Let O be the circumcenter of acute triangle ABC, and let AO meet BC at point D. From point D, draw two rays perpendicular to AB and AC respectively, meeting AB, AC at E, F respectively, and meeting the circumcircle of ABC at K, L. Prove: if K, E, F, L are concyclic, then AB=AC.
Solution
Since ∠EKL=∠DFE=∠DAE, KL⊥AD. Hence AO bisects KL, so K, L are symmetric with respect to AO. Hence ∠ADK=∠ADL, ∠BAD=∠CAD. Hence AB=AC.
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