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Angle PDE

Source: Serbia MO 2018 P1

April 2, 2018
geometryincentermoving pointsSerbiaTriangle

Problem Statement

Let △ABC\triangle ABC be a triangle with incenter II. Points PP and QQ are chosen on segmets BIBI and CICI such that 2∠PAQ=∠BAC2\angle PAQ=\angle BAC. If DD is the touch point of incircle and side BCBC prove that ∠PDQ=90\angle PDQ=90.
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