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Ad > be

Source: Problem 1, Polish NO 1999

October 13, 2005
inequalitiestriangle inequalitygeometry solvedgeometry

Problem Statement

Let DD be a point on the side BCBC of a triangle ABCABC such that AD>BCAD > BC. Let EE be a point on the side ACAC such that AEEC=BDAD−BC\frac{AE}{EC} = \frac{BD}{AD-BC}. Show that AD>BEAD > BE.
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