A rectangle $ABCD$ and its circumcircle are given. Let $E$ be an arbitrary point on the minor arc $BC$. The tangent to the circle at $B$ meets $CE$ at point $G$. The segments $AE$ and $BD$ meet at point $K$. Prove that $GK$ and $AD$ are perpendicular.
Just a straighforward application of Pascal's theorem on the hexagon \(BBDCEA\), which implies that \(GK\) is collinear with \(DC \cap BA\), i.e. the point at infinity on \(BA\), which implies that \(GK || BA \perp DA\) which proves the problem.
Complex trivialises this problem.
Here is the solution I found after submitting the solution to the jury :
$BD$ is the diameter, thus $GB\perp BD$. Also $AE\perp EC$, thus $BGEK$ is cyclic. Thus, \[\angle EKG=\angle EBG=\angle EAB\implies KG||AB.\]Proved