A trapezoid $ABCD$ is bicentral. The vertex $A$, the incenter $I$, the circumcircle $\omega$ and its center $O$ are given and the trapezoid is erased. Restore it using only a ruler.
Problem
Source: XVII Sharygin Correspondence Round, P21
Tags: geometry, trapezoid
amar_04
02.03.2021 06:37
First we introduce our three main weapons. HarmonicaConj123 - Creates the harmonic conjugate of a point $X$ WRT three collinear points $A,B,X$ only using a straightedge.
This is how the weapon works. Suppose $A,B,X$ lie on a line $\ell$, Take any point $P$ on the plane. Let $Q$ be a point on line $PX$. Let $M=AQ\cap PB$ and $N=BQ\cap PA$. Let $MN\cap \ell=Y$. It's well known that $(A,B;X,Y)=-1$.
Parallello456- If $\ell$ is a line passing though center $O$ of a circumcircle $\omega$ and $P$ is some arbitary point on the plane. Then this weapon creates a line through $P$ parallel to $\ell$ using only a straightedge.
This is how it woks. Let $Q$ be a point on $PX$. Let $QO\cap PY=T$ and $XT\cap QY=R$. By ceva $PR\parallel XY$ and we are done.
Midsegmento789- If $\ell$ is a line through the center $O$ of a circumcircle $\omega$. $X,Y$ are two points on $\ell$, then it creates the midpoint of $XY$ only using a straightedge.
This is how it works. Let $P$ be any point on the plane. From Parallelo456 we can create a line $\tau$ tough $P$ parallel to $\ell$. $T$ be another point in the plane. Let $M=TX\cap\ell_1$ and $N=TY\cap\ell_1$. Let $XN\cap YM=K$. Then by ceva $TK$ passes through the midpoint of $XY$.
Let $OI\cap\odot(ABC)=U,V$. Let $EI\cap AD=F$. Clearly $IB=IF$ which further implies that $\Delta AIB\cong\Delta AIF\implies AI\perp BI$. Hence, $\measuredangle TIB=\measuredangle IAB\implies TA\cdot TB=TU\cdot TV=TI^2$, also $(U,V;I,I^*)=-1$. So by Mc'Laurin's theorem we indeed get that $T$ is the midpoint of $\overline{II^*}$.
Now it's the construction time. Let $U,V=OI\cap\odot(ABCD)$. Using HarmonicaConj123 we can create the inverse $I^*$ of $I$ WRT $\odot(ABCD)$ a $(U,V;I,I^*)=-1$. Using Midsegmento789 we can construct the midpoint $T$ of $II^*$. Using Parallello456 we can create a line $\tau$ through $A$ parallel to $OI$. Let $\tau\cap\odot(ABC)=K$. Hence, $B=AT\cap\odot(ABCD)$ ; $D=KO\cap\odot(ABCD)$ and $C=TD\cap\odot(ABCD)$ and by our $\textbf{LEMMA}$ we can assure that $ABCD$ is bicentric. Thus we have restored $ABCD$! $\quad\blacksquare$
Euler365
02.03.2021 09:42
amar_04 wrote: First we introduce our three main weapons. HarmonicaConj123 - Creates the harmonic conjugate of a point $X$ WRT three collinear points $A,B,X$ only using a straightedge.
This is how the weapon works. Suppose $A,B,X$ lie on a line $\ell$, Take any point $P$ on the plane. Let $Q$ be a point on line $PX$. Let $M=AQ\cap PB$ and $N=BQ\cap PA$. Let $MN\cap \ell=Y$. It's well known that $(A,B;X,Y)=-1$.
Parallello456- If $\ell$ is a line passing though center $O$ of a circumcircle $\omega$ and $P$ is some arbitary point on the plane. Then this weapon creates a line through $D$ parallel to $\ell$ using only a straightedge.
This is how it woks. Let $Q$ be a point on $PX$. Let $QO\cap PY=T$ and $XT\cap QY=R$. By ceva $PR\parallel XY$ and we are done.
Midsegmento789- If $\ell$ is a line through the center $O$ of a circumcircle $\omega$. $X,Y$ are two points on $\ell$, then it creates the midpoint of $XY$ only using a straightedge.
This is how it works. Let $P$ be any point on the plane. From Parallelo456 we can create a line $\tau$ tough $P$ parallel to $\ell$. $T$ be another point in the plane. Let $M=TX\cap\ell_1$ and $N=TY\cap\ell_1$. Let $XN\cap YM=K$. Then by ceva $TK$ passes through the midpoint of $XY$.
Let $OI\cap\odot(ABC)=U,V$. Let $EI\cap AD=F$. Clearly $IB=IF$ which further implies that $\Delta AIB\cong\Delta AIF\implies AI\perp BI$. Hence, $\measuredangle TIB=\measuredangle IAB\implies TA\cdot TB=TU\cdot TV=TI^2$, also $(U,V;I,I^*)=-1$. So by Mc'Laurin's theorem we indeed get that $T$ is the midpoint of $\overline{II^*}$.
Now it's the construction time. Let $U,V=OI\cap\odot(ABCD)$. Using HarmonicaConj123 we can create the inverse $I^*$ of $I$ WRT $\odot(ABCD)$ a $(U,V;I,I^*)=-1$. Using Midsegmento789 we can construct the midpoint $T$ of $II^*$. Using Parallello456 we can create a line $\tau$ through $A$ parallel to $OI$. Let $\tau\cap\odot(ABC)=K$. Hence, $B=AT\cap\odot(ABCD)$ ; $D=KO\cap\odot(ABCD)$ and $C=TD\cap\odot(ABCD)$ and by our $\textbf{LEMMA}$ we can assure that $ABCD$ is bicentric. Thus we have restored $ABCD$! $\quad\blacksquare$ I think so that your solution is incomplete because you have only dealt with the case $O\neq I$. If $O$ and $I$ coincide, then we are no more able to draw line $OI$ and hence we have to do something else to construct $ABCD$ in that case.
Dr_Vex
02.03.2021 09:52
Euler365 wrote: I think so that your solution is incomplete because you have only dealt with the case $O\neq I$. If $O$ and $I$ coincide, then we are no more able to draw line $OI$ and hence we have to do something else to construct $ABCD$ in that case. But in question it is given that $O, I$ are not erased...So ig it means they are distinct
amar_04
02.03.2021 10:02
I guess they do meant distinct, but still the case when $O$ coincides with $I$ is not very hard to deal with. By some congruency arguments we can get that it can only happen iff $ABCD$ is a square. Hence, $AO\cap\odot(ABCD)=C$. Using Parallello456 we can draw a line parallel to $AC$ which intersects $\odot(ABCD)$ at $P,Q$. Let $AQ\cap BP=K$. Hence, $OK\cap\odot(ABCD)=C,D$.
mathaddiction
05.03.2021 16:54
We will show the following well-known lemma: Lemma. Given a fixed circle $c$ with center $O$ we can perform the following constructions with only a ruler: (a) Suppose $A,M,B$ are known points such that $M$ is the midpoint of $AB$ then given any point $C$ we can construct the line through $C$ parallel to $AB$ (b) Given any line $l$ and a point $A$ we can construct the line through $A$ parallel to $l$ (c) Given any line $l$ and a point $A$ we can construct the line through $A$ perpendicular to $l$ (d) Given fixed points $A$, $P,Q$ and any ray $l$ through $A$, construct a point $S$ on $l$ such that $AS=PQ$ Proof. (a)[asy][asy] size(6cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -2.0154591035286664, xmax = 5.0341556660074005, ymin = -2.5182845613178473, ymax = 2.845306772165453; /* image dimensions */ /* draw figures */draw((1.8907270901183728,2.2719498809265755)--(0,0), linewidth(0.8)); draw((1.8907270901183728,2.2719498809265755)--(2,0), linewidth(0.8)); draw((2,0)--(1.2943520041588128,1.5553291096821276), linewidth(0.8)); draw((1,0)--(1.8907270901183728,2.2719498809265755), linewidth(0.8)); draw((0,0)--(2,0), linewidth(0.8)); draw((0,0)--(1.9251940638896252,1.5553291096821276), linewidth(0.8)); /* dots and labels */dot((0,0),dotstyle); label("$A$", (-0.11443939039534504,0.17482669895435837), NE * labelscalefactor); dot((2,0),dotstyle); label("$B$", (2.0751279577671413,-0.14200991990119521), NE * labelscalefactor); dot((1,0),dotstyle); label("$M$", (0.9209374176504819,-0.18727229402341716), NE * labelscalefactor); dot((1.2943520041588128,1.5553291096821276),dotstyle); label("$C$", (1.2094850526796468,1.6288804676307382), NE * labelscalefactor); dot((1.8907270901183728,2.2719498809265755),dotstyle); label("$D$", (1.9110518515740869,2.3304472665251783), NE * labelscalefactor); dot((1.4635573099257893,1.1823812105552765),linewidth(4pt) + dotstyle); label("$E$", (1.2660630203324241,1.1649411328779633), NE * labelscalefactor); dot((1.9251940638896252,1.5553291096821276),linewidth(4pt) + dotstyle); label("$F$", (1.9449986321657533,1.6005914838043496), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Let $D$ be a point on line $AC$, construct $E=BC\cap DM$ and $F=AE\cap DB$, by Ceva's theorem we have $CF\|AB$ as desired [asy][asy] size(4cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.84126408162143, xmax = 5.61925111929918, ymin = -3.4050906151833695, ymax = 4.553632025645474; /* image dimensions */ /* draw figures */draw(circle((-1.080179964436492,1.1787306526357748), 1.3161641579061605), linewidth(0.8)); draw((-2.310556680685093,1.646129012357054)--(0.15019675181210912,0.7113322929144956), linewidth(0.8)); draw((-0.7307750149439005,-0.09020759323236517)--(-1.4295849139290842,2.4476688985039154), linewidth(0.8)); draw((-1.96592489110787,2.1522521035718842)--(0.7835715400614015,-0.8697169469024508), linewidth(0.8)); draw((-1.4295849139290842,2.4476688985039154)--(0.7331998777776746,0.07055408239378388), linewidth(0.8)); draw((-2.310556680685093,1.646129012357054)--(0.8339432023451283,-1.8099879761986855), linewidth(0.8)); draw((0.7331998777776746,0.07055408239378388)--(0.8339432023451283,-1.8099879761986855), linewidth(0.8)); /* dots and labels */dot((-1.919707669165273,-0.6682302977675434),dotstyle); label("$A$", (-1.886126560976121,-0.5842775272946652), NE * labelscalefactor); dot((0.7835715400614015,-0.8697169469024508),dotstyle); label("$B$", (0.8171526482505538,-0.7857641764295726), NE * labelscalefactor); dot((0.7331998777776746,0.07055408239378388),dotstyle); label("$C$", (0.766780985966827,0.15450685286666196), NE * labelscalefactor); dot((-1.080179964436492,1.1787306526357748),dotstyle); label("$O$", (-1.04659885624734,1.2626834231086528), NE * labelscalefactor); dot((-0.1944350377651139,0.20520920169966544),linewidth(4pt) + dotstyle); label("$E$", (-0.1566994892348322,0.2720407315286913), NE * labelscalefactor); dot((-1.96592489110787,2.1522521035718842),linewidth(4pt) + dotstyle); label("$F$", (-2.1547754264893313,2.253326114688614), NE * labelscalefactor); dot((-1.4295849139290842,2.4476688985039154),linewidth(4pt) + dotstyle); label("$G$", (-1.3992004922334282,2.5135797031545364), NE * labelscalefactor); dot((0.15019675181210912,0.7113322929144956),linewidth(4pt) + dotstyle); label("$H$", (0.18750686970396804,0.7757573543659598), NE * labelscalefactor); dot((-2.310556680685093,1.646129012357054),linewidth(4pt) + dotstyle); label("$I$", (-2.507377062475419,1.6992378295676187), NE * labelscalefactor); dot((-0.7307750149439005,-0.09020759323236517),linewidth(4pt) + dotstyle); label("$J$", (-0.6939972202612521,-0.021793965126382025), NE * labelscalefactor); dot((0.8339432023451283,-1.8099879761986855),linewidth(4pt) + dotstyle); label("$K$", (0.8675243105342807,-1.742825759820383), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Let $B$ be a point on $l$ and suppose $BO$ intersect the circle at $E$ and $F$. Pick a point $C$ sufficiently close to $B$ and construct the line through $C$ parallel to $FE$, suppose this line intersect the circle at $G,H$. Suppose $GO,HO$ intersect the circle again at $J,I$. and $JI$ intersect $l$ at $K$. (c) [asy][asy] size(6cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -5.135098778276503, xmax = 5.325416422644107, ymin = -3.472252831561673, ymax = 4.48646980926717; /* image dimensions */ /* draw figures */draw(circle((-1.080179964436492,1.1787306526357748), 1.3161641579061605), linewidth(0.8)); draw((-0.04126286796201817,1.9867772832270307)--(-2.1190970609109656,0.37068402204451933), linewidth(0.8)); draw((-2.1190970609109656,0.37068402204451933)--(0.03911969488126377,0.48630277681910283), linewidth(0.8)); draw((0.7835715400614015,-0.8697169469024508)--(0.7331998777776746,0.07055408239378388), linewidth(0.8)); /* dots and labels */dot((-1.919707669165273,-0.6682302977675434),dotstyle); label("$A$", (-1.8861265609761202,-0.5842775272946662), NE * labelscalefactor); dot((0.7835715400614015,-0.8697169469024508),dotstyle); label("$B$", (0.8843148646288572,-0.987250825564481), NE * labelscalefactor); dot((0.7331998777776746,0.07055408239378388),dotstyle); label("$C$", (0.8591290334869938,0.17969268400852442), NE * labelscalefactor); dot((-1.080179964436492,1.1787306526357748),dotstyle); label("$O$", (-1.3572241069969881,1.3550314706288176), NE * labelscalefactor); dot((-0.04126286796201817,1.9867772832270307),dotstyle); label("$E$", (0.061577713994651825,2.0686300196482814), NE * labelscalefactor); dot((0.03911969488126377,0.48630277681910283),linewidth(4pt) + dotstyle); label("$F$", (0.011206051710924966,0.154506852866661), NE * labelscalefactor); dot((-2.1190970609109656,0.37068402204451933),linewidth(4pt) + dotstyle); label("$H$", (-2.3646573526715255,0.1629021299139488), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Let $E$ be a point on the circle, let $F$ be the point on the circle such that $EF\|l$(b) and let $EO$ meet the circle again at $H$, by $(b)$ we can construct a line through $A$ parallel to $HF$, hence perpendicular to $l$ as desired. $\blacksquare$ (d) [asy][asy] size(6cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.572615216108219, xmax = 4.796513968664981, ymin = -3.9004119609733543, ymax = 4.058310679855493; /* image dimensions */ /* draw figures */draw(circle((-1.080179964436492,1.1787306526357748), 1.3161641579061605), linewidth(0.8)); draw((0.8675243105342839,0.8681054018861248)--(0.7751762630141179,-0.5590916961528035), linewidth(0.8)); draw((-1.919707669165273,-0.6682302977675434)--(-2.0120557166854387,-2.0954273958064715), linewidth(0.8)); draw((-1.919707669165273,-0.6682302977675434)--(-0.5399200992771378,-0.2919245968889645), linewidth(0.8)); draw((-1.080179964436492,1.1787306526357748)--(-1.16516580190032,-0.13468683544157078), linewidth(0.8)); draw((-1.080179964436492,1.1787306526357748)--(0.18960759792119808,1.525036351460596), linewidth(0.8)); draw((-1.16516580190032,-0.13468683544157078)--(0.18960759792119808,1.525036351460596), linewidth(0.8)); draw((-2.0120557166854387,-2.0954273958064715)--(-0.5399200992771378,-0.2919245968889645), linewidth(0.8)); /* dots and labels */dot((-1.919707669165273,-0.6682302977675434),dotstyle); label("$A$", (-1.886126560976119,-0.584277527294668), NE * labelscalefactor); dot((-1.080179964436492,1.1787306526357748),dotstyle); label("$O$", (-1.3572241069969868,1.3550314706288171), NE * labelscalefactor); dot((0.8675243105342839,0.8681054018861248),dotstyle); label("$P$", (0.9011054187234352,0.952058172359002), NE * labelscalefactor); dot((0.7751762630141179,-0.5590916961528035),dotstyle); label("$Q$", (0.8087573712032692,-0.4751389256799264), NE * labelscalefactor); dot((-2.0120557166854387,-2.0954273958064715),linewidth(4pt) + dotstyle); label("$B$", (-2.3646573526715247,-2.204565997421216), NE * labelscalefactor); dot((-0.5399200992771378,-0.2919245968889645),dotstyle); label("$S$", (-0.5093011252209175,-0.20649006016671634), NE * labelscalefactor); dot((-1.16516580190032,-0.13468683544157078),linewidth(4pt) + dotstyle); label("$E$", (-1.3740146610915625,-0.005003411031808795), NE * labelscalefactor); dot((0.18960759792119808,1.525036351460596),linewidth(4pt) + dotstyle); label("$F$", (0.22108797789312223,1.5900992279528758), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Let $B$ be the point such that $ABQP$ is a parallelogram. Construct points $E,F$ on the circle such that $OE\|AB$ and $OF\|l$, then construct a line $BS$ through $B$ such that $EF\|BS$ and $S$ lies on $l$, then $AS=AB=PQ$ as desired. $\blacksquare$ Now we return to the original problem. We will do the following operations: 1. Construct a line through $A$ perpendicular to $OI$ to meet the circle again at $B$. 2. Let $M=OI\cap AB$, and let the line through $M$ perpendicular to $IB$ meet $IB$ at $N$ 3. again at $C$. 4. Construct a point $D$ on $\omega$ such that $CD\|AB$. Then we can restore the trapezoid. [asy][asy] size(8cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -15.31205739443475, xmax = 16.02643494519567, ymin = -12.300410867908075, ymax = 14.130986585801567; /* image dimensions */ /* draw figures */draw(circle((0,0), 5), linewidth(0.8)); draw(circle((0.07388272429308466,2.2015658801529825), 7.948115596144877), linewidth(0.8)); draw((-7.2677527464153355,5.246714504591108)--(-3.526188442333636,-4.884478794057333), linewidth(0.8)); draw((-3.526188442333636,-4.884478794057333)--(3.1907855545822263,-5.109894868303306), linewidth(0.8)); draw((3.1907855545822263,-5.109894868303306)--(7.603155634166744,4.747659157769531), linewidth(0.8)); draw((7.603155634166744,4.747659157769531)--(-7.2677527464153355,5.246714504591108), linewidth(0.8)); draw((0,0)--(0.1677014438757035,4.997186831180319), linewidth(0.8)); draw((0,0)--(7.603155634166744,4.747659157769531), linewidth(0.8)); draw((0.1677014438757035,4.997186831180319)--(4.563674964070192,-2.0427605885954767), linewidth(0.8)); /* dots and labels */dot((0,0),dotstyle); label("$I$", (-0.4513666319587587,0.3019009982066507), NE * labelscalefactor); dot((0.1677014438757035,4.997186831180319),dotstyle); label("$M$", (-0.14467320158871014,5.5435705354402085), NE * labelscalefactor); dot((4.563674964070192,-2.0427605885954767),dotstyle); label("$N$", (4.9575902309311894,-2.6256272007801766), NE * labelscalefactor); dot((-3.526188442333636,-4.884478794057333),linewidth(4pt) + dotstyle); label("$D$", (-4.354737563941195,-5.748323946366126), NE * labelscalefactor); dot((3.1907855545822263,-5.109894868303306),linewidth(4pt) + dotstyle); label("$C$", (3.5077667419091414,-5.748323946366126), NE * labelscalefactor); dot((-7.2677527464153355,5.246714504591108),linewidth(4pt) + dotstyle); label("$A$", (-8.2302272749809,5.655095419211135), NE * labelscalefactor); dot((7.603155634166744,4.747659157769531),linewidth(4pt) + dotstyle); label("$B$", (8.024524534631675,5.06958977941377), NE * labelscalefactor); dot((0.07388272429308466,2.2015658801529825),linewidth(4pt) + dotstyle); label("$O$", (0.18990144972407014,2.420873789854259), NE * labelscalefactor); dot((2.365688203972948,1.477213121292422),linewidth(4pt) + dotstyle); label("$P$", (2.2252305785434836,0.6922380914048943), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy]
PROF65
06.03.2021 02:44
generally consider that $ABCD $ is just bicentral .Remark that if $AI,CI$ cuts $(O)$ at $K,L$ then it s easy to show by angle chase $K,O,L$ are collinear . so to construct $C$ it suffices to draw the line $AI$ that intersects $(O)$ at $K$ then draw $OK$ to get $L$ and draw $LI$ to find $C$ finally draw OI to get $D$ and draw the perpendicular....
nichas
19.02.2022 14:30
mathaddiction wrote: We will show the following well-known lemma: Lemma. Given a fixed circle $c$ with center $O$ we can perform the following constructions with only a ruler: (a) Suppose $A,M,B$ are known points such that $M$ is the midpoint of $AB$ then given any point $C$ we can construct the line through $C$ parallel to $AB$ (b) Given any line $l$ and a point $A$ we can construct the line through $A$ parallel to $l$ (c) Given any line $l$ and a point $A$ we can construct the line through $A$ perpendicular to $l$ (d) Given fixed points $A$, $P,Q$ and any ray $l$ through $A$, construct a point $S$ on $l$ such that $AS=PQ$ Proof. (a)[asy][asy] size(6cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -2.0154591035286664, xmax = 5.0341556660074005, ymin = -2.5182845613178473, ymax = 2.845306772165453; /* image dimensions */ /* draw figures */draw((1.8907270901183728,2.2719498809265755)--(0,0), linewidth(0.8)); draw((1.8907270901183728,2.2719498809265755)--(2,0), linewidth(0.8)); draw((2,0)--(1.2943520041588128,1.5553291096821276), linewidth(0.8)); draw((1,0)--(1.8907270901183728,2.2719498809265755), linewidth(0.8)); draw((0,0)--(2,0), linewidth(0.8)); draw((0,0)--(1.9251940638896252,1.5553291096821276), linewidth(0.8)); /* dots and labels */dot((0,0),dotstyle); label("$A$", (-0.11443939039534504,0.17482669895435837), NE * labelscalefactor); dot((2,0),dotstyle); label("$B$", (2.0751279577671413,-0.14200991990119521), NE * labelscalefactor); dot((1,0),dotstyle); label("$M$", (0.9209374176504819,-0.18727229402341716), NE * labelscalefactor); dot((1.2943520041588128,1.5553291096821276),dotstyle); label("$C$", (1.2094850526796468,1.6288804676307382), NE * labelscalefactor); dot((1.8907270901183728,2.2719498809265755),dotstyle); label("$D$", (1.9110518515740869,2.3304472665251783), NE * labelscalefactor); dot((1.4635573099257893,1.1823812105552765),linewidth(4pt) + dotstyle); label("$E$", (1.2660630203324241,1.1649411328779633), NE * labelscalefactor); dot((1.9251940638896252,1.5553291096821276),linewidth(4pt) + dotstyle); label("$F$", (1.9449986321657533,1.6005914838043496), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Let $D$ be a point on line $AC$, construct $E=BC\cap DM$ and $F=AE\cap DB$, by Ceva's theorem we have $CF\|AB$ as desired [asy][asy] size(4cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.84126408162143, xmax = 5.61925111929918, ymin = -3.4050906151833695, ymax = 4.553632025645474; /* image dimensions */ /* draw figures */draw(circle((-1.080179964436492,1.1787306526357748), 1.3161641579061605), linewidth(0.8)); draw((-2.310556680685093,1.646129012357054)--(0.15019675181210912,0.7113322929144956), linewidth(0.8)); draw((-0.7307750149439005,-0.09020759323236517)--(-1.4295849139290842,2.4476688985039154), linewidth(0.8)); draw((-1.96592489110787,2.1522521035718842)--(0.7835715400614015,-0.8697169469024508), linewidth(0.8)); draw((-1.4295849139290842,2.4476688985039154)--(0.7331998777776746,0.07055408239378388), linewidth(0.8)); draw((-2.310556680685093,1.646129012357054)--(0.8339432023451283,-1.8099879761986855), linewidth(0.8)); draw((0.7331998777776746,0.07055408239378388)--(0.8339432023451283,-1.8099879761986855), linewidth(0.8)); /* dots and labels */dot((-1.919707669165273,-0.6682302977675434),dotstyle); label("$A$", (-1.886126560976121,-0.5842775272946652), NE * labelscalefactor); dot((0.7835715400614015,-0.8697169469024508),dotstyle); label("$B$", (0.8171526482505538,-0.7857641764295726), NE * labelscalefactor); dot((0.7331998777776746,0.07055408239378388),dotstyle); label("$C$", (0.766780985966827,0.15450685286666196), NE * labelscalefactor); dot((-1.080179964436492,1.1787306526357748),dotstyle); label("$O$", (-1.04659885624734,1.2626834231086528), NE * labelscalefactor); dot((-0.1944350377651139,0.20520920169966544),linewidth(4pt) + dotstyle); label("$E$", (-0.1566994892348322,0.2720407315286913), NE * labelscalefactor); dot((-1.96592489110787,2.1522521035718842),linewidth(4pt) + dotstyle); label("$F$", (-2.1547754264893313,2.253326114688614), NE * labelscalefactor); dot((-1.4295849139290842,2.4476688985039154),linewidth(4pt) + dotstyle); label("$G$", (-1.3992004922334282,2.5135797031545364), NE * labelscalefactor); dot((0.15019675181210912,0.7113322929144956),linewidth(4pt) + dotstyle); label("$H$", (0.18750686970396804,0.7757573543659598), NE * labelscalefactor); dot((-2.310556680685093,1.646129012357054),linewidth(4pt) + dotstyle); label("$I$", (-2.507377062475419,1.6992378295676187), NE * labelscalefactor); dot((-0.7307750149439005,-0.09020759323236517),linewidth(4pt) + dotstyle); label("$J$", (-0.6939972202612521,-0.021793965126382025), NE * labelscalefactor); dot((0.8339432023451283,-1.8099879761986855),linewidth(4pt) + dotstyle); label("$K$", (0.8675243105342807,-1.742825759820383), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Let $B$ be a point on $l$ and suppose $BO$ intersect the circle at $E$ and $F$. Pick a point $C$ sufficiently close to $B$ and construct the line through $C$ parallel to $FE$, suppose this line intersect the circle at $G,H$. Suppose $GO,HO$ intersect the circle again at $J,I$. and $JI$ intersect $l$ at $K$. (c) [asy][asy] size(6cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -5.135098778276503, xmax = 5.325416422644107, ymin = -3.472252831561673, ymax = 4.48646980926717; /* image dimensions */ /* draw figures */draw(circle((-1.080179964436492,1.1787306526357748), 1.3161641579061605), linewidth(0.8)); draw((-0.04126286796201817,1.9867772832270307)--(-2.1190970609109656,0.37068402204451933), linewidth(0.8)); draw((-2.1190970609109656,0.37068402204451933)--(0.03911969488126377,0.48630277681910283), linewidth(0.8)); draw((0.7835715400614015,-0.8697169469024508)--(0.7331998777776746,0.07055408239378388), linewidth(0.8)); /* dots and labels */dot((-1.919707669165273,-0.6682302977675434),dotstyle); label("$A$", (-1.8861265609761202,-0.5842775272946662), NE * labelscalefactor); dot((0.7835715400614015,-0.8697169469024508),dotstyle); label("$B$", (0.8843148646288572,-0.987250825564481), NE * labelscalefactor); dot((0.7331998777776746,0.07055408239378388),dotstyle); label("$C$", (0.8591290334869938,0.17969268400852442), NE * labelscalefactor); dot((-1.080179964436492,1.1787306526357748),dotstyle); label("$O$", (-1.3572241069969881,1.3550314706288176), NE * labelscalefactor); dot((-0.04126286796201817,1.9867772832270307),dotstyle); label("$E$", (0.061577713994651825,2.0686300196482814), NE * labelscalefactor); dot((0.03911969488126377,0.48630277681910283),linewidth(4pt) + dotstyle); label("$F$", (0.011206051710924966,0.154506852866661), NE * labelscalefactor); dot((-2.1190970609109656,0.37068402204451933),linewidth(4pt) + dotstyle); label("$H$", (-2.3646573526715255,0.1629021299139488), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Let $E$ be a point on the circle, let $F$ be the point on the circle such that $EF\|l$(b) and let $EO$ meet the circle again at $H$, by $(b)$ we can construct a line through $A$ parallel to $HF$, hence perpendicular to $l$ as desired. $\blacksquare$ (d) [asy][asy] size(6cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.572615216108219, xmax = 4.796513968664981, ymin = -3.9004119609733543, ymax = 4.058310679855493; /* image dimensions */ /* draw figures */draw(circle((-1.080179964436492,1.1787306526357748), 1.3161641579061605), linewidth(0.8)); draw((0.8675243105342839,0.8681054018861248)--(0.7751762630141179,-0.5590916961528035), linewidth(0.8)); draw((-1.919707669165273,-0.6682302977675434)--(-2.0120557166854387,-2.0954273958064715), linewidth(0.8)); draw((-1.919707669165273,-0.6682302977675434)--(-0.5399200992771378,-0.2919245968889645), linewidth(0.8)); draw((-1.080179964436492,1.1787306526357748)--(-1.16516580190032,-0.13468683544157078), linewidth(0.8)); draw((-1.080179964436492,1.1787306526357748)--(0.18960759792119808,1.525036351460596), linewidth(0.8)); draw((-1.16516580190032,-0.13468683544157078)--(0.18960759792119808,1.525036351460596), linewidth(0.8)); draw((-2.0120557166854387,-2.0954273958064715)--(-0.5399200992771378,-0.2919245968889645), linewidth(0.8)); /* dots and labels */dot((-1.919707669165273,-0.6682302977675434),dotstyle); label("$A$", (-1.886126560976119,-0.584277527294668), NE * labelscalefactor); dot((-1.080179964436492,1.1787306526357748),dotstyle); label("$O$", (-1.3572241069969868,1.3550314706288171), NE * labelscalefactor); dot((0.8675243105342839,0.8681054018861248),dotstyle); label("$P$", (0.9011054187234352,0.952058172359002), NE * labelscalefactor); dot((0.7751762630141179,-0.5590916961528035),dotstyle); label("$Q$", (0.8087573712032692,-0.4751389256799264), NE * labelscalefactor); dot((-2.0120557166854387,-2.0954273958064715),linewidth(4pt) + dotstyle); label("$B$", (-2.3646573526715247,-2.204565997421216), NE * labelscalefactor); dot((-0.5399200992771378,-0.2919245968889645),dotstyle); label("$S$", (-0.5093011252209175,-0.20649006016671634), NE * labelscalefactor); dot((-1.16516580190032,-0.13468683544157078),linewidth(4pt) + dotstyle); label("$E$", (-1.3740146610915625,-0.005003411031808795), NE * labelscalefactor); dot((0.18960759792119808,1.525036351460596),linewidth(4pt) + dotstyle); label("$F$", (0.22108797789312223,1.5900992279528758), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Let $B$ be the point such that $ABQP$ is a parallelogram. Construct points $E,F$ on the circle such that $OE\|AB$ and $OF\|l$, then construct a line $BS$ through $B$ such that $EF\|BS$ and $S$ lies on $l$, then $AS=AB=PQ$ as desired. $\blacksquare$ Now we return to the original problem. We will do the following operations: 1. Construct a line through $A$ perpendicular to $OI$ to meet the circle again at $B$. 2. Let $M=OI\cap AB$, and let the line through $M$ perpendicular to $IB$ meet $IB$ at $N$ 3. again at $C$. 4. Construct a point $D$ on $\omega$ such that $CD\|AB$. Then we can restore the trapezoid. [asy][asy] size(8cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -15.31205739443475, xmax = 16.02643494519567, ymin = -12.300410867908075, ymax = 14.130986585801567; /* image dimensions */ /* draw figures */draw(circle((0,0), 5), linewidth(0.8)); draw(circle((0.07388272429308466,2.2015658801529825), 7.948115596144877), linewidth(0.8)); draw((-7.2677527464153355,5.246714504591108)--(-3.526188442333636,-4.884478794057333), linewidth(0.8)); draw((-3.526188442333636,-4.884478794057333)--(3.1907855545822263,-5.109894868303306), linewidth(0.8)); draw((3.1907855545822263,-5.109894868303306)--(7.603155634166744,4.747659157769531), linewidth(0.8)); draw((7.603155634166744,4.747659157769531)--(-7.2677527464153355,5.246714504591108), linewidth(0.8)); draw((0,0)--(0.1677014438757035,4.997186831180319), linewidth(0.8)); draw((0,0)--(7.603155634166744,4.747659157769531), linewidth(0.8)); draw((0.1677014438757035,4.997186831180319)--(4.563674964070192,-2.0427605885954767), linewidth(0.8)); /* dots and labels */dot((0,0),dotstyle); label("$I$", (-0.4513666319587587,0.3019009982066507), NE * labelscalefactor); dot((0.1677014438757035,4.997186831180319),dotstyle); label("$M$", (-0.14467320158871014,5.5435705354402085), NE * labelscalefactor); dot((4.563674964070192,-2.0427605885954767),dotstyle); label("$N$", (4.9575902309311894,-2.6256272007801766), NE * labelscalefactor); dot((-3.526188442333636,-4.884478794057333),linewidth(4pt) + dotstyle); label("$D$", (-4.354737563941195,-5.748323946366126), NE * labelscalefactor); dot((3.1907855545822263,-5.109894868303306),linewidth(4pt) + dotstyle); label("$C$", (3.5077667419091414,-5.748323946366126), NE * labelscalefactor); dot((-7.2677527464153355,5.246714504591108),linewidth(4pt) + dotstyle); label("$A$", (-8.2302272749809,5.655095419211135), NE * labelscalefactor); dot((7.603155634166744,4.747659157769531),linewidth(4pt) + dotstyle); label("$B$", (8.024524534631675,5.06958977941377), NE * labelscalefactor); dot((0.07388272429308466,2.2015658801529825),linewidth(4pt) + dotstyle); label("$O$", (0.18990144972407014,2.420873789854259), NE * labelscalefactor); dot((2.365688203972948,1.477213121292422),linewidth(4pt) + dotstyle); label("$P$", (2.2252305785434836,0.6922380914048943), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy]