MathLabs

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 1 of 6: Use the diameter
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ON diameter of ΓON\text{ diameter of }\Gamma
Detailed analysis

Since ONON is a diameter, OB=OCOB=OC and triangles OBN,OCNOBN,OCN are congruent right triangles. Hence ∠NOC=12∠BOC=∠BAC\angle NOC=\frac12\angle BOC=\angle BAC.