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Problem 1

Let ABCABC be a triangle with ∠BAC≠90∘\angle BAC\ne90^\circ. Let OO be the circumcenter of triangle ABCABC and let Γ\Gamma be the circumcircle of triangle BOCBOC. Suppose that Γ\Gamma intersects segment ABAB at P≠BP\ne B and segment ACAC at Q≠CQ\ne C. Let ONON be a diameter of Γ\Gamma. Prove that quadrilateral APNQAPNQ is a parallelogram.
Step 3 of 6: Cyclic angle at QQ
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∠NQC=∠NOC\angle NQC=\angle NOC
Detailed analysis

The points N,C,Q,ON,C,Q,O are concyclic on Γ\Gamma, so angles subtending chord NCNC are equal: ∠NQC=∠NOC=∠BAC\angle NQC=\angle NOC=\angle BAC.