In convex quadrilateral $ABCD$, $\angle CAB = \angle BCD$. $P$ lies on line $BC$ such that $AP = PC$, $Q$ lies on line $AP$ such that $AC$ and $DQ$ are parallel, $R$ is the point of intersection of lines $AB$ and $CD$, and $S$ is the point of intersection of lines $AC$ and $QR$. Line $AD$ meets the circumcircle of $AQS$ again at $T$. Prove that $AB$ and $QT$ are parallel.
Problem
Source: PMO
Tags: geometry, PMO
AnSoLiN
20.03.2021 01:33
$ADRQ$ is cyclic because $\angle RDQ = \angle RCA = \angle BCD - \angle ACP = \angle CAB - \angle CAP = \angle PAB = \angle RAQ$ $TAQS$ is cyclic, so $\angle QTA = \angle QSA$ $ADRQ$ is cyclic, so $\angle QSA = \angle RQD = \angle RAD$ And finally $\angle RAD = \angle TAB$, so $\angle QTA = \angle TAB$ which means $AB\parallel QT$
geometry6
20.03.2021 15:29
Nice angle chasing exercise. The problem follows just by seeing that $RQAD$ is cyclic. [asy][asy]/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -25.67, xmax = 9.37, ymin = -6.01, ymax = 16.43; /* image dimensions */ pen wwccff = rgb(0.4,0.8,1); pen qqwuqq = rgb(0,0.39215686274509803,0); draw(arc((-7,4.18),0.6,-103.40959586709407,-35.34502290434381)--(-7,4.18)--cycle, linewidth(0.8) + qqwuqq); draw(arc((1.46,-1.82),0.6,112.2313151801026,180.23149683933005)--(1.46,-1.82)--cycle, linewidth(0.8) + qqwuqq); draw(arc((-7,4.18),0.9,-70.92154264801765,-35.34502290434381)--(-7,4.18)--cycle, linewidth(0.8) + red); draw(arc((1.46,-1.82),0.9,144.6549770956562,180.23149683933005)--(1.46,-1.82)--cycle, linewidth(0.8) + red); /* draw figures */ draw((-7,4.18)--(-8.44,-1.86), linewidth(0.8) + wwccff); draw((-8.44,-1.86)--(1.46,-1.82), linewidth(0.8)); draw((1.46,-1.82)--(-0.6,3.22), linewidth(0.8)); draw((-0.6,3.22)--(-7,4.18), linewidth(0.8)); draw((1.46,-1.82)--(-7,4.18), linewidth(0.8) + red); draw((-7,4.18)--(-4.9159300237086,-1.8457613334291259), linewidth(0.8)); draw((-8.64014893080143,8.922233284256333)--(-7,4.18), linewidth(0.8)); draw((-0.6,3.22)--(-8.64014893080143,8.922233284256333), linewidth(0.8) + red); draw((-4.786756548223351,13.463326700507613)--(-7,4.18), linewidth(0.8) + wwccff); draw((-4.786756548223351,13.463326700507613)--(-0.6,3.22), linewidth(0.8)); draw((-4.786756548223351,13.463326700507613)--(-10.536124542633779,6.687889746548778), linewidth(0.8)); draw((-10.536124542633779,6.687889746548778)--(-7,4.18), linewidth(0.8) + red); draw(circle((-8.026335900001953,6.479779056828449), 2.5184020905843396), linewidth(0.8)); draw((-7,4.18)--(-9.682254784403371,4.582338217660505), linewidth(0.8)); draw((-9.682254784403371,4.582338217660505)--(-8.64014893080143,8.922233284256333), linewidth(0.8) + wwccff); /* dots and labels */ dot((-7,4.18),dotstyle); label("$A$", (-6.89,4.49), NE * labelscalefactor); dot((-8.44,-1.86),dotstyle); label("$B$", (-8.33,-1.57), NE * labelscalefactor); dot((1.46,-1.82),dotstyle); label("$C$", (1.57,-1.51), NE * labelscalefactor); dot((-0.6,3.22),dotstyle); label("$D$", (-0.47,3.53), NE * labelscalefactor); dot((-4.9159300237086,-1.8457613334291259),linewidth(4pt) + dotstyle); label("$P$", (-4.79,-1.6), NE * labelscalefactor); dot((-8.64014893080143,8.922233284256333),linewidth(4pt) + dotstyle); label("$Q$", (-9.14,9.14), NE * labelscalefactor); dot((-4.786756548223351,13.463326700507613),linewidth(4pt) + dotstyle); label("$R$", (-4.67,13.7), NE * labelscalefactor); dot((-10.536124542633779,6.687889746548778),linewidth(4pt) + dotstyle); label("$S$", (-11.24,6.71), NE * labelscalefactor); dot((-9.682254784403371,4.582338217660505),linewidth(4pt) + dotstyle); label("$T$", (-9.56,4.82), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy]
rctmathadventures
21.04.2021 05:45
$\angle RAQ = \angle BAP =\angle BAP - \angle PAC = \angle PCD - \angle PCA = \angle ACD = \angle QDR$ so $A,D,R,Q$ are cyclic. So, $\angle QTA = \angle QSA = \angle RQD = \angle RAD$ which gives $AB || QT$
rafaello
01.08.2021 16:24
[asy][asy]import olympiad;import geometry; size(10cm); defaultpen(fontsize(10pt));pen med=mediummagenta;pen light=pink;pen deep=deepmagenta; pair O,A,E,C,B,P,a,D,R,Q,S,T; O=(0,0);A=dir(120);E=dir(185);C=dir(330);B=intersectionpoint(line(C,E),perpendicular(A,line(A,O)));P=intersectionpoint(line(C,E),perpendicular(0.5*A+0.5*C,line(A,C)));a=dir(15);D=intersectionpoint(line(C,A+C-E),line(a,A));R=extension(A,B,C,D);Q=extension(A,P,D,A+D-C);S=extension(A,C,Q,R);guide w=circumcircle(A,Q,S);T=intersectionpoints(w,A--2A-D)[0]; draw(anglemark(B,A,C),magenta);draw(anglemark(D,C,B),magenta);draw(w,deep);draw(B--R,deep);draw(T--D,deep);draw(R--C,deep);draw(Q--D,med);draw(C--S,med);draw(R--S,deep);draw(A--Q,deep);draw(A--P--C,light, StickIntervalMarker(2,1,light));draw(B--P,deep);draw(circumcircle(A,D,R),deep+dotted);draw(Q--T,deep+dashed); dot("$A$",A,dir(A)); dot("$C$",C,dir(C)); dot("$B$",B,dir(B)); dot("$P$",P,dir(P)); dot("$D$",D,dir(D)); dot("$R$",R,dir(R)); dot("$Q$",Q,dir(Q)); dot("$S$",S,dir(S)); dot("$T$",T,dir(T)); [/asy][/asy] The angle condition implies that $\triangle BAC\sim\triangle BCR$. Hence, $\measuredangle AQD=\measuredangle PAC=\measuredangle ACP=\measuredangle ARD$, therefore $AQRD$ is cyclic quadrilateral. Now $\measuredangle SQT=\measuredangle SAT=\measuredangle QDA=\measuredangle QRA$, hence $QT\parallel AB$.