Wut

According to me this was too trivial for a p2. [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 = -2.5225738482195497, xmax = 40.07349417975526, ymin = -14.831491880613632, ymax = 13.298508888509993; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen qqffff = rgb(0,1,1); /* draw figures */ draw(circle((7.96,-2.72), 5), linewidth(0.8) + rvwvcq); draw((4.230390387440957,0.6101669834870429)--(3.778089507537813,-5.460734323686751), linewidth(0.8) + dtsfsf); draw((3.778089507537813,-5.460734323686751)--(12.629934335312443,-4.506536678568321), linewidth(0.8) + dtsfsf); draw((12.629934335312443,-4.506536678568321)--(11.974914977673556,0.26000968489240606), linewidth(0.8) + dtsfsf); draw((11.974914977673556,0.26000968489240606)--(4.230390387440957,0.6101669834870429), linewidth(0.8) + dtsfsf); draw((4.230390387440957,0.6101669834870429)--(12.629934335312443,-4.506536678568321), linewidth(0.8) + qqffff); draw((3.778089507537813,-5.460734323686751)--(11.974914977673556,0.26000968489240606), linewidth(0.8) + qqffff); draw(circle((6.528437027146399,-2.6133441412573584), 3.9587930395023667), linewidth(0.8) + rvwvcq); draw((19.4895199580039,-0.07975213641252826)--(19.988838783417158,-3.713272739620765), linewidth(0.8) + dtsfsf); draw(circle((11.71234345859784,-2.9995611659865227), 8.30721126925852), linewidth(0.8) + rvwvcq); draw((10.149711433212543,-1.013839841299819)--(10.446936307051281,-3.1767318612134416), linewidth(0.8) + dtsfsf); draw((12.629934335312443,-4.506536678568321)--(19.988838783417158,-3.713272739620765), linewidth(0.8) + dtsfsf); draw((3.778089507537813,-5.460734323686751)--(19.4895199580039,-0.07975213641252826), linewidth(0.8) + green); draw((11.974914977673556,0.26000968489240606)--(19.4895199580039,-0.07975213641252826), linewidth(0.8) + dtsfsf); draw((4.230390387440957,0.6101669834870429)--(19.988838783417158,-3.713272739620765), linewidth(0.8) + green); /* dots and labels */ dot((4.230390387440957,0.6101669834870429),dotstyle); label("$A$", (4.337052691350419,0.879265677756297), NE * labelscalefactor); dot((3.778089507537813,-5.460734323686751),dotstyle); label("$B$", (3.894496140410421,-5.178227113234927), NE * labelscalefactor); dot((12.629934335312443,-4.506536678568321),dotstyle); label("$C$", (12.74562715921038,-4.237794442487431), NE * labelscalefactor); dot((11.974914977673556,0.26000968489240606),dotstyle); label("$D$", (12.081792332800385,0.5473482645512985), NE * labelscalefactor); dot((10.446936307051281,-3.1767318612134416),dotstyle); label("$E$", (10.56050418894414,-2.9101247896674365), NE * labelscalefactor); dot((10.149711433212543,-1.013839841299819),linewidth(4pt) + dotstyle); label("$F$", (10.256246560172892,-0.7803213882686957), NE * labelscalefactor); dot((19.988838783417158,-3.713272739620765),linewidth(4pt) + dotstyle); label("$P$", (20.10312981858785,-3.490980262776184), NE * labelscalefactor); dot((19.4895199580039,-0.07975213641252826),linewidth(4pt) + dotstyle); label("$Q$", (19.605253698780352,0.13245149804505033), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Observe that $\angle APB=\angle ACB-\angle PAC=\angle ADB-\angle DBQ=\angle AQB$ since $\angle PAC \equiv \angle FAE = \angle FBE \equiv \angle DBQ$. By this we have that $AQPB$ is cyclic. Now notice that by converse reims theorem on $\odot(ABCD)$ and $\odot(AQPB)$ we get $PQ \parallel CD$ hence done.$\blacksquare$

Wow! So hard! Sharygin 2020 CR P2 wrote: Let $ABCD$ be a cyclic quadrilateral. A circle passing through $A$ and $B$ meets $AC$ and $BD$ at points $E$ and $F$ respectively. The lines $AF$ and $BC$ meet at point $P$, and the lines $BE$ and $AD$ meet at point $Q$. Prove that $PQ$ is parallel to $CD$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8cm); 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 = -9.86, xmax = 9.86, ymin = -5.44, ymax = 5.44; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw(circle((0.5625856138776734,-1.2316808733363245), 3.001136305869245), linewidth(0.4) + red); draw((0.02,1.72)--(-2.0174141346235697,-2.7647889661138754), linewidth(0.4) + wrwrwr); draw((3.18,-2.7)--(-1.7,0.74), linewidth(0.4) + wrwrwr); draw(circle((-0.04960592944703176,-0.15722224627663797), 1.878512269677814), linewidth(0.4) + green); draw((-2.0174141346235697,-2.7647889661138754)--(-1.34,4.84), linewidth(0.4) + wrwrwr); draw((-1.34,4.84)--(1.350100642194333,-1.4100709444976445), linewidth(0.4) + wrwrwr); draw((-1.4176473561196692,-1.4445726268015309)--(-2.22,4.82), linewidth(0.4) + wrwrwr); draw((3.18,-2.7)--(-2.22,4.82), linewidth(0.4) + wrwrwr); draw((3.18,-2.7)--(-2.0174141346235697,-2.7647889661138754), linewidth(2) + wrwrwr); draw((-1.34,4.84)--(-2.22,4.82), linewidth(2) + wrwrwr); /* dots and labels */ dot((-1.7,0.74),dotstyle); label("$A$", (-2.1,0.9), NE * labelscalefactor); dot((0.02,1.72),dotstyle); label("$B$", (0.1,1.92), NE * labelscalefactor); dot((3.18,-2.7),dotstyle); label("$C$", (3.36,-2.94), NE * labelscalefactor); dot((-2.0174141346235697,-2.7647889661138754),dotstyle); label("$D$", (-2.44,-2.96), NE * labelscalefactor); dot((-1.4176473561196692,-1.4445726268015309),dotstyle); label("$F$", (-1.78,-1.76), S * labelscalefactor); dot((1.350100642194333,-1.4100709444976445),linewidth(4pt) + dotstyle); label("$E$", (1.6,-1.5), NE * labelscalefactor); dot((-1.34,4.84),dotstyle); label("$Q$", (-1.24,4.9), NE * labelscalefactor); dot((-2.22,4.82),dotstyle); label("$P$", (-2.52,4.92), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Observe that $\angle DAC=\angle DBC$ and $\angle FAE=\angle FBE$. So, this implies that $\angle DAF=\angle DAC-\angle FAE=\angle DBC-\angle FBE=\angle EBC=\angle PAQ=\angle PBQ\implies PABQ$ is a cyclic quadrilateral. Hence, $\angle PQA=\angle PQD=\angle PBA=\angle ADC=\angle QDC\implies PQ\|DC$. $\blacksquare$

so here is my solution-

Very trivial problem,also it looks to me that i saw this before somewhere.

Apply pascal's theorem on hexagon $DAFEBC$ which means $P,Q,EF \cap CD$ concur. Since $EF$ is antiparallel to $AB$ which is antiparallel to $CD \implies EF || CD$ . We conclude $P_{\infty,CD} \in PQ$ so we are done.

hellomath010118 wrote: Apply pascal's theorem on hexagon $DAFEBC$ How? They are not on a common circle, right?

GeoMetrix wrote: According to me this was too trivial for a p2 What isn't trivial for you? Almost every sharygin problem I go to, I see you saying "Trivial" "Nice and easy" and stuff like that But ok yeah I agree this is indeed trivial xD $$\angle DAF = \angle DAC-\angle FAE = \angle DBC - \angle EBF = \angle EBC \ \text{Which means } \angle DAF = \angle PAQ = \angle PBQ = \angle EBC \implies \odot(APQB)$$And we are done by double reims theorem which gives $CD \parallel EF \parallel PQ \ \blacksquare$

Claim 1: $EF\parallel CD$ Proof: $\angle BDC=\angle BAC=\angle FEC$ Claim 2: $A,B,P,Q$ is a cyclic quadrilateral. Proof: We define $X=AP\cup BQ$. Due to claim $1$, we have that $\frac{XE}{XQ}=\frac{XF}{XP}$. As points $A,B,F,E$ are concyclic, $XA.XE=XB.XE\implies XA.XP=XB.XQ$. Our claim is thus proved. Now from claim $2$, we have that $\angle BQP=\angle BAP=\angle BEF$. So, $EF\parallel PQ$. From this, and claim $1$, we have that $PQ\parallel CD$ as desired