IMO Shortlist 1990 problem 12


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2. travnja 2012.
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Let ABC be a triangle, and let the angle bisectors of its angles CAB and ABC meet the sides BC and CA at the points D and F, respectively. The lines AD and BF meet the line through the point C parallel to AB at the points E and G respectively, and we have FG = DE. Prove that CA = CB.

Original formulation:

Let ABC be a triangle and L the line through C parallel to the side AB. Let the internal bisector of the angle at A meet the side BC at D and the line L at E and let the internal bisector of the angle at B meet the side AC at F and the line L at G. If GF = DE, prove that AC = BC.
Izvor: Međunarodna matematička olimpijada, shortlist 1990