In quadrilateral $ABCD$, a circle $\omega$ is inscribed. A point $K$ is chosen on diagonal $AC$. Segment $BK$ intersects $\omega$ at a unique point $X$, and segment $DK$ intersects $\omega$ at a unique point $Y$. It turns out that $XY$ is the diameter of $\omega$. Prove that it is perpendicular to $AC$.
Proposed by Tseren Frantsuzov
this might be a weird solution:
Claim. $K$ should be the tangent point of excircles of triangles $ABC$ and $ADC$ (tangents to the side $AC$).
Assume the contrary, we have two such points $K_1$ and $K_2$; they make up diameters $X_1Y_1$ and $X_2Y_2$, respectively. Use the directed angles to conclude that the segments $X_1Y_1$ and $X_2Y_2$ cannot intersect each other.
So, it suffices to show that the point $K$, that is the tangent point of excircles $ABC$ and $ADC$, satisfies the condition, i.e. for such $K$, the segment $XY$ be the diameter of $\omega$ (note that: applying Pitot's theorem, it's not difficult to see that the excircles tangent to the $AC$ at one point).
Using the homotheties (centred at $B$ and $C$), one can show that tangent lines to the circle $\omega$ at points $X$ and $Y$ are parallel to $AC$. Therefore, $XY$ should be a diameter of $\omega$, moreover, it is perpendicular to $AC$.