Note that the tangency is not necesary

Let $ \angle ACB = \alpha $. Since $QA$ is tangent to the circumcircle of $ABC$ then $\angle QAB = \alpha $. Since $NL$ is parallel to $AC$ $=> \angle BLN=\alpha$ Therefore quadrilateral $QALB$ is cyclic. Similarly quadrilateral $APCL $ is cyclic. Therefore $\angle AQB+\angle ALB= 180$ and $\angle APC+\angle ALC= 180$ but since $\angle ALB+\angle ALC= 180$ $=>$ $\angle AQB+\angle APC= 180$ $=>$ $CP$ is parallel to $BQ$

Concyclicboy wrote: Note that the tangency is not necesary Since $QN\parallel AM,NA\parallel MP,AQ\parallel PA$, then $\triangle QNA\sim\triangle AMP$, so $\frac{QN}{NA}=\frac{AM}{MP}$. But $AN=NB$ and $AM=MC$ so $\frac{QN}{NB}=\frac{MC}{MP}$. We also have $BN\parallel MP$ and $QN\parallel MC$ so $\angle QNB=\angle MPC$, which implies that $\triangle QNB\sim\triangle CMP$ and hence $BQ\parallel CP$.

By Pappus theorem for lines $BLC, QAP,$ intersections $BA \cap LP, CA \cap LQ, BQ \cap CP$ are collinear. $BA \parallel LP, CA \parallel LQ$ $\Longrightarrow$ $BA \cap LP, CA \cap LQ$ are at infinity $\Longrightarrow$ $BQ \cap CP$ is also at infinity $\Longrightarrow$ $BQ \parallel CP.$

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(20cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(13); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -21.136434331797066, xmax = 16.638904599304784, ymin = -5.019263026194824, ymax = 15.911990348743599; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); draw((-3.022849680408068,3.231073665836452)--(-5.54,-3.24)--(3.48,-3.2)--cycle, linewidth(3.2) + rvwvcq); /* draw figures */ draw((-3.022849680408068,3.231073665836452)--(-5.54,-3.24), linewidth(3.2) + rvwvcq); draw((-5.54,-3.24)--(3.48,-3.2), linewidth(3.2) + rvwvcq); draw((3.48,-3.2)--(-3.022849680408068,3.231073665836452), linewidth(3.2) + rvwvcq); draw(circle((-1.0386658641600006,-1.2658476319198442), 4.915209718300423), linewidth(3.6)); draw((2.4122018834508143,5.62919020811688)--(3.48,-3.2), linewidth(2.8) + dtsfsf); draw((-17.812124479205174,-3.2944218380452557)--(2.4122018834508143,5.62919020811688), linewidth(3.2)); draw((-17.812124479205174,-3.2944218380452557)--(-5.54,-3.24), linewidth(3.2) + rvwvcq); draw((-5.54,-3.24)--(-6.1554459324972735,1.8488731836229884), linewidth(3.2) + dtsfsf); draw((xmin, -0.9889623752509822*xmin-4.238631246508512)--(xmax, -0.9889623752509822*xmax-4.238631246508512), linewidth(2.4) + linetype("4 4")); /* line */ draw((-1.03,-3.22)--(2.4122018834508143,5.62919020811688), linewidth(2.8)); draw((2.4122018834508143,5.62919020811688)--(xmin, -8.268595038030467*xmin + 25.574710732346027), linewidth(2) + linetype("4 4")); /* ray */ draw((xmin, 2.570793494321591*xmin + 11.002195958541614)--(xmax, 2.570793494321591*xmax + 11.002195958541614), linewidth(3.2) + rvwvcq); /* line */ /* dots and labels */ dot((-3.022849680408068,3.231073665836452),dotstyle); label("$A$", (-3.6111541391711226,3.836267247817586), NE * labelscalefactor); dot((-5.54,-3.24),dotstyle); label("$B$", (-6.490749647853477,-2.9137523316743903), NE * labelscalefactor); dot((3.48,-3.2),dotstyle); label("$C$", (3.912950254482771,-3.6259103607033603), NE * labelscalefactor); dot((-1.03,-3.22),dotstyle); label("$L$", (-1.629497014916599,-3.9665076789346068), NE * labelscalefactor); dot((0.228575159795966,0.01553683291822594),dotstyle); label("$M$", (0.8785377829680321,-0.09608360812498748), NE * labelscalefactor); dot((-4.281424840204034,-0.004463167081774078),dotstyle); label("$N$", (-3.827897887136461,0.02776996214092034), NE * labelscalefactor); dot((2.4122018834508143,5.62919020811688),dotstyle); label("$P$", (2.891158299789032,5.632144016673249), NE * labelscalefactor); dot((-6.1554459324972735,1.8488731836229884),dotstyle); label("$Q$", (-6.243042507321661,2.6286949377249846), NE * labelscalefactor); dot((-17.812124479205174,-3.2944218380452557),dotstyle); label("$D$", (-18.473582571080048,-2.8208621539749594), NE * labelscalefactor); dot((1.3444037669016282,14.458380416233759),dotstyle); label("$E$", (1.4668422417310936,14.766344823783951), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Define: $AB \cap CP=E$ and $BC \cap PQ=D$, then, $LM||AB$ $\implies$ $CP=PE$, then, $$-1=(A,C;M, \infty_{AC}) \overset{L}{=} (A,D;P,Q) \overset{B}{=} (E,C;P,BQ \cap PC) \implies BQ||PC$$

We have: $\widehat{BAQ}$ = $\widehat{ACB}$ = $\widehat{BLN}$ So: $A$, $L$, $B$, $Q$ lie on a circle Similarly: $A$, $L$, $C$, $P$ lie on a circle Then: $\widehat{BQA}$ + $\widehat{CPA}$ = $180^o$ $-$ $\widehat{ALB}$ + $180^o$ $-$ $\widehat{ALC}$ = $180^o$ or $BQ$ $\parallel$ $CP$

Jutaro wrote: Let $ABC$ be a triangle and $L$, $M$, $N$ be the midpoints of $BC$, $CA$ and $AB$, respectively. The tangent to the circumcircle of $ABC$ at $A$ intersects $LM$ and $LN$ at $P$ and $Q$, respectively. Show that $CP$ is parallel to $BQ$. Let $ML\cap QA,QB=\{P,R\}$ and $QL\cap PC=S$. So, $-1=(AB;N\infty_{AB})\overset{Q}{=}(PR;L\infty_{PR})$ and $-1=(AC;M\infty_{AC})\overset{P}{=}(QS;L\infty_{QT})$. Hence, $QL=LS$ and $PL=LR$. So, $PQRS$ is a parallelogram. Hence, $QB\|AC$.

$\angle QAB=\angle C=\angle BLC$. So $ABLQ$ and similarly $ALCP$ are cyclic and we are done by Reim's theorem.

Since $M, N, L$ are midpoints, implies that $LQ \parallel CA$. $\Rightarrow \angle ACB= \angle QLB$. Let $\Omega$ the circumcircle of $\Delta ABC$, implies that $\angle ACB= \angle BAQ= \angle QLB$. So, $ALBQ$ is cyclic. We know that $AB \parallel PL$. So, $\angle BAQ= \angle LPA \Rightarrow \angle ACB= \angle LPA$. Hence, $APCL$ is cyclic. $\Rightarrow \angle PCL= \angle LAQ =\angle QBD$. Therefore $CP \parallel BQ$ omccp2.jpg

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