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.