it's just length chasing first note $AL/LC=AE*AB/BC*CE$ also by Menelaus theorem $CR/RQ * QD/DP * PA/AC=1$ also since we are supposed to prove $AL/LC * CR/RQ * QE/EA=1$ by putting $CR/RQ$ and $AL/LC$ we get that we need to chase $AB/BC * AE/EC * QE/EA * PD/QD * AC/PA=1$. now since $QE/EC=QC/AC$ , $QC=QD$ we are only left with $AB/BC * PD/PA=1$ but since $PB=PD$ and $PB/PA=BC/AB$ we are done .

Let $R'=AD \cap BE$. Note that the tangents at $B$ and $D$ concur on $AC$ at $P$, so $ABCD$ is harmonic, hence the tangents at $C$ and $A$ concur on $BD$ at $X$, say. Now apply Pascal's Theorem to hexagon $AAEBDD$ to find that $X=AA \cap BD$, $Q=AE \cap DD$ and $R'=BE \cap AD$ are collinear. Now note that $X$ and $Q$ both lie on the tangent at $C$, hence $R'$ also lies on the tangent at $C$. It follows that $R'=CC \cap AD \cap BE=CC \cap AD=R$. So $R'$ and $R$ are in fact the same point. Since $R'=R$ lies on $BE$ by definition, it follows that $B, E, R$, are indeed collinear, and thus the problem is solved.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(300); 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 = -2.500000000000003, xmax = 6.000000000000000, ymin = -4.500000000000005, ymax = 5.500000000000006; /* image dimensions */ /* draw figures */ draw(circle((0.8800000000000010,2.380000000000003), 2.812220560462430)); draw((-0.1722700811190205,4.987932544578062)--(-0.6572410236823310,0.02511687852757574)); draw((-0.6572410236823310,0.02511687852757574)--(2.741979059560311,0.2724852402778953)); draw((2.741979059560311,0.2724852402778953)--(3.611687492680652,3.048182552166493)); draw((-0.6572410236823310,0.02511687852757574)--(5.049468200679748,2.311140695053324), red); draw((-0.1722700811190205,4.987932544578062)--(5.049468200679748,2.311140695053324)); draw((-0.1722700811190205,4.987932544578062)--(5.302813708552788,-3.865539330369592)); draw((-0.6572410236823310,0.02511687852757574)--(5.302813708552788,-3.865539330369592)); draw((5.049468200679748,2.311140695053324)--(5.302813708552788,-3.865539330369592)); draw((5.049468200679748,2.311140695053324)--(2.741979059560311,0.2724852402778953)); draw((-0.1722700811190205,4.987932544578062)--(4.015433061238433,1.397575668944976)); draw((3.611687492680652,3.048182552166493)--(5.302813708552788,-3.865539330369592)); /* dots and labels */ dot((-0.1722700811190205,4.987932544578062),dotstyle); label("$A$", (-0.3684982006817434,5.129612282793771), NE * labelscalefactor); dot((-0.6572410236823310,0.02511687852757574),dotstyle); label("$B$", (-0.8960041163806602,-0.3165298738815641), NE * labelscalefactor); dot((2.741979059560311,0.2724852402778953),dotstyle); label("$C$", (2.882078793354825,0.1254345419742877), NE * labelscalefactor); dot((3.611687492680652,3.048182552166493),dotstyle); label("$D$", (3.780264541707035,3.048102453279115), NE * labelscalefactor); dot((5.302813708552788,-3.865539330369592),dotstyle); label("$P$", (5.391296122084808,-4.151640450179118), NE * labelscalefactor); dot((4.015433061238433,1.397575668944976),dotstyle); label("$Q$", (4.150944374360328,1.194703290012639), NE * labelscalefactor); dot((5.049468200679748,2.311140695053324),dotstyle); label("$R$", (5.106157789274582,2.392284287815592), NE * labelscalefactor); dot((3.618028027981951,1.738293681912910),dotstyle); label("$E$", (3.346612375615787,1.907549122038207), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] We use complex numbers. Let $\omega$ be the unit circle, and let the lowercase letter of a point be its complex coordinate. Since $P$ lies on the intersection of the tangents to $\omega$ at $B$ and $D$, we have $p = \frac{2bd}{b+d}$. In addition, $P$ lies on chord $AC$, so $p = a+c-ac\bar{p}$. This implies that $\frac{2bd}{b+d} = a+c - \frac{2ac}{b+d}$, or $ab+bc+cd+da=2ac+2bd$. $Q$ lies on the tangent at $C$, and lies on $PD$, so $q = \frac{2cd}{c+d}$. $R$ lies on chord $AD$ and on the tangent at $C$. Therefore we have $r = 2c - c^2\bar{r}$ and $r = a+d-ad\bar{r}$. Solving for $r$ yields \[r = a + d - ad \frac{2c-r}{c^2} \implies r = \frac{ac^2+c^2d-2acd}{c^2-ad}\] $A, Q, E$ are collinear, so we have $\frac{a-q}{\bar{a}-\bar{q}} = \frac{a-e}{\bar{a}-\bar{e}}$, or \[\frac{a-\frac{2cd}{c+d}}{\frac{1}{a}-\frac{2}{c+d}} = -ae \implies e = \frac{ac + ad - 2cd}{2a-c-d}\] We must prove that $B, E, R$ are collinear, or that \[\frac{b-e}{\bar{b}-\bar{e}} = \frac{b-r}{\bar{b}-\bar{r}} \implies -be = \frac{b-r}{\frac{1}{b} - \bar{r}}\] or \begin{align*}\frac{abc+abd-2bcd}{c+d-2a} = \frac{b-\frac{ac^2+c^2d-2acd}{c^2-ad}}{\frac{1}{b}- \frac{\frac{1}{ac^2}+\frac{1}{c^2d}-\frac{2}{acd}}{\frac{1}{c^2}-\frac{1}{ad}}} = \frac{b - \frac{ac^2+c^2d-2acd}{c^2-ad}}{\frac{1}{b} - \frac{d+a-2c}{ad-c^2}} = \frac{b(c^2-ad)-(ac^2+c^2d-2acd)}{\frac{c^2-ad}{b} + (d+a-2c)}\end{align*} Cross-multiplying, we have \begin{align*}(ac+ad-2cd)(c^2-ad+bd+ab-2bc) = (c+d-2a)(bc^2-abd - ac^2 - c^2d + 2acd) \\ \implies (a-c)(c-d)(ab+bc+cd+da-2ac-2bd) = 0\end{align*} which is true.

robinpark wrote: Cross-multiplying, we have \begin{align*}(ac+ad-2cd)(c^2-ad+bd+ab-2bc) = (c+d-2a)(bc^2-abd - ac^2 - c^2d + 2acd) \\ \implies (a-c)(c-d)(ab+bc+cd+da-2ac-2bd) = 0\end{align*} which is true. FTFY

Let $BC$ meet $AD$ at $M$. Taking the perspective of $C$, and using the fact that $CBAD$ is a harmonic quadrilateral, gives $(R, A; D, M)$ is harmonic. But, $ECAD$ is also an harmonic quadrilateral, and taking the perspective of $B$ means $(R', A; D, M)$ is harmonic, where $R' = BE \cap AD$. But, the means $R' = R$, so done $\blacksquare$. Note we have multiple things to be rectified. For instance, the cross-ratio of $4$ points on a conic $w$ from a point is invariant. Also, the construction of an harmonic quad is: Take a point $P$ and circle $(O)$, let the tagents to $P$ be $B, C$ and let a line $\ell$ through $P$ intersect $(O)$ at $A, D$. Then $ABCD$ is harmonic.

Set $T = \overline{BD} \cap \overline{CR}$, $K = \overline{AC} \cap \overline{BD}$, $Z = \overline{AB} \cap \overline{CR}$ and $E' = \overline{BR} \cap \omega$, where $E' \neq A$. Note that since $ABCD$ is harmonic, we have $T,K,B,D$ collinear and with \[ -1 = (T,K;B,D) \stackrel{A}{=} (T,C; Z,R) \stackrel{B}{=} (D,C; A,E'). \] But $DACE$ is harmonic; therefore $E = E'$. $\blacksquare$ [asy][asy]/* DRAGON 0.0.9.6 Homemade Script by v_Enhance. */ import olympiad; import cse5; size(11cm); real lsf=0.8000; real lisf=2011.0; defaultpen(fontsize(10pt)); /* Initialize Objects */ pair P = (4.0, -6.0); pair O = (4.0, 2.0); pair D = (7.794019966423534, -0.7329512659808339); path DP = D--P; pair A = (0.7383826268204343, 5.350442096025524); pair C = (2)*(foot(O,A,P))-A; pair B = 2*foot(D,relpoint(O--P,0.5-10/lisf),relpoint(O--P,0.5+10/lisf))-D; pair T = IntersectionPoint(Line(B,D,lisf),Line(O,midpoint(A--C),lisf)); pair Q = IntersectionPoint(DP,Line(T,C,lisf)); pair Z = IntersectionPoint(Line(A,B,lisf),Line(T,C,lisf)); pair R = IntersectionPoint(Line(A,D,lisf),Line(T,C,lisf)); pair E_prime = (foot(O,A,Q))*(2)-A; path BD = T--D; path AP = A--P; pair K = IntersectionPoint(BD,AP); /* Draw objects */ draw(DP, rgb(0.6,0.0,0.6)); draw(CirclebyPoint(O,D), rgb(0.8,0.0,0.8) + linewidth(1.2)); draw(B--P, rgb(0.6,0.0,0.6)); draw(BD, rgb(0.8,0.0,1.0) + dashed); draw(T--A, rgb(1.0,0.6,1.0)); draw(T--R, rgb(1.0,0.6,1.0)); draw(A--R, rgb(1.0,0.6,1.0)); draw(A--Z, rgb(1.0,0.0,0.8)); draw(AP, rgb(0.8,0.0,1.0) + dashed); draw(B--R, rgb(1.0,0.0,0.8)); draw(A--Q, rgb(0.8,0.0,0.4) + linetype("4 4")); /* Place dots on each point */ dot(P); dot(D); dot(A); dot(C); dot(B); dot(T); dot(Q); dot(Z); dot(R); dot(E_prime); dot(K); /* Label points */ label("$P$", P, lsf * dir(0)); label("$D$", D, lsf * dir(0)); label("$A$", A, lsf * dir(135)); label("$C$", C, lsf * dir(45)); label("$B$", B, lsf * dir(225)); label("$T$", T, lsf * dir(135)); label("$Q$", Q, lsf * dir(-60)); label("$Z$", Z, lsf * dir(-90)); label("$R$", R, lsf * dir(-90)); label("$E'$", E_prime, lsf * dir(70)); label("$K$", K, lsf * dir(45)); [/asy][/asy]

I believe this was too easy for an apmo 5... Also, inversion about C works, it transforms the problem into a problem about congruent triangles.

Conceptual/symmetric solution: Work in the projective plane, and focus on $\triangle{ACD}$; then $ABCD$, $ACED$ are harmonic. We need to show the symmedian point $K = AE\cap BD$ of $\triangle{ACD}$ lies on the polar of $R$ (since $K$ lies on the polar of $BE\cap AD$ by, for instance, Brokard's theorem). But this is trivial, since letting $T = AK\cap (ACD)$, we have $ACDT$ harmonic, and thus $K$ lies on the polar $CT$ of $R = CC\cap AD\cap TT$, as desired. Note that if we view this from the perspective of the triangle with sides $AA,CC,DD$, we get a well-known rephrasing of the problem: Quote: The Gergonne point of a triangle $XYZ$ (with intouch triangle $UVW$) lies on the polar $XU$ of $YZ\cap VW$, etc. Also, note that the problem is purely projective. Master geometer geoishard took advantage of this by taking a projective transformation that (1) keeps $(ABCD)$ a circle, and (2) sends $AC\cap BD$ to the center of the circle $(ABCD)$. Then $ABCD$ becomes a square and we can coordinate bash (or anything else). rrusczyk, WOOT USAMO Strategy Session wrote: The only students I've seen use that stuff [inversion/projective] successfully in the olympiad are students who were master geometers and could have nearly as easily solved the problems without them.

Wow, your solution is so elegant. I also have a solution using harmonic conjugate, but much longer: Denote $I$ as the intersection of $AC, BD$, $J$ as the intersection of $AE, CD$, $CE$ intersects $PD$ at $K$ and $CK$ intersects $AD$ at $L$. We will prove following results: *$IJ, EC, PQ$ are concurrent: From the given conditions, we have $ABCD, ACED$ are harmonic quadrilaterals. Hence, $(ACIP)=(AEJQ)=-1\Rightarrow IJ, CE, PQ$ are concurrent at $K$. *$I, J, K, R$ are collinear and harmonic: $ABCD$ is harmonic $\Rightarrow D(RJIK)=D(ACBD)=-1(1)$ $ACED$ is harmonic $\Rightarrow C(IKJR)=C(AEDC)=-1(2)$ From $(1), (2)$ we have $C(IKJR)=D(IKJR) \Rightarrow$ $I, J, K, R$ are collinear. *Final touch: $D(CKEL)=D(CDEA)=-1(3)$ due to the fact that $ACED$ is harmonic Denote $T$ as the intersection of $RC$ and $BD$, we have $C(TIBD)=C(CABD)=-1$ Hence $R(TIBD)=-1(4)$ But from $(3)$, we also have $R(CKEL)=-1(5)$ From $(4), (5)$ we have $B, E, R$ are collinear

This problem was proposed by our professor who showed that the problem was equivalent to $2+2 = 2*2$ using properties of complete quadrilateral and one of our TST problems as a lemma.

fattypiggy123 wrote: ... one of our TST problems as a lemma. Come on now, you can't leave us hanging like that. Care to share the problem?

Here my solution: I used from poles and polers theory. (1) Nice lemma: $ \Gamma $ - circle and points $A,B,C,D$ lie on the circle $ \Gamma $. If $P$ be intersection point of tangents $ \Gamma $ at the points $B,C$, then $P,A,C$ are collinear if and only if $AB\cdot CD=AD\cdot BC$. Proof: Hint: $ \frac{PC}{PA}=\frac{BC^2}{AB^2}=\frac{CD^2}{AD^2} $. Let $F$ lie on the circle $ \omega $ and $RF$ be tangent to $ \omega $. So, we get from lemma \[ AF\cdot CD=AC\cdot FD, \] \[AC\cdot DE = CE\cdot AD \] and \[ AD\cdot BC=AB\cdot CD .\] $ \Longrightarrow $ \[ \frac{AB}{BC}\cdot \frac{CE}{ED}\cdot \frac{DF}{FA}=1 \] and for $ \triangle ACD $ from theorem Ceva , we have $AE$, $BD$ and $CF$ are concurrent. (at point $X$) Hence, from (1) line $CF$ be poler line of point $R$ with regard to circle $ \omega $ $ \Rightarrow $ it pass at $X$ $ \Rightarrow $ $R$, $B$ and $E$ are collinear. [ it's fact and lemma about poles and polers ].

hatchguy wrote: I believe this was too easy for an apmo 5... Also, inversion about C works, it transforms the problem into a problem about congruent triangles. Not really, without projective it is pretty hard.

v_Enhance wrote: fattypiggy123 wrote: ... one of our TST problems as a lemma. Come on now, you can't leave us hanging like that. Care to share the problem? http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=531674

Let $BR \cap \omega=E',CE' \cap AD=S$. It's well-known that $AC$ is the $A-$symmedian wrt $\triangle ABD \Rightarrow (B,A;D,C)=-1$ Now $E'(B,A;D,C) \cap AD \Rightarrow (R,A;D,S)=-1.$ $C(R,A;D,S) \cap \omega \Rightarrow (C,A;D,E')=-1 \Rightarrow AE'$ is the $A-$symmedian wrt $\triangle ACD \Rightarrow A,E',Q$ are collinear hence $E=E'$, done!

It should be very simple by drawing tangent $RF$ with $F$ on the circle. We then have three harmonic quadrilaterals $ABCD$, $ACED$ and $ACDF$ with equations $AB*CD = AD*BC$, $AC*ED = AD*CE$ and $AC*DF = AF*CD$. From there we get $BC*EF = BF*CE$ or $BCEF$ is yet another harmonic quadrilateral and thus point $R$ lies on the extension of the diagonal $BE$.

For me the main thing is to prove that $Y=AB\cap DE$, $X=BD\cap AE$, and $C$ are collinear. Easy to see that: $ \frac{\sin\angle BYX}{\sin\angle EYX}=\frac{AX\cdot\sin\angle YAX}{DX\cdot\sin\angle YDX}=\frac{AX}{DX}=\frac {AB}{DE}$ as $\bigtriangleup AXB\sim\bigtriangleup DXE$ and similar proof yields $\frac{\sin\angle BYC}{\sin\angle EYC}=\frac{AC\cdot BC}{CD\cdot CE}=\frac{AC\cdot BA}{AD\cdot CE}$ so $\frac{\sin\angle BYX}{\sin\angle EYX}:\frac{\sin\angle BYC}{\sin\angle EYC}=\frac {AB}{DE}:\frac{AC\cdot BA}{AD\cdot CE}=\frac {1}{DE}:\frac{AC}{AD\cdot CE}=1:1$ since $\frac{AC}{AD}=\frac{CE}{DE}$ and so $Y,X,C$ collinear. After that if $AD\cap BE=R'$ then $\bigtriangleup XYR'$ are self polar w.r.t. $\omega$, or $XY$=polar of $R'$, so $R'\in CQ\Rightarrow R=R'$.

$ ABCD $ be a quadrilateral inscribed in a circle $\omega \cdot \cdot \cdot (1) $ $ PB $ and $ PD $ are tangent to $\omega \cdot \cdot \cdot (2) $ $ QC $ and $ QD $ are tangent to $\omega \cdot \cdot \cdot (3) $ Let $ S $ be a point such that $ S=AB\cap CQ $ and let $T$ be a point such that $BD\cap CQ$, let $U$ be a point such that $AC\cap BT$, let $W$ be a point such that $BE\cap CQ$. \[(S,R;C,T)\\\stackrel{A}{=} (B,D;U,T)\\\stackrel{C}{=} C(B,D;A,C)\\=B(B,D;A,C) (\because (1)) \\\stackrel{B}{=}(P,U;A,C) \\= -1 (\because (2)) \\= (A,E;V,Q) (\because (3)) \\\stackrel{D}{=}D(A,E;C,D)\\=B(A,E;C,D) (\because (1)) \\\stackrel{B}{=}(S,W;C,T) \\\therefore (S,R;C,T)=(S,W;C,T) \\\therefore R=W \] Hence, by $W=BE\cap CQ$, $B,E,R$ are collinear. $(Q,E,D)$

hyperspace.rulz wrote: Let $R'=AD \cap BE$. Note that the tangents at $B$ and $D$ concur on $AC$ at $P$, so $ABCD$ is harmonic, hence the tangents at $C$ and $A$ concur on $BD$ at $X$, say. Now apply Pascal's Theorem to hexagon $AAEBDD$ to find that $X=AA \cap BD$, $Q=AE \cap DD$ and $R'=BE \cap AD$ are collinear. Now note that $X$ and $Q$ both lie on the tangent at $C$, hence $R'$ also lies on the tangent at $C$. It follows that $R'=CC \cap AD \cap BE=CC \cap AD=R$. So $R'$ and $R$ are in fact the same point. Since $R'=R$ lies on $BE$ by definition, it follows that $B, E, R$, are indeed collinear, and thus the problem is solved. Me too!

Take a homography fixing $\omega$ that sends $ABCD$ to a square. Thus it suffices to prove the problem when $ABCD$ is a square. From Power of a Point on $R$ with respect to $\omega$ we have that $CDR$ is an isosceles right triangle, and thus $\tan \angle ARB = \tfrac{1}{2}$. Since $\overline{AQ}$ is an $A$-symmedian in $\triangle ADC$, it follows that $\tan \angle DAE = \tfrac{1}{3}$. By the inscribed angle theorem, $\angle REA = 135^{\circ}$. It suffices to show that $\angle ARB + \angle DAE + \angle REA = 180^{\circ}$, or $\angle ARB + \angle DAE = 45^{\circ}$. But this is simple complex numbers: since $(2+i)(3+i)=5+5i$ has an argument of $45^{\circ}$, we are done.

Let $F = BC\cap AD$ and phantom $R' = BE\cap AD$, then we have $$(DF;AR') \overset{B}{=} (DC;AE) = -1 = (DB;AC)\overset{C}{=}(DF'AR)$$ so $R=R'$.

Note that $ABCD$ is harmonic. Redefine $E$ to be the second interesction of $BR$ and $(ABCD)$. Let $K = BD \cap AC$, and let $T = AA \cap BD$. Note that $TC$ must also be tangent to $(ABCD)$, so $T$, $C$, $Q$ and $R$ are collinear. Then \[ (AE;CD) \stackrel{B}{=} (AB\cap TR, R; CT) \stackrel{A}{=} (BD; KT) \stackrel{A}{=} (AC;BD) = -1 \]So $(AE;CD)$ is harmonic, which implies the conclusion. $\blacksquare$

If we suppose $E' = BR \cap (ABC)$, it suffices to show $ACE'D$ is harmonic. Noting $ABCD$ is harmonic, we get \[-1 = (A,C;B,D) \overset{E'}{=} (A, CE \cap AR; R, D) \overset{C}{=} (AE';CD).~\blacksquare\]

Let $R' = AD \cap BE$. $$-1=(A,E;C,D) \stackrel{B}{=} (A,R';AD \cap BC,D) \stackrel{C}{=} (A,CR' \cap \omega;B,D)$$Since $(A,C;B,D)=-1$, $CR'$ is tangent to $\omega$ at $C$, so $R'=R$. $\square$

Trig bash. We wish to show that $\frac{\sin CBE}{\sin EBD} = \frac{\sin CBR}{\sin RBD}$. Notice that $AE$ is the $A$-symmedian of $\triangle ADC$. By LoS in $\triangle BCR$ and $\triangle BDR$ we have $RHS = \frac{CR}{RD} \cdot \frac{\frac{BR}{\sin BCR}}{\frac{BR}{\sin BDR}} = \frac{CR}{RD} \cdot \frac{AB}{BC}.$ By the symmedian line we have $LHS = \frac{\sin CAE}{\sin EAD} = \frac{AC}{AD}.$ By LoS in $\triangle CDR$ we have $\frac{CR}{RD} = \frac{\sin CDR}{\sin DCR} = \frac{AC}{CD}.$ Now multiplying the equations gives that the desired result is equivalent to $AD \cdot BC = CD \cdot AB$. This follows since $AP$ is the $A$-symmedian of $\triangle ADB$, finishing.

Consider a homography which fixes $\omega$ and takes $\overline{AC} \cap \overline{BD}$ to the center of $\omega$. Then $ABCD$ becomes a square since it must be a rectangle and also harmonic, so $R$ must be the reflection of $A$ across $D$ and $Q$ must be the midpoint of $\overline{CR}$. Now, redefine $E = \overline{BR} \cap \omega$, so it suffices to show that $E$ lies on $AQ$. Since $\overline{BD}$ is a diameter of $\omega$, we have $\angle RED = 90^{\circ} = \angle RQD$, and $RQED$ is cyclic. Thus, $$\angle REQ = \angle RDQ = 45^{\circ} = \angle AEB,$$which implies that $A, E$, and $Q$ are collinear, as desired.

Since $\overline{PB},\overline{PD}$ are tangents and $P,A,C$ are collinear, we have that $(A,C;B,D)=-1$. Similarly $\overline{QD},\overline{QC}$ are tangents, and $Q,E,A$ are collinear, we have that $(C,D;E,A)=-1$. Let $S$ be the intersection of lines $AD,BC$. Projecting $(A,C;B,D)$ onto line $AD$ taking perspective at $C$, $A,B,C,D$ go to $A,S,R,D$. Hence, $(A,R;S,D)=(A,C;B,D)=-1$. Let $B,E'$ be the intersections of line $BR$ and $(ABCD)$. Projecting $(A,R;S,D)$ onto the circumcircle $(ABCD)$ taking perspective at $B$, gives $(A,E'C,D)=-1=(A,E;C,D)\Longrightarrow E'\equiv E$. Hence, $B,E,R$ are collinear. $\blacksquare$

Attachments:

Denote $BD \cap CR = T$, $AC \cap BD = K$, $AB \cap CR = M$ and $BR \cap \omega = E'$. Now ABCD is harmonic, from PD and PB being tangents, T, K, B, D are on one line and (T,K; B,D) = -1. Also (T,K ; B, D) = (T,C; M, R) = (D, C; A, E') (projecting through A and B respectively). From QD and and QC being tangents we have that DCEA is harmonic quadrilateral $\Rightarrow$ E'=E $\Rightarrow$ B, E, R lie on one line and we are ready.

Since the conditions are purely projective, we take a homography that sends $ABCD$ to a square, with the rest of the diagram being straightforward to construct. So then we coordbash by (WLOG) letting $AC \cap BD = (0, 0)$ and $A = (0, 1)$ and $B = (1, 0)$. It follows that $R = (-2, 1)$ so line $BR$ has equation $y = \frac{1}{3}x - \frac{1}{3}$. Now, by PoP we get that $QE \cdot QA = QC^2 = 1$ and $QA = \sqrt{5}$ so $QE = \frac{\sqrt{5}}{5}$. We then draw the altitude from $E$ to $RC$ and use similar triangles to find that $E = (-\frac{4}{5}, -\frac{3}{5})$ which clearly lies on line $BR$ as desired.

$(A,C;B,D)=-1=(D,C;E,A)$ so project through $R$ back onto $\omega.$

Observe since $(AC;BD)=-1$, and all other points can be defined in terms of tangencies, and intersections this problem is purely projective. Now affine transform $ABCD$ to the cartesian plane with $A=(1,1)$, $B=(-1,1)$, $C=(-1,-1)$, $D=(1,-1)$. Now note that $P=\infty_{AC}$, $Q=(0,-2)$, $R=(1,-3)$, $E=(\tfrac{1}{5}, -\tfrac{7}{5})$. Note that $B$, $E$, $R$ all satisfy the linear equation $y=-2x-1$, therefore they are collinear and we are finished.

Let $R'$ be intersection of $AD$ and $BE$. We wish to show $R'C$ tangent to $(ABCD)$. Let $X$ be on $BD$ with $XA$ tangent. Then Pascal on $AAEBDD$ gives $X, Q, R'$ collinear. But by La Hire we know $XC$ is tangent. Also $QC$ is tangent. So $R'C$ is tangent and we are done.

Let $DC$ meet $AQ$ at $F$ and $AC$ meet $BD$ at $G$. Define $E$ to be the intersection of $BR$ with $AQ$. By Pascal (converse) on $BEACCD$, it suffices to prove that $G$, $F$ and $R$ are collinear. We will do this using Menelaus in $\triangle QAC$. It suffices for \[\frac{AG}{GC}\cdot\frac{CR}{RQ}\cdot\frac{QF}{FA}=1\]We compute each ratio one at a time \begin{align*} \frac{AG}{GC}&=\frac{DA}{DC}\cdot\frac{BA}{BC}\\ \frac{CR}{QQ}&=\frac{\sin(\angle RDC)}{\sin(\angle RDQ)}\cdot\frac{DC}{DQ}=\frac{\sin(\angle ADC)}{\sin(\angle DBA)}\cdot\frac{DC}{DQ}=\frac{AC}{DA}\cdot\frac{DC}{DQ}\\ \frac{QF}{FA}&=\frac{\sin(\angle QCF)}{\sin(\angle DCA)}\cdot\frac{CQ}{CA}=\frac{\sin(\angle DAC)}{\sin(\angle DCA)}\cdot\frac{CQ}{CA}=\frac{DC}{DA}\cdot\frac{CQ}{CA} \end{align*} Therefore, the product is \[\frac{DA}{DC}\cdot\frac{BA}{BC}\cdot\frac{AC}{DA}\cdot\frac{DC}{DQ}\cdot\frac{DC}{DA}\cdot\frac{CQ}{CA}=\frac{BA}{BC}\cdot\frac{DC}{AD}=1\]Since $ABCD$ is harmonic. $\blacksquare$

We have the harmonic bundles \[(D, C; A, E) = -1 = (A, C; B, D) \overset{E}{=} (Q, C; BE \cap CC, DE \cap CC)\]\[\overset{D}{=} (D, C;D(BE \cap CC) \cap \omega, E),\] which gives $D(BE \cap CC) \cap \omega = A \implies BE \cap CC = R$, so $B$, $E$, $R$ collinear. $\blacksquare$