### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

 What if applet does not run?

A few words

I am grateful to Professor W. McWorter for bringing this problem to my attention. The problem has been posted to the sci.math newsgroup on November 17, 2005. It was found in an old geometry book:

 Chords AB and CD are parallel, whereas P and Q are additional points on the same circle. Let X be the intersection of BP and CQ, Y the intersection of AQ and DP. Prove that XY is parallel to AB and CD.

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

 What if applet does not run?

An elegant proof was posted the same day:

∠AQC = ∠BPD, as inscribed angles subtending by equal arcs AC and BD. (The latter are enclosed by parallel chords.) This makes the quadrilateral PXYQ cyclic. In its circumcircle, the angles XYP and XQP subtend the same chord XP and are therefore equal. On the other hand, for the same reason but in the given circle angles CQP and CDP are also equal. By transitivity, ∠CDP = ∠XYP. It follows that XY||CD.

I wish to note here that, nice as the problem is, it is only a special case of Pascal's theorem. The points on the given circle form a hexagon, ABPDCQ, whose opposite sides, in general, intersect in points Z, X and Y:

 Z = AB∩CD, X = BP∩CQ, Y = AQ∩DP.

By Pascal's theorem the three points are collinear. This is true even if one of them lies at infinity.