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2017 Team #5: Collinearity with circumcenter

Source:

February 19, 2017
geometrycircumcircle

Problem Statement

Let ABCABC be an acute triangle. The altitudes BEBE and CFCF intersect at the orthocenter HH, and point OO denotes the circumcenter. Point PP is chosen so that ∠APH=∠OPE=90∘\angle APH = \angle OPE = 90^{\circ}, and point QQ is chosen so that ∠AQH=∠OQF=90∘\angle AQH = \angle OQF = 90^{\circ}. Lines EPEP and FQFQ meet at point TT. Prove that points AA, TT, OO are collinear.
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