A, B, C, D are four “consecutive" points on the circumference of a circle and P, Q, R, S are points on the circumference which are respectively the midpoints of the arcs AB, BC, CD, DA. Prove that PR is perpendicular to QS.

**Solution by Steve Dinh, a.k.a. Vo Duc Dien**

Let I be the intersection of PR and QS. We have ∠PSQ subtending arcs PB and BQ; ∠SPQ subtending arcs SD and DR.

But since arc PB = ˝ arc AB, arc BQ = ˝ arc BC, arc SD = ˝ arc AD and arc DR = ˝ arc DC,

or ∠PSQ and ∠SPR combined to cut ˝ the circle. Therefore,

∠PSQ + ∠SPR = 90° and ∠SIP = 180° - ∠PSQ - ∠SPR = 90°,

or PR is perpendicular to QS.