MathDB
AL = AD iff <KCE = <ALE , starting with a cyclic ABCD

Source: KJMO 2012 p5

May 3, 2019
geometryequal anglesequal segmentstangentcyclic quadrilateral

Problem Statement

Let ABCDABCD be a cyclic quadrilateral inscirbed in a circle OO (AB>ADAB> AD), and let EE be a point on segment ABAB such that AE=ADAE = AD. Let AC∩DE=FAC \cap DE = F, and DE∩O=K(≠D)DE \cap O = K(\ne D). The tangent to the circle passing through C,F,EC,F,E at EE hits AKAK at LL. Prove that AL=ADAL = AD if and only if ∠KCE=∠ALE\angle KCE = \angle ALE.
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