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Show that $ca = cb$.

Source: IMO ShortList 1990, Problem 12 (IRE 1)

November 2, 2005
geometryincentertrigonometryangle bisectorIMO Shortlist

Problem Statement

Let ABC ABC be a triangle, and let the angle bisectors of its angles CAB CAB and ABC ABC meet the sides BC BC and CA CA at the points D D and F F, respectively. The lines AD AD and BF BF meet the line through the point C C parallel to AB AB at the points E E and G G respectively, and we have FG \equal{} DE. Prove that CA \equal{} CB. Original formulation: Let ABC ABC be a triangle and L L the line through C C parallel to the side AB. AB. Let the internal bisector of the angle at A A meet the side BC BC at D D and the line L L at E E and let the internal bisector of the angle at B B meet the side AC AC at F F and the line L L at G. G. If GF \equal{} DE, prove that AC \equal{} BC.
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