By an easy angle chasing you get $ \angle PAQ$ = $ \angle PNQ$

$\angle OAB=\angle OBA=\angle OBP=\angle ONP$ (because $O$ is on the perpendicular bisector of $AB$ $\triangle ABC$ is isosceles) Similarly, $\angle OAQ=\angle ONQ$, so $\angle PNQ=\angle ONP+\angle ONQ=\angle OAP+\angle OAQ=\angle PAQ$. Also, since $O$ is on the perpendicular bisector of $BC$ and $N$ is diametrically opposite $O$, $N$ bisects $\overarc{BC}$, so $\angle BPN=\angle CQN$, so $\angle APN=\angle AQN$, and $APNQ$ is a parallelogram.

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.APMO2010Problem1 Vo Duc Dien

Dear Mathlinkers, 1. NB and NC are two tangents to (O). 2. By a special case of the Rem's theorem applied to these two circles, NB// AC and ND//AB. ...and we are done. Sincerely Jean-Louis

note that , $\angle BOC= 2A$ . so , $\angle OBC=\angle OCB= \pi/2-A$ . as $ON$ is a diameter , we have , $\angle OBN = \pi/2$ .hence $\angle CBN =A$ . so , $ \angle NQC= A$ . so , $AP$ is parallel to $QN$ . similarly , $PN$ is parallel to $AQ$ . hence done

Another fairly easy way that hasn't been mentioned yet WLOG $\angle{B} > \angle{C}$. Let $PO \cap AC = P'$. Since $OC = OB$ this means $\angle{P'PB} = \angle{CBO}$ because they subtend arcs of the same length. $\angle{CBO} = 90-\angle{A} = \angle{P'PB}$ and thus $\angle{PP'A} = 180-\angle{A}-\angle{CBO} = 90$. This implies $PN||AC$ Now let $AO \cap PC = A'$. Notice that $\angle{PQC} = \angle{PBC} = 180 - \angle{B}$ by cyclic quad BPQC. Then $\angle{AQA'} = \angle{B}$ and $\angle{A'AQ} = \angle{OAC} = \angle{B}$ and thus $AA'Q = 90$ which implies that $O$ is the orthocenter of $APQ$. Finally letting $QO \cap AP = Q'$ we have $QQ'$ is a height which implies $QN||AP$ from $\angle{OQ'P} = \angle{OQN}$

We have $$\angle APN = \angle APO + \angle OPN = \angle OCB + \angle OPN = 90-\angle A + 90 = 180-\angle A$$and similarly we have $$\angle AQN = 180-\angle A$$so we are done.

Let $D$ be the perpendicular from $O$ to $BC$, which lies on $ON$ by the definition of the circumcenter. Since $OB=OC$, we have that $\Delta OBC$ is isosceles. This means that $\angle OBC \cong \angle OCB$. From here we can conclude that $\angle DOB \cong \angle DOC$. This implies that $\overarc{BN}\cong \overarc{CN}$. Furthermore, we have that $\angle BPN\cong \angle NQC$. Finally, it follows that $\angle APN=180-\angle BPN=180-\angle NQC=\angle AQN$. Note that $\angle PAQ = \frac{\angle BOC}{2} =\angle DOC$. Note that $\angle BPC=2\angle DOC$ since $\overarc{BN}=\overarc{CN}$. We then have that $\angle PNQ=\angle BPC-\angle NQC=\angle BPC-\angle DOC=2\angle DOC-\angle DOC=\angle DOC$. Thus, $\angle PAQ\cong \angle PNQ$. Combining the above 2 facts gives the desired. Q.E.D.

simply angle chasing $\angle BOC= 2\angle BAC$ then $\angle OBC=\angle OCB=90-\angle BAC$. Since $ON$ is diameter $\angle BAC=\angle CBN=\angle CQN$ and $\angle CBN=\angle BCN=\angle BPN$ and so $APNQ$ is parallelogram.

$\angle BPN=\angle BON=\angle BAC \implies AQ \parallel PN$,$\angle AQN=180^{\circ}-\angle CON=180^{\circ}-\angle BAC \implies AP \parallel NQ$,hence $APNQ$ is a parallelogram.

Note that $\triangle OBC$ is isosceles, it follows that $N$ is the mid-point of arc $BC$ not containing $O$. Thus, $NB=NC$ This implies that $\angle NPB = \angle NQC$. Thus, $$ 180^{\circ}-\angle NPB = 180^{\circ}- \angle NQC \iff \angle APN = \angle AQN$$Now, it is trivial to observe that $\angle PNQ = \angle PAQ = \angle BAC$. Thus, $APNQ$ is a $\parallel ^{gm}$.

Woah... is this the easiest APMO Geo ever? Let $\angle BNO= \theta$. As $OB=OC$, $\angle ONC = \theta$. As $ON$ is diameter, $\angle BOC = 90 - \theta => \angle BPC = 90 - \theta$, and similarly, $\angle NQC = 90 - \theta$. As we know, $\angle OPN = \angle OQN = 90$, $\angle APO = \angle AQO = \theta => \angle APN =\angle AQN =90 + \theta$. As $\angle BAC = \frac12 \angle BOC = 90 - \theta$, we can conclude $PN \parallel AQ$ and $AP \parallel QN$, thus $APNQ$ is a parallelogram.

My Solution: $\angle BOC=2\angle A=\angle BQC\Rightarrow \angle BQC=2\angle A=\angle A+\angle QBA\Rightarrow \angle QBA=\angle A\Rightarrow \triangle QAB$ is isosceles.Next notice that:$\angle PNQ=\angle PNB+\angle BNQ=\angle BPQ+\angle BQP=180^{\circ}-\angle PBQ=\angle ABQ=\angle BAQ=\angle A\Rightarrow \angle PNQ=\angle BAQ$.$\angle BAC=\alpha$ , $\angle ABC=\beta$ , $\angle ACB=\gamma$.$\angle OBC=\angle OCB=\frac{180^{\circ}-2\alpha}{2}=90^{\circ}-\alpha$.$\angle QNO=\angle QCO=\gamma -(90^{\circ}-\alpha)=\alpha+\gamma-90^{\circ}$.$\angle BNO=\angle BCO=90^{\circ}-\alpha$.$\angle BNQ=\angle BNO+\angle QNO=\alpha+\gamma-90^{\circ}+90^{\circ}-\alpha=\gamma$.$\angle BQC=2\angle \alpha$ ; $\angle PQC =\angle PBC=180^{\circ}-\beta\Rightarrow \angle BQP=2\alpha -(180^{\circ}-\beta)=2\alpha +\beta-180^{\circ}$.$=\angle QNC=180^{\circ}-2\alpha-\gamma$.$\angle NQC=180^{\circ}-\angle QNC -\angle QCN=180^{\circ}-(180^{\circ}-2\alpha -\gamma)-(\alpha+\gamma)=\alpha$.$angle PQN =\angle BQC-\angle BQP=2\alpha-\alpha-(2\alpha+\beta-180^{\circ}=180^{\circ}-\alpha-\beta\Rightarrow \angle BNQ =\angle PNQ =180^{\circ}-\alpha -\beta=\gamma\Rightarrow \triangle APQ \sim \triangle NPQ$,but $PQ$ is the common side,so $\triangle APQ \cong \triangle NPQ\Rightarrow QN\parallel AP\Rightarrow$ $ANPQ$ is parallelogram. $Q.E.D$

Since $ON$ is perpendicular bisector of $BC$, $\angle BON = \angle NOC = \angle BAC$. Since $ONPB$ is cyclic, $\angle BPN = 180 - \angle BON = 180 - \angle BAC = 180 - \angle PAQ$ and so $PN || AQ$. Also, $\angle NQC = \angle NOC = \angle BAC = \angle PAQ$ and so $AP || QN$. Therefore, $ANPQ$ is a parallelogram

Goutham wrote: Let $ABC$ be a triangle with $\angle BAC \neq 90^{\circ}.$ Let $O$ be the circumcenter of the triangle $ABC$ and $\Gamma$ be the circumcircle of the triangle $BOC.$ Suppose that $\Gamma$ intersects the line segment $AB$ at $P$ different from $B$, and the line segment $AC$ at $Q$ different from $C.$ Let $ON$ be the diameter of the circle $\Gamma.$ Prove that the quadrilateral $APNQ$ is a parallelogram. Angle chase gooo brrrr. Since $NOQC$ cyclic and $O$ circumcenter. $$\angle BAC=\frac{\angle BOC}{2}=\angle NOC=\angle NQC \implies \angle PAQ=\angle NQC \implies AQ \parallel PN$$That also means that $PNCQ$ is an isosceles trapecium that gives $\widehat{CN}=\widehat{CP}$ and that means: $$\angle CQN=\angle QCP=\angle QPN=\angle PNQ \implies AP \parallel QN$$That means $APNQ$ parallelogram. Thus we are done

Woah!

Diagram for reference

imagine holding down shift when caps is on lmfao

Notice $$\measuredangle NPA=\measuredangle CQN=\measuredangle AQN$$and \begin{align*}\measuredangle PAQ&=\measuredangle BAO+\measuredangle OAC\\&=\measuredangle OBA+\measuredangle ACO\\&=\measuredangle ONB+\measuredangle QNO\\&=\measuredangle QNP.\end{align*}$\square$

$\angle BAC = \alpha \implies$ $\angle OBC = \angle OCB =90 - \alpha \implies \angle OQA = 90 -\alpha$ $\angle OQN = 90 \implies \angle NQA = 180 - \alpha$ $\implies$ $\angle PAQ+ \angle NQA = 180$ $PA$ and $QN$ parallel.$\blacksquare$

Let $\angle BAC=\alpha$. Note that since $\angle BOC=2\alpha$, $\angle BON=\angle CON=\alpha$. From cyclic $PONB$, $\angle BPN=\alpha$, and from cyclic $QONC$, $\angle CQN=\alpha$. Therefore, $AQ\parallel PN$ and $AP\parallel QN$, so we are done.

Solved on djmathman's 100 Geometry Problems. Remark: For some reason I couldn't even draw/use ggb to find a diagram where it intersected line SEGMENT, so instead I just used one where P,Q were both outside AB and AC, but made sure the angles still matched. After posting my solution I'll look around to see if there was an actual good diagram that had this. So make sure NOT to reference my diagram, it will confuse you a LOT. We have that <BOC=2A, hence the arc BNC measures 4A. By well known property of angles formed by secants outside of a triangle, we have that A=(4A-arcPQ)/2 (the diagram doesn't show that, but my remark about the config makes this true). This yields arc PQ=2A, or <PNQ=A. We also have that since O and the center of the circle T both lie on perp. bisector of BC, they bisect arc BC and hence <NQC=<NOC=<NOB=<NPB$\implies$<NQA=180-A=<NPA, as desired. $\blacksquare$ Also apologies for no latex but I find that writing up solutions takes a really long time especially when typing \angle so this time I had no latex (and some of my previous posts have minimal latex too).

note that BOCN is a kite so BON=NOC so BPN=NQC so APN=AQN. And notice that BQC=BOC=2*<A so <A=ABQ=PBQ=PNQ thus APNQ is a parallelogram.

Note $\angle BOC = 2A$, so $\angle BON = \angle CON = A$. Then \[\angle APN = 180 - \angle BON = 180 - A,\]\[\angle AQN = 180 -\angle CON = 180 - A,\] making $APNQ$ a parallelogram.

The diagram is given below.

\[\angle CQN=\angle CON=\frac{1}{2}\angle BOC=\angle BAC\implies AP\parallel NQ\]Similar calculations give $AQ\parallel PN$ which implies the desired result.

Denote by $\widehat{XY}$ the arc measure of arc $XY$. Observe that $\angle BPC = 2A, \angle PBC = B \implies \angle PCB = C - A \implies \widehat{PO} = \widehat{BO} - \widehat{BP} = \frac{\pi}{2} - 2A - 2C + 2A = \frac{\pi}{2} - 2C$. Similarly, $\widehat{OQ} = \frac{\pi}{2} - 2B$, so $\widehat{PQ} = 2A = \widehat{CN} \implies \widehat{QN} = \widehat{CP} \implies QN = CP = PA$. Similarly we obtain $AQ = PN$. Since opposite sides of quadrilateral $APNQ$ are equal, it is indeed a parallelogram. $\square$

Let $\angle BAC=\alpha, \angle ABC=\beta, \angle ACB=\gamma$. Firstly, observe that $BC$ is the radical axis of $(ABC)$ and $BOC$. Therefore, as $ON$ is the diameter of $(BOC)$, we get that $ON\perp BC$. Next, we get that $\angle BAC=\angle BON=\angle CON$, as $\angle BOC=2\angle BAC$. Quadrilateral $OBPN$ is cyclic $\implies \angle BPN=\angle APN=180^{\circ}-\alpha$. $$\therefore \angle CAP+\angle APN=180^{\circ}\implies AC\parallel NP\implies \boxed{AQ\parallel NP}$$ $$\angle QCO=\angle QNO=\angle ACB-\angle OCB=\gamma-(90^{\circ}-\alpha)$$$$\angle ABO=\angle PNO=\angle ABC-\angle OBC=\beta-(90^{\circ}-\alpha)$$ $$\implies \angle PNQ=\angle QNO+\angle PNO= 2\alpha+\beta+\gamma-180^{\circ}=\alpha$$$$\implies \angle APN+\angle PNQ=180^{\circ}\implies \boxed{AP\parallel NQ}$$ From the above two conditions, we get that $APNQ$ is a parallelogram, and we're done!