Problem 1

Let

ABC

be a triangle with

\angle BAC\ne90^\circ

. Let

O

be the circumcenter of triangle

ABC

and let

\Gamma

be the circumcircle of triangle

BOC

. Suppose that

\Gamma

intersects segment

AB

P\ne B

and segment

AC

Q\ne C

. Let

ON

be a diameter of

\Gamma

. Prove that quadrilateral

APNQ

is a parallelogram.

Step 5 of 6: Second parallelism

PN\parallel AC

Detailed analysis

The identical cyclic-angle argument with $B$ and $C$ interchanged gives $PN\parallel AC$ . Since $A,Q,C$ are collinear, $PN\parallel AQ$ .