Problem 1217

Proof: Let $a=CE$,
$$\begin{array}{l} b=EA, c=AC, \\ A(1,0,0), \\ \quad C(0,1,0), E(0,0,1), B\left(x_{1}: y_{1}: z_{1}\right), \\ \quad D\left(x_{2}: y_{2}: z_{2}\right), F\left(x_{3}: y_{3}: z_{3}\right). \end{array}$$

Since the circle $\Gamma$ is the circumcircle of $\triangle ACE$, its equation is
$$-a^{2} y z - b^{2} z x - c^{2} x y = 0.$$

Since points $B$, $D$, and $F$ lie on the circle $\Gamma$, we have
$$-a^{2} y_{i} z_{i} - b^{2} z_{i} x_{i} - c^{2} x_{i} y_{i} = 0 \quad (i=1,2,3).$$

Also, $l_{AB}: z_{1} y - y_{1} z = 0$, $l_{ED}: y_{2} x - x_{2} y = 0$, so the intersection of $AB$ and $ED$ is the point $P\left(\frac{x_{2}}{y_{2}}: 1: \frac{z_{1}}{y_{1}}\right)$.
Similarly, the intersection of $CD$ and $AF$ is the point $Q\left(\frac{x_{2}}{z_{2}}: \frac{y_{3}}{z_{3}}: 1\right)$,
and the intersection of $EF$ and $CB$ is the point $R\left(1: \frac{y_{3}}{x_{3}}: \frac{z_{1}}{x_{1}}\right)$.
$$\begin{array}{l} \text{By } -a^{2} y_{i} z_{i} - b^{2} z_{i} x_{i} - c^{2} x_{i} y_{i} = 0 \\ \Rightarrow a^{2} \cdot \frac{1}{x_{i}} + b^{2} \cdot \frac{1}{y_{i}} + c^{2} \cdot \frac{1}{z_{i}} = 0 \end{array}$$
$\Rightarrow$ the system of linear equations in $x$, $y$, and $z$
$$\left\{\begin{array}{l} x \cdot \frac{1}{x_{1}} + y \cdot \frac{1}{y_{1}} + z \cdot \frac{1}{z_{1}} = 0, \\ x \cdot \frac{1}{x_{2}} + y \cdot \frac{1}{y_{2}} + z \cdot \frac{1}{z_{2}} = 0, \\ x \cdot \frac{1}{x_{3}} + y \cdot \frac{1}{y_{3}} + z \cdot \frac{1}{z_{3}} = 0 \end{array}\right.$$

has a non-zero solution $\left(a^{2}, b^{2}, c^{2}\right)$
$$\Rightarrow \left|\begin{array}{ccc} \frac{1}{x_{1}} & \frac{1}{y_{1}} & \frac{1}{z_{1}} \\ \frac{1}{x_{2}} & \frac{1}{y_{2}} & \frac{1}{z_{2}} \\ \frac{1}{x_{3}} & \frac{1}{y_{3}} & \frac{1}{z_{3}} \end{array}\right| = 0 \Rightarrow \left|\begin{array}{ccc} 1 & \frac{y_{3}}{x_{3}} & \frac{z_{1}}{x_{1}} \\ \frac{x_{2}}{y_{2}} & 1 & \frac{z_{1}}{y_{1}} \\ \frac{x_{2}}{z_{2}} & \frac{y_{3}}{z_{3}} & 1 \end{array}\right| = 0$$
$\Rightarrow$ points $P$, $Q$, and $R$ are collinear.