IMO shortlist 1994/G1 solution

35th IMO 1994 shortlist

Problem G1

C and D are points on a semicircle. The tangent at C meets the extended diameter of the semicircle at B, and the tangent at D meets it at A, so that A and B are on opposite sides of the center. The lines AC and BD meet at E. F is the foot of the perpendicular from E to AB. Show that EF bisects angle CFD.

Solution

Let O be the center of the circle. Let the tangents meet at P. Let H be the foot of the perpendicular from P to AB. The key is that AC and BD intersect on AH.

PAH and OAD are similar, so AH/AD = HP/DO. PBH and OBC are similar, so BH/BC = HP/CO. But BO = CO, so BH/BC = AH/AD. Also PC = PD (equal tangents), so (AH/HB)(BC/CP)(PD/DA) = 1. Hence by Ceva's theorem applied to the triangle ABC, the lines AC, BD and PH are concurrent. So H = F. ∠PDO = ∠PFO = ∠PCO, so D, F and O lie on the circle diameter PO. So ∠DFP = ∠DOP = ∠COP = ∠CFP.

35th IMO shortlist 1994