C}-k_{F C}}{1+k_{B C} k_{F C}}=-\frac{1}{2 y_{2}}, \tan \measuredangle F C A=\frac{k_{A C}-k_{F C}}{1+k_{A C} k_{F C}}=-\frac{1}{2 y_{2}}$
$\Rightarrow \angle F B C=\angle F C A$, similarly $\angle F A C=\angle F C B$ (15 points)
Thus, $\angle A F B=\measuredangle C B F+\measuredangle A C B+\measuredangle F A C=\measuredangle A C F+\angle A C B+\measuredangle F C B=2 \angle A C B=\angle A D B$ indicating that the circumcircle of $\triangle A B D$ always passes through the fixed point $F\left(\frac{1}{4}, 0\right)$ (20 points)