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Prove that AE = EC + CB

Source: INAMO 2002 - Problem 4

August 5, 2011
geometrycircumcircletrigonometryangle bisectorgeometry unsolved

Problem Statement

Given a triangle ABCABC where AC>BCAC > BC, DD is located on the circumcircle of ABCABC such that DD is the midpoint of the arc ABAB that contains CC. EE is a point on ACAC such that DEDE is perpendicular to ACAC. Prove that AE=EC+CBAE = EC + CB.
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