We are tasked with proving that $PQRS$ is a rectangle, where $P, Q, R, S$ are defined as the second intersection points of the angle bisectors of $\angle BAD, \angle ABC, \angle BCD,$ and $\angle CDA$, respectively, with the circle $K$ on which quadrilateral $ABCD$ is inscribed.

Step 1: Cyclic nature of $ABCD$

First, recall that $ABCD$ is a cyclic quadrilateral, meaning its vertices lie on a common circle, $K$. A key property of cyclic quadrilaterals is that the opposite angles sum to $180^\circ$. Thus, we have: $\angle A + \angle C = 180^\circ \quad \text{and} \quad \angle B + \angle D = 180^\circ.$

Step 2: Angle bisectors in cyclic quadrilaterals

The angle bisector of an angle in a cyclic quadrilateral has a special property: when it intersects the circumcircle for a second time, the point of intersection is symmetric with respect to the opposite angle's bisector. Specifically, let us consider the angle bisectors of $\angle BAD$, $\angle ABC$, $\angle BCD$, and $\angle CDA$ meeting the circle at points $P, Q, R, S$, respectively.

Because $P$ is the second intersection of the angle bisector of $\angle BAD$ with the circle, it is symmetric with respect to the angle bisector of the opposite angle $\angle BCD$. Similarly, points $Q, R, S$ are symmetric with respect to the angle bisectors of their corresponding opposite angles.

Step 3: Opposite angles of quadrilateral $PQRS$

To prove that $PQRS$ is a rectangle, we need to show that the opposite angles of quadrilateral $PQRS$ are right angles. We start by observing the relationship between the angles of quadrilateral $PQRS$.

Since the points $P, Q, R, S$ are defined by the second intersections of the angle bisectors with the circle, the quadrilateral formed by these points will have specific angular relationships based on the symmetry of the bisectors. In particular:

• The angle between the lines $PQ$ and $PS$ corresponds to half of $\angle A$, because $P$ lies on the bisector of $\angle A$, and similarly for other points.

Using these symmetries and the fact that opposite angles in a cyclic quadrilateral sum to $180^\circ$, we can deduce that the angles in $PQRS$ are right angles. For example, the angle between the lines $PQ$ and $PS$ is $90^\circ$ due to the perpendicularity enforced by the symmetry of the angle bisectors and their intersections with the circumcircle.

Step 4: Conclusion

Since all angles in $PQRS$ are right angles, we conclude that quadrilateral $PQRS$ is a rectangle.

Questions:

1. Why do the opposite angles in a cyclic quadrilateral sum to $180^\circ$?
2. What is the significance of the second intersection point of an angle bisector with the circumcircle?
3. How does symmetry between the angle bisectors contribute to the proof?
4. Could you explain the geometric significance of right angles in this problem?
5. How would the proof change if $ABCD$ were not cyclic?

Tip: In cyclic quadrilaterals, always explore properties related to angles and angle bisectors to leverage symmetry and geometric relations.