IMO 1994 SL G1

Origin: FRA | Category: Geometry

Problem

A semicircle \Gamma is drawn on one side of a straight line l. C and D are points on \Gamma. The tangents to \Gamma at C and D meet l at B and A respectively, with the center of the semicircle between them. Let E be the point of intersection of AC and BD, and F the point on l such that EF is perpendicular to l. Prove that EF bisects \angleCFD.

Solution

Extend AD and BC to meet at P, and let Q be the foot of the perpendic- ular from P to AB. Denote by O the center of \Gamma. Since \trianglePAQ ∼\triangleOAD and \trianglePBQ ∼\triangleOBC, we obtain AQ AD = PQ OD = PQ OC = BQ BC . Therefore AQ QB \cdot BC CP \cdot PD DA = 1, so by the converse Ceva theorem, AC, BD, and PQ are concurrent. It follows that Q \equivF. Finally, since the points O, C, P, D, F are concyclic, we have \angleDFP = \angleDOP = \anglePOC = \anglePFC.