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Geometry Difficulty 4.5 AIME Prove it Taiwan

Let OO be the circumcenter of acute triangle ABCABC, and let AOAO meet BCBC at point DD. From point DD, draw two rays perpendicular to ABAB and ACAC respectively, meeting ABAB, ACAC at EE, FF respectively, and meeting the circumcircle of ABCABC at KK, LL. Prove: if KK, EE, FF, LL are concyclic, then AB=ACAB = AC.

Solution

Since EKL=DFE=DAE\angle EKL = \angle DFE = \angle DAE, KLADKL \perp AD. Hence AOAO bisects KLKL, so KK, LL are symmetric with respect to AOAO. Hence ADK=ADL\angle ADK = \angle ADL, BAD=CAD\angle BAD = \angle CAD. Hence AB=ACAB = AC.

Figure 1

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from zh; metadata (topic, difficulty) added by this project.