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convex ABCD, AB=CB, <ABC +2<CDA = \pi, AE=EC, prove <CDE=<BDA

Source: IMAR 2009 p3

September 26, 2018
geometryAngle Chasingconvex quadrilateralanglesequal angles

Problem Statement

Consider a convex quadrilateral ABCDABCD with AB=CBAB=CB and ∠ABC+2∠CDA=π\angle ABC +2 \angle CDA = \pi and let EE be the midpoint of ACAC. Show that ∠CDE=∠BDA\angle CDE =\angle BDA.
Paolo Leonetti
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