IMO 2002 SL G3

Origin: KOR | Category: Geometry

Problem

The circle S has center O, and BC is a diameter of S. Let A be a point of S such that ∡AOB < 120◦. Let D be the midpoint of the arc AB that does not contain C. The line through O parallel to DA meets the line AC at I. The perpendicular bisector of OA meets S at E and at F. Prove that I is the incenter of the triangle CEF.

Solution

Since \angleBCA = 1 2\angleBOA = \angleBOD, the lines CA and OD are parallel, so that ODAI is a parallelogram. It follows that AI = OD = OE = AE = AF. Hence

\angleIFE = \angleIFA−\angleEFA = \angleAIF −\angleECA = \angleAIF −\angleACF = \angleCFI. Also, from AE = AF we get that CI bisects \angleECF. Therefore I is the incenter of \triangleCEF.