Let $ABCD$ be a quadrilateral. The circumcircle of the triangle $ABC$ intersects the sides $CD$ and $DA$ in the points $P$ and $Q$ respectively, while the circumcircle of $CDA$ intersects the sides $AB$ and $BC$ in the points $R$ and $S$. The lines $BP$ and $BQ$ intersect the line $RS$ in the points $M$ and $N$ respectively. Prove that the points $M, N, P$ and $Q$ lie on the same circle.
Problem
Source: 2017 NZOMC Camp Selections p6
Tags: geometry, Concyclic
EulersTurban
11.01.2021 01:15
[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); 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 = -16.744665054318084, xmax = 8.405153023802612, ymin = -5.099629674611138, ymax = 10.614897176815852; /* image dimensions */ pen ffxfqq = rgb(1,0.4980392156862745,0); /* draw figures */ draw(circle((-6.829916054095756,2.290102603766537), 4.0518924714466245), linewidth(0.8) + red); draw((-7.16428024939307,6.3281755581680565)--(-10.66791615783826,0.9910307028163324), linewidth(0.8) + blue); draw((-10.66791615783826,0.9910307028163324)--(-4.132636743121845,-0.7335569205116056), linewidth(0.8) + blue); draw((-4.132636743121845,-0.7335569205116056)--(0.0748431114652912,5.556788014622477), linewidth(0.8) + blue); draw((0.0748431114652912,5.556788014622477)--(-7.16428024939307,6.3281755581680565), linewidth(0.8) + blue); draw((-7.16428024939307,6.3281755581680565)--(-4.132636743121845,-0.7335569205116056), linewidth(0.8) + blue); draw(circle((-3.7950774624979946,3.5929767549151945), 4.33968200592364), linewidth(0.8) + red); draw((-10.66791615783826,0.9910307028163324)--(-5.652142523051047,6.167044969824654), linewidth(0.8) + blue); draw((-10.66791615783826,0.9910307028163324)--(-3.0055088891965647,0.9515427667650074), linewidth(0.8) + blue); draw((-7.644552164097669,5.596569636286439)--(-5.636258296676332,-0.33676789943472807), linewidth(0.8) + blue); draw(circle((-4.528718906432124,3.457857027408358), 2.9328791195190744), linewidth(0.8) + linetype("4 4") + ffxfqq); draw((-5.652142523051047,6.167044969824654)--(-3.0055088891965647,0.9515427667650074), linewidth(0.8) + blue); /* dots and labels */ dot((-7.16428024939307,6.3281755581680565),dotstyle); label("$A$", (-7.0956825629083795,6.48901198883659), NE * labelscalefactor); dot((-10.66791615783826,0.9910307028163324),dotstyle); label("$B$", (-10.596931746333025,1.1631681605286979), NE * labelscalefactor); dot((-4.132636743121845,-0.7335569205116056),dotstyle); label("$C$", (-4.071129277696479,-0.5627997467933041), NE * labelscalefactor); dot((0.0748431114652912,5.556788014622477),dotstyle); label("$D$", (0.13694485825051342,5.716435877940074), NE * labelscalefactor); dot((-3.0055088891965647,0.9515427667650074),linewidth(4pt) + dotstyle); label("$P$", (-2.936921795742016,1.0809792125609834), NE * labelscalefactor); dot((-5.652142523051047,6.167044969824654),linewidth(4pt) + dotstyle); label("$Q$", (-5.583405920302429,6.2917585137140755), NE * labelscalefactor); dot((-7.644552164097669,5.596569636286439),linewidth(4pt) + dotstyle); label("$R$", (-7.572378461121125,5.732873667533617), NE * labelscalefactor); dot((-5.636258296676332,-0.33676789943472807),linewidth(4pt) + dotstyle); label("$S$", (-5.5669681307088865,-0.20116837573536087), NE * labelscalefactor); dot((-7.271885149125896,4.495555873247706),linewidth(4pt) + dotstyle); label("$M$", (-7.21074709006318,4.6315417647662445), NE * labelscalefactor); dot((-6.077679730422079,0.9673750870090035),linewidth(4pt) + dotstyle); label("$N$", (-6.010788449734545,1.0974170021545264), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] May I say that this is quite a quick exercise $\color{black}\rule{25cm}{1pt}$ Let $\alpha = \angle BAC, \beta = \angle CBA, \gamma = \angle BCA$ and $x = \angle PBA$. Since $BCQA$ is cyclic we have that $\angle AQB = \angle BCA = \gamma$ and since $BPQA$ is cyclic we must have that $\angle DQP = \angle ABP = x$. This implies that $\angle MQP = 180-x-\gamma$ But notice since we have that $ARSC$ is cyclic that $\angle BSN = \alpha$ and that we must have that $\angle NBS = \beta -x$. Thus this implies that $\angle MNP =\angle BNS = 180-(\beta-x+\alpha)=180-180+\gamma +x=\gamma +x$ Thus we have that $MNPQ$ is a cyclic quadrilateral.