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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 2 of 6: Central angle
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∠BOC=2∠BAC\angle BOC=2\angle BAC
Detailed analysis

The central angle ∠BOC\angle BOC and inscribed angle ∠BAC\angle BAC subtend chord BCBC, so ∠BOC=2∠BAC\angle BOC=2\angle BAC.