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Angle ODQ=90

Source: Singapore MO 2011 open round 2 Q1

July 2, 2011
geometrycircumcirclegeometric transformationreflectiongeometry proposed

Problem Statement

In the acute-angled non-isosceles triangle ABCABC, OO is its circumcenter, HH is its orthocenter and AB>ACAB>AC. Let QQ be a point on ACAC such that the extension of HQHQ meets the extension of BCBC at the point PP. Suppose BD=DPBD=DP, where DD is the foot of the perpendicular from AA onto BCBC. Prove that ∠ODQ=90∘\angle ODQ=90^{\circ}.
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