Cyclic quadrilateral angle bisectors meet the circle\nABCD is a quadrilateral inscribed in a circle of radius r. The internal bisectors of ∠DAB and ∠BCD meet the circumcircle again at X and Y, respectively. Find the length of XY.

Introduction / Context:
This is a well-known geometry fact for a cyclic quadrilateral: the internal angle bisectors at opposite vertices (A and C) meet the circumcircle again at diametrically opposite points. Consequently, the chord joining those second intersection points is a diameter of the circle, whose length equals 2r, where r is the circumradius.

Given Data / Assumptions:

• ABCD cyclic with circumradius r.
• The internal angle bisectors of ∠DAB and ∠BCD intersect the circumcircle again at X and Y.
• X and Y are antipodal (diametrically opposite) points.

Concept / Approach:
In a circle, the locus of points subtending equal angles at chord endpoints ties to angle-bisector properties. In a cyclic quadrilateral, the bisectors of ∠A and ∠C meet the circumcircle at endpoints of a diameter (this can be reasoned using directed arcs or properties of supplementary arcs in opposite angles). Hence XY is a diameter.

Step-by-Step Solution:

Verification / Alternative check:
Constructive proofs via inscribed angle theorem or using equal subtended angles at the circumference also arrive at XY as a diameter directly.

Why Other Options Are Wrong:
π(r r) and π(r r)/2 resemble area-like forms, not a length; (r + 2) is dimensionally inconsistent. Only 2r is a length equal to the diameter.

Common Pitfalls:
Treating the bisectors as intersecting at arbitrary points on the circle leads to overlooking the supplementary relationship of opposite angles that forces a diameter.

Final Answer:
2r