Problem 1490

It is useful for this task to analyze what happens when you slide points. You can, for example, slide line segment EF without affecting $K$. Therefore, $L$ can move independently of $K$. Apparently, $L$ always moves along a line through D that is perpendicular to KD. The angle $\angle L D S$ is apparently independent of the position of line segment EF. Since we are dealing with a circle, this suggests that $\angle L D S$ might be equal to a certain inscribed angle. Then $\angle K D S$ would also be equal to an inscribed angle, and these two inscribed angles together would be $90^{\circ}$. In this way, you can get ideas to ultimately arrive at the following proof.
The configuration is fixed by the given order of points $A$ through $F$ on the circle and by the given positions of points $P$ and $Q$ on line $A D$. Therefore, we do not need to distinguish between different configurations. (The points $P$ and $Q$ can be swapped on line $A D$, but this is not relevant.)
First, note that $S$ is the center of the circle, since $\angle A B D=90^{\circ}$, line $A D$ is a diameter, and $S$ is the midpoint of $A D$. We will now prove that $\angle K D S=\angle K B S$. We know that $K S$ is a bisector, so $\angle B S K=\angle K S D$. Furthermore, $|S D|=|S B|$, since both lengths are equal to the radius of the circle. Since also $|S K|=|S K|$, we find $\triangle D S K \cong \triangle B S K$ (SAS). Therefore, $\angle K D S=\angle K B S$.
Since $B K$ is also a bisector, $\angle K B S=\angle C B K=\frac{1}{2} \angle C B S$. Thus, $\angle K D S=$ $\frac{1}{2} \angle C B S$.
Analogously, we derive that $\angle L D S=\frac{1}{2} \angle Q E S$, which implies that $\angle L D S=\frac{1}{2} \left(180^{\circ}-\angle F E S\right)=90^{\circ}-\frac{1}{2} \angle F E S$. Therefore, $\angle L D S+\angle K D S=90^{\circ}-\frac{1}{2} \angle F E S+\frac{1}{2} \angle C B S$. Furthermore, note that $\triangle S B C \cong \triangle S E F$ (SSS), so $\angle F E S=\angle C B S$. Now we can conclude that $\angle K D L=\angle L D S+\angle K D S=90^{\circ}$.