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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 4 of 6: First parallelism
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QN∥ABQN\parallel AB
Detailed analysis

Because A,Q,CA,Q,C are collinear and ∠NQC=∠BAC\angle NQC=\angle BAC, the line QNQN makes with ACAC the same angle that ABAB makes with ACAC. Thus QN∥ABQN\parallel AB, hence QN∥APQN\parallel AP.