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SHARYGIN 2013 Problem 7

    Let BDBD be a bisector of triangle ABCABC. Points IaI_a, IcI_c are the incenters of triangles ABDABD, CBDCBD respectively. The line IaIcI_aI_c meets ACAC in point QQ. Prove that ∠DBQ=90∘\angle DBQ = 90^\circ.

    Official text Sharygin Geometry Olympiad 2013 official paper, page 2. Quoted with source.

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