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DE + EF + FD >= 1/2 (AB + BC + CA) if DC + CE = EA + AF = FB + BD

Source: Indonesia INAMO Shortlist 2008 G9

August 25, 2021
geometryGeometric Inequalitiesperimetersemiperimeter

Problem Statement

Given a triangle ABCABC, the points DD, EE, and FF lie on the sides BCBC, CACA, and ABAB, respectively, are such that DC+CE=EA+AF=FB+BD.DC + CE = EA + AF = FB + BD. Prove that DE+EF+FD≥12(AB+BC+CA).DE + EF + FD \ge \frac12 (AB + BC + CA).
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