Asian Pacific Mathematics Olympiad

Let the line $EF$ intersect the line $BC$ at $P$ and the circumcircle of $\triangle ACD$ at $G$ distinct from $F$. We will prove that $P$ is the fixed point.
First, notice that $\triangle BED$ is isosceles with $EB = ED$. This implies $\angle EBC = \angle EDP$.
Then, $\angle DAG = \angle DFG = \angle EBC = \angle EDP$ which implies $AG \parallel DC$. Hence, $AGCD$ is an isosceles trapezoid.
Also, $AG \parallel DC$ and $AE = ED$. This implies $\triangle AEG \cong \triangle DEP$ and $AG = DP$.
Since $B$ is the foot of the perpendicular from $A$ onto the side $CD$ of the isosceles trapezoid $AGCD$, we have $PB = PD + DB = AG + DB = BC$, which does not depend on the choice of $D$. Hence, the initial statement is proven.

Set up a coordinate system where $BC$ is along the positive $x$-axis, $BA$ is along the positive $y$-axis, and $B$ is the origin. Take $A = (0, a)$, $B = (0, 0)$, $C = (c, 0)$, $D = (-d, 0)$ where $a, b, c, d > 0$. Then $E = \left(-\frac{d}{2}, \frac{a}{2}\right)$. The general equation of a circle is
$$x^2 + y^2 + 2fx + 2gy + h = 0$$
Substituting the coordinates of $A, D, C$ into (1) and solving for $f, g, h$, we find that the equation of the circumcircle of $\triangle ADC$ is
$$x^2 + y^2 + (d - c)x + \left(\frac{cd}{a} - a\right)y - cd = 0.$$
Similarly, the equation of the circumcircle of $\triangle BDE$ is
$$x^2 + y^2 + dx + \left(\frac{d^2}{2a} - \frac{a}{2}\right)y = 0$$
Then (3)-(2) gives the equation of the line $DF$ which is
$$cx + \frac{a^2 + d^2 - 2cd}{2a}y + cd = 0.$$
Solving (3) and (4) simultaneously, we get
$$F = \left(\frac{c(d^2 - a^2 - 2cd)}{a^2 + (d - 2c)^2}, \frac{2ac(c - d)}{a^2 + (d - 2c)^2}\right),$$
and the other solution $D = (-d, 0)$.
From this we obtain the equation of the line $EF$ which is $ax + (d - 2c)y + ac = 0$. It passes through $P(-c, 0)$ which is independent of $d$.