A quadrilateral ABCD has all four of its vertices on one circle, named in order round it. The internal bisector of ∠DAB meets the circle again at P, and the internal bisector of ∠BCD meets the circle again at Q. What can be said about the chord PQ?

A bisector of an inscribed angle always meets the far arc at its midpoint, so P is the midpoint of arc BCD and Q is the midpoint of arc DAB. Travelling from P to Q the short way, through D, the journey is arc PD + arc DQ = ½(arc BCD) + ½(arc DAB) = ½ × 360° = 180°. A chord subtending a semicircle is a diameter, so PQ is a diameter. Nothing forces PQ to be parallel to BD, to bisect it, or to match its length — those all fail as soon as the quadrilateral is drawn lopsidedly.