Problem 1157

Solution: From the assumption that the circle $\Gamma$ intersects both of the line segments $A B$ and $A C$, it follows that the 4 points $N, C, Q, O$ are located on $\Gamma$ in the order of $N, C, Q, O$ or in the order of $N, C, O, Q$. The following argument for the proof of the assertion of the problem is valid in either case. Since $\angle N Q C$ and $\angle N O C$ are subtended by the same arc $\overparen{N C}$ of $\Gamma$ at the points $Q$ and $O$, respectively, on $\Gamma$, we have $\angle N Q C=\angle N O C$. We also have $\angle B O C=2 \angle B A C$, since $\angle B O C$ and $\angle B A C$ are subtended by the same arc $\overparen{B C}$ of the circum-circle of the triangle $A B C$ at the center $O$ of the circle and at the point $A$ on the circle, respectively. From $O B=O C$ and the fact that $O N$ is a diameter of $\Gamma$, it follows that the triangles $O B N$ and $O C N$ are congruent, and therefore we obtain $2 \angle N O C=\angle B O C$. Consequently, we have $\angle N Q C=\frac{1}{2} \angle B O C=\angle B A C$, which shows that the 2 lines $A P, Q N$ are parallel.

In the same manner, we can show that the 2 lines $A Q, P N$ are also parallel. Thus, the quadrilateral $A P N Q$ is a parallelogram.