Denote by $\omega$ the circumcircle of $ABCD$. Let $\{T\} = DQ \cap \omega$. By converse of Reim's Theorem on the parallel lines $PK \mid \mid CD$ and circle $\omega$ we have that $BDTK$ is cyclic. By converse of Reim's Theorem on the parallel lines $LQ \mid \mid BD$ and circle $\omega$ we have that $CQTL$ is cyclic. Now because $\angle{ACT}=\angle{ABT}$ we have that the lines tangent to the circumcircles of $QCT$ and $BDT$ at $T$ coincide, thus the circumcircles of the triangles $BKP$ and $CLQ$ are tangent at $T$.

Pretty easy problem compared to the ones from the last year. A bit too easy. Let $DP$ intersect $(ABCD)$ in $D$ and $S$. We can easily observ that $S$ is on both circumcircles of $BKP$ and $CLQ$. Indeed, $\angle PSB=180-\angle BCB=\angle PKB$, since $PK\parallel CD$, so $P,B,K,S$ are concyclic, and $\angle QSC=\angle DBC=\angle QLC$, so $Q,C,L,S$ are concyclic. Now, for an easier explanation, if we let $d$ be the tangent in $S$ at $(BKP)$, and $R$ a point on $d$ such that $L,R$ are on opposite sides wrt $SB$, we'll have $\angle RSP=\angle ABS=\angle ACS=\angle QLS$, so $d$ is tangent in $S$ at $(CLQ)$. This concludes the tangency of $BKP$ and $CLQ$.

The intersection of $PD$ and $(ABCD)$ is $X$. $\measuredangle PKB = \measuredangle PXB$, so $PKBX, CQLX$ are cyclic. Now $XK, XB$ are isogonal in $XLC$ so we are done.

Too easy.Let $T=DP\cap \odot (ABCD) $.Easy angle-chasing implies that $CLTQ$,$BKTP $ are cyclic.Again chasing some angles $\angle LTK=\angle BAC=\angle CTB$ which yields that $TK,TB $ are isogonals wrt $TLC $ which finishes problem.

Letting $O=DP\cap BC$ (assuming it exists), then Power of a Point wrt secants $OD,OC$ shows that $(BKPT)$ and $(CLQT)$ are cyclic, where $T=(ABCD)\cap OD$. (Since the diagram varies continuously as $P$ varies continuously, this proves they are cyclic even if $DP$ and $BC$ are parallel.) To show tangency, note that by angle chasing $PTK\sim ATC$ and $ATB\sim QTL$. Hence there exist spiral similarities $\phi,\psi$ centered at $T$ with $\phi(PTK)=ATC$ and $\psi(ATB)=QTL$. Then $\phi\circ \psi$ maps $P\to Q$, hence it's a homothety, and circles $(PTK)$ and $(QTL)$ are homothetic (with center $T$), as desired.

Let $E=PD\cap\left(ABCD\right)$. Then \[\measuredangle{PEB}=\measuredangle{DEB}=\measuredangle{DCB}=\measuredangle{PKB},\]so $EPBK$ is cyclic. Similarly, $EQCL$ is cyclic. Let $\ell_B,\ell_C$ be the tangents to $\left(EPBK\right),\left(EQCL\right)$ at $E$. Then \[\measuredangle{\left(\ell_B,ED\right)}=\measuredangle{EBP}=\measuredangle{EBA}=\measuredangle{ECA}=\measuredangle{ECQ}=\measuredangle{\left(\ell_C,ED\right)},\]so $\ell_B=\ell_C$ and hence $\left(BKP\right)$ and $\left(CLQ\right)$ are tangent.

asdf took way too long (and Geogebra help) for me to realize that phantom pointing basically kills this problem; time to draw better diagrams I guess Let $X = PQD\cap\odot(ABCD)$. Note that \[\angle PXB \equiv\angle DXB = \angle DAB = \angle PKB,\]so $PBKX$ is cyclic. Similarly, $QCLX$ is cyclic. We can thus get rid of $K$ and $L$, since it suffices to show that the circles $\odot(BPX)$ and $\odot(CQX)$ are tangent to each other. But upon letting $O_B$ and $O_C$ be their respective centers, we obtain \[\angle O_BXP = 90^\circ - \angle XBP = 90^\circ - \angle XCQ = O_CXQ,\]so $X$, $O_B$, and $O_C$ are collinear, implying the tangency. $\blacksquare$

$PD \cap \odot ABCD = \{ D,X \}$ By Reim's theorem, $X \in \odot BKP$ and $X \in \odot QLC$ Angle chasing yields, $\angle AXC = \angle PXK$ and $\angle AXB = \angle QXL$ which implies $K$ and $B$ are isogonal in $\triangle XLC$ it leads us to our conclusion.

huh no inversion?

also @post 2 I think you meant $BPTK$ cyclic not $BDTK$ cyclic

Nice Problem Here is a boring solution. Denote the intersection of $PQ$ with $(ABCD)$ as $G$, other than $D$. We claim that the tangency point is $G$. Claim 01. $G$ lies on both $(BKP)$ and $(CLQ)$. In other words, $BKGP$ and $CLGQ$ are both cyclic. Proof. Notice that \[ \measuredangle BKP = \measuredangle BCD = \measuredangle BAD = \measuredangle BGD \equiv \measuredangle BGP\]Similarly, \[ \measuredangle LQG = \measuredangle BDG = \measuredangle BCG \equiv \measuredangle LCG \] Claim 02. Let $GK$ intersects $(CQL)$ at $H$. Then , $HQ \parallel PK$. Proof. We'll prove this by phantom point. Notice that \[ \measuredangle GHQ = \measuredangle GKP = \measuredangle GBP = \measuredangle GBA = \measuredangle GCA = \measuredangle GCQ \]which is what we wanted. Therefore, $G$ sends $KP$ to $HQ$, which makes $G$ the tangency point of the two circle.

Storage. RMM 2018 Day 1 P1 wrote: Let $ABCD$ be a cyclic quadrilateral an let $P$ be a point on the side $AB.$ The diagonals $AC$ meets the segments $DP$ at $Q.$ The line through $P$ parallel to $CD$ mmets the extension of the side $CB$ beyond $B$ at $K.$ The line through $Q$ parallel to $BD$ meets the extension of the side $CB$ beyond $B$ at $L.$ Prove that the circumcircles of the triangles $BKP$ and $CLQ$ are tangent . Let $DP\cap\odot(ABC)=X$. Then $\angle KPX=\angle CDX=\angle XBK\implies X,K,B,P$ are concyclic. And $\angle XQL=\angle XDB=\angle XCL\implies L,C,Q,X$ are concyclic, Now $$\angle LXK=\angle LXQ-\angle KXP=(180^\circ-\angle BCA)-(180^\circ-\angle ABL)=\angle ABL-\angle BCA=\angle BAC=\angle BXC$$So, $\{XK,XB\}$ are isogonal WRT $\triangle LXC\implies\odot(BKP)$ and $\odot(CLQ)$ are tangent to each other at $X$. $\blacksquare$

1)Let $(CLQ)n(ABCD)=X$ ,prove that $(BPXK)$-cyclic and $(BPXK)$ tangent to $(CLQ)$. 2)Prove that $X-Q-D$ collnear $\angle CLQ=\angle CXQ$ but $\angle CXD=\angle CAD$,which implies that $\angle CXD=\angle CXQ$,which implies the desired collinearity 3)Prove that $(BKXP)-cyclic$ $\angle DCY=\angle DAB=\angle BKP$ and its obvious that $\angle BXP=\angle BAD$,which implies the desired claim 4)Let $\ell$ be a line that passes through $X$ and tangent to $(CQL)$.Prove that $\ell$ is tangent to $(KBPX)$ at $X$. $\angle QLX=\angle QX=\angle QCX=\angle ACX=\angle ABX=\angle PKX$

Let $PQ \cap \odot{ABCD}=E$. I claim that $E$ is the point of tangency. First I will prove that $EQCL$ is cyclic $$\angle{QLC}=\angle{DBC}=\angle{DEC}$$Similarly, $$\angle{BKP}=180-\angle{DCB}=180-(180-\angle{DEB})=\angle{DEB}$$Hence $EPBK$ is cyclic. To finish, we present the following claim : Claim : $\angle{PBE}=\angle{QLE}$ Proof. $$\angle{QLE}=\angle{QCE}=\angle{ABE}=\angle{PBE}$$Now, draw a tangent to $\odot{BKP}$ at $E$. Let $G$ be an arbitrary point on the tangent inside $\odot{ABC}$. We see that $$\angle{GEQ}=\angle{GEP}=\angle{EBP}=\angle{ELQ}$$So we are done $\blacksquare$

Nice diagram = Problem done! (And this time it's on paper!) Let $\odot(ABC) \cap \odot(LQC) = X$ We claim that $X$ is the desired tangency point Claim: $D-Q-P-F$ are collinear

Hence One of $\odot(CLQ) \cap \odot(BKP)=X$ Now draw the line that is tangent to $\odot(LQC)$ at $T$ and name it $\ell$ and let a point on $\ell$ on the same side of $F$ as $L$ be $Y$ We know that $$\angle YFL= \angle XCL = \angle XQL = \angle XDB$$and after some easy angle chasing we get $\angle XPK = \angle XDC = \angle YXL + \angle CXB$ Therefore it suffices to show that $XK,XB$ are isogonal in $\triangle XLC$ This is true after some angle chasing which I am too lazy to write, just assume variables and calculate each angle, which finally gives the desired conclusion $\blacksquare$

The line $DP$ meets the circumcircle of triangle $CLQ$ at $R$. We get $\angle RDB = \angle RQL = \angle RCL = \angle RCB \Rightarrow$ $R$ lie on $\odot(ABCD)$. Since $PK \parallel CD$, by the Reim's Theorem we get that $RBKP$ is cyclic. Finally, let $X$ be the intersection of the tangent to $\odot(CLQ)$ at $R$ with the line $CD$. Thus $\angle XRQ = \angle RCQ = \angle RCA = \angle RBA = \angle RBP \Rightarrow$ $XR$ is tangent to$\odot (RBP)$. Therefore, the circumcircles of the triangles $BKP$ and $CLQ$ are tangent to the line $XR$ at $R$ $\blacksquare$

Let $DQ$ meet $ABCD$ at $S$. $\angle QLC = \angle DBC = \angle QSC \implies SLCQ$ is cyclic. $\angle BKP = \angle 180 - \angle BCD = \angle BSP \implies BKSP$ is cyclic. Let $L_1$ be line tangent to $SLCQ$ at $S$ we have $\angle SCQ = \angle SCA = \angle SBA = \angle SBP \implies L_1$ is tangent to $BKSP$ as well so our circles are tangent at $S$. we're Done.

Let $E= DQ \cap \odot(ABCD) \ne D$. We will prove that $E$ is the common point of tangency. [asy][asy] 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 = 1.9, xmax = 11.2, ymin = -2.5862804185721906, ymax = 5.031020679728067; /* image dimensions */ pen cqcqcq = rgb(0.7529411764705882,0.7529411764705882,0.7529411764705882); /* draw figures */ draw(circle((5.852809635608784,1.367535756966066), 3.5453838571445635), linewidth(0.8) + blue); draw((3.741438563859456,4.215668284058398)--(9.,3.), linewidth(0.8) + green); draw((9.,3.)--(6.68,-2.08), linewidth(0.8) + green); draw((6.68,-2.08)--(2.373492429373072,0.6862880245710273), linewidth(0.8) + green); draw((3.741438563859456,4.215668284058398)--(2.373492429373072,0.6862880245710273), linewidth(0.8) + green); draw((9.,3.)--(2.373492429373072,0.6862880245710273), linewidth(0.8) + green); draw(circle((8.003294213009466,3.084537618648431), 1.0002844769300214), linewidth(0.8) + red); draw((9.,3.)--(9.306489730729668,3.6711068241839295), linewidth(0.8) + green); draw((9.306489730729668,3.6711068241839295)--(4.739858471858444,2.07662144581455), linewidth(0.8) + green); draw((8.719877477199834,2.3866282690410148)--(7.070615206872309,3.4460329938583647), linewidth(0.8) + green); draw((4.739858471858444,2.07662144581455)--(8.141511444765205,4.075226788687628), linewidth(0.8) + green); draw((8.141511444765205,4.075226788687628)--(9.,3.), linewidth(0.8) + green); draw((2.373492429373072,0.6862880245710273)--(4.739858471858444,2.07662144581455), linewidth(0.8) + green); draw(circle((7.702470722917133,0.9283479023436838), 3.1773579402953915), linewidth(0.8) + red); draw((3.741438563859456,4.215668284058398)--(6.68,-2.08), linewidth(0.8) + green); draw((8.141511444765205,4.075226788687628)--(6.68,-2.08), linewidth(0.8) + green); /* dots and labels */ dot((3.741438563859456,4.215668284058398),dotstyle); label("$A$", (3.551178767049365,4.389270037746726), NE * labelscalefactor); dot((9.,3.),dotstyle); label("$B$", (9.14557023301717,2.8686000382692027), NE * labelscalefactor); dot((6.68,-2.08),dotstyle); label("$C$", (6.606469866917069,-2.4049161067078986), NE * labelscalefactor); dot((2.373492429373072,0.6862880245710273),dotstyle); label("$D$", (2.0584109693971078,0.5666683876839601), NE * labelscalefactor); dot((7.070615206872309,3.4460329938583647),dotstyle); label("$P$", (6.941296288820379,3.6219594875516457), NE * labelscalefactor); dot((4.739858471858444,2.07662144581455),linewidth(4.pt) + dotstyle); label("$Q$", (4.471951427283468,2.101289488074122), NE * labelscalefactor); dot((8.719877477199834,2.3866282690410148),linewidth(4.pt) + dotstyle); label("$K$", (8.852597113851774,2.198947194462587), NE * labelscalefactor); dot((9.306489730729668,3.6711068241839295),linewidth(4.pt) + dotstyle); label("$L$", (9.368787847619377,3.7893726985033), NE * labelscalefactor); dot((8.141511444765205,4.075226788687628),linewidth(4.pt) + dotstyle); label("$E$", (8.196895370957792,4.18000352405716), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Claim: $E \in \odot(PBK)$ and $E \in \odot(CQL)$. Proof: For $PBKE$ we have \[ \measuredangle DEB = \measuredangle DCB = \measuredangle PKB. \]For $CQLE$ we have \[ \measuredangle ECL = \measuredangle ECB = \measuredangle EDB = \measuredangle EQL. \]This shows the claim. $\square$ To finish it off, it suffices to show that $\measuredangle EBP = \measuredangle ECQ$ by considering a tangent to either $\odot(PBKE)$ or $\odot(CQLE)$ at $E$. This part is obviously true as \[\measuredangle EBP = \measuredangle EBA = \measuredangle ECA = \measuredangle ECQ.\]So, we are done. $\blacksquare$

Let $T=(ABC)\cap\overline{PQ}.$ Note $$\measuredangle TPK=\measuredangle PDC=\measuredangle TBC=\measuredangle TBK$$and $$\measuredangle CKQ=\measuredangle CBD=\measuredangle CTD$$so $T$ lies on $(BKP)$ and $(CLQ).$ Let $\ell$ be the line tangent to $(CLQ)$ at $T$; we claim $\ell$ is tangent to $(BKP)$ at $T.$ Indeed, $$\measuredangle (\overline{DT},\ell)=\measuredangle QCT=\measuredangle ACT=\measuredangle ABT.$$$\square$

. This follows by $$\measuredangle XBP=\measuredangle XBA=\measuredangle XCA=\measuredangle XCQ.~\blacksquare$$

Let $E = \overline{DP} \cap (ABC)$. I claim that $E$ lies on both $(BKP)$ and $(CLQ)$. This is because $$\measuredangle PKB = \measuredangle DCK = \measuredangle DEB$$and similarly $\measuredangle QLC = \measuredangle DBC = \measuredangle DEC$. Thus it suffices to show that the tangent at $E$ to $(BKP)$ is also tangent to $(CLQ)$, which is equivalent to $\measuredangle EBP = \measuredangle ECQ$. This is evident.

Let $X = DP\cap \omega$. Then, note that Claim 1: $X\in (PBK)$ Proof: Angle chase with the following directed angles: \[\angle PKB = \angle DCB = \angle DAB = \angle DXB = \angle PXB\] Claim 2: $X\in (QLC)$. Proof: Angle chase: \[\angle XQL = \angle XDB = \angle XCB = \angle XCL\] Thus, $X = (PKB)\cap (QLC)$. Let $\ell_1$ be the tangent to $(PKB)$ at $X$ and $\ell_2$ be the tangent to $(QLC)$ at $X$. Then, we have \[\angle (\ell_1, XP) = \angle XBP = \angle XBA = \angle XCA = \angle XCQ = \angle (\ell_2, XQ)\]Since $X,P,Q$ are collinear, this means that $\ell_1$ and $\ell_2$ are the same line, which means that $(PKB)$ and $(QLC)$ share a tangent line with the same orientation at $X$, and therefore the two circles are tangent. $\blacksquare$.

Let $T= \odot(PKB)\cap \odot(ABCD), S=\odot(QLC)\cap \odot(ABCD$ Since $PK//DC\Rightarrow \angle PKB=\angle PTB=\angle DAB$ So $T,P,D$ is collinear Similarly, $S,Q,D$ is collinear It shows that$T=S$ Finally,suppose $l_{1},l_{2}$ through point $T$,and tangent to $\odot(PKB)$,$\odot(QLC)$ respectively $\measuredangle (l_{1},TK) = \angle TDC= \angle TDB+ \angle BDC= \angle LTK+ \angle TCL=\measuredangle (l_{2},TK)\Rightarrow l_{1}=l_{2}$ Which meas that $\odot(PKB)$and $\odot(QLC)$ are tangent.

Let $E = (ABCD) \cap DP$. Then \[\angle EPK = \angle EDC = 180^\circ - \angle EBC = \angle EBK,\]so $(EPBK)$ is cyclic. Similarly \[\angle EQL = \angle EDB = \angle ECB = \angle ECL\]gives $(EQCL)$ cyclic. Now invert at $E$; in the inverted diagram, we have $ABCD$ collinear, $PEQD$ collinear, $(ABEP)$ cyclic, and $(ACQE)$ cyclic. Then $\angle ACQ = 180^\circ - \angle AEQ = \angle AEP = \angle ABP$, so $BP \parallel CQ$. Inverting back gives $(EPB) = (BKP)$ tangent to $(ECQ) = (CLQ)$, as desired.

Let $DP$ meets $(ABCD)$ at $T$. Then Reim implies $BKPT$ and $CLQT$ are cyclic. Now by homothety centered $T$, it suffices to show that $\angle TBP = \angle TCQ$, which follows from $\angle TBP = \angle TBA = \angle TCA = \angle TCQ$. $\blacksquare$

All directed angles Rating (MOHs): 0

Angle-Chase: Let $X=DQ \cap (ABC)$. Then: $$\angle XQL = \angle XDB = \angle XCB = \angle XCL \implies X \in (CQL)$$$$\angle PXB = \angle DXB = 180^\circ - \angle DCB = 180^\circ - \angle BKP \implies X \in (BKP).$$Let $T$ be a point on $CD$ such that $TX$ is tangent. Then: $$\angle TXQ=\angle TXP = \angle XBP = \angle XBA = \angle XCA = \angle XCQ$$which implies $XT$ is also tangent to $(CQL)$ and we are done.

Let $X=(DP)\cap (ABCD)$ other than $D.$ $(PK)\parallel CD$ so by Reim's theorem, we have $XPBK$ is cyclic. Similarily, $(QL) \parallel (BD)$ implies, by Reim's theorem, that $(XQCL)$ is cyclic. Thus, it suffices to prove that $\angle XBP=\angle XCQ.$ Indeed, \[\angle XBP= \angle XBA=\angle XCA=\angle XCQ.\] [asy][asy] 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 = -28.672699506166545, xmax = 30.304799479546876, ymin = -13.213124589216047, ymax = 15.349276752032559; /* image dimensions */ pen ffdxqq = rgb(1,0.8431372549019608,0); pen ffqqtt = rgb(1,0,0.2); pen wwqqcc = rgb(0.4,0,0.8); pen ffzzqq = rgb(1,0.6,0); pen ccffww = rgb(0.8,1,0.4); draw(circle((-4.374514807706929,4.554578946863126), 7.441362978795981), linewidth(0.8) + ffdxqq); draw(circle((-10.265245927810327,1.1410398077282977), 2.971023698496136), linewidth(0.8) + ffzzqq); draw(circle((-7.480815345801481,-3.611376349266896), 8.479064456814383), linewidth(0.8) + ccffww); /* draw figures */ draw((-8.56658899673541,10.702781690419702)--(0.584132797532879,-0.9939070546732326), linewidth(0.8)); draw((-8.56658899673541,10.702781690419702)--(-8.787977427241902,-1.4366839156862143), linewidth(0.8)); draw((2.9926105352026013,3.5060813722090263)--(-8.694970695175076,3.6631852259780002), linewidth(0.8)); draw((0.584132797532879,-0.9939070546732326)--(2.9926105352026013,3.5060813722090263), linewidth(0.8) + ffqqtt); draw((0.584132797532879,-0.9939070546732326)--(-8.787977427241902,-1.4366839156862143), linewidth(0.8)); draw((-8.787977427241902,-1.4366839156862143)--(2.9926105352026013,3.5060813722090263), linewidth(0.8) + wwqqcc); draw((-15.75661080980647,-1.7659106896656425)--(-8.787977427241902,-1.4366839156862143), linewidth(0.8)); draw((-15.75661080980647,-1.7659106896656425)--(-2.9993751879215687,3.586625329960073), linewidth(0.8) + wwqqcc); draw((-8.694970695175076,3.6631852259780002)--(-11.49291255479232,-1.5644761264366276), linewidth(0.8) + ffqqtt); draw((-11.767160966015583,3.7044814464317337)--(-8.787977427241902,-1.4366839156862143), linewidth(0.8)); draw((-8.694970695175076,3.6631852259780002)--(-11.767160966015583,3.7044814464317337), linewidth(0.8)); /* dots and labels */ dot((-8.56658899673541,10.702781690419702),linewidth(4pt) + dotstyle); label("$A$", (-9.721160237851564,10.678938154341909), NE * labelscalefactor); dot((-8.787977427241902,-1.4366839156862143),linewidth(4pt) + dotstyle); label("$B$", (-9.25798616204753,-2.5601208457233233), NE * labelscalefactor); dot((0.584132797532879,-0.9939070546732326),linewidth(4pt) + dotstyle); label("$C$", (0.9704413452915136,-1.9425554113179482), NE * labelscalefactor); dot((2.9926105352026013,3.5060813722090263),linewidth(4pt) + dotstyle); label("$D$", (3.1319203657103305,3.808522696582109), NE * labelscalefactor); dot((-8.694970695175076,3.6631852259780002),linewidth(4pt) + dotstyle); label("$P$", (-9.605366718900555,4.117305413784797), NE * labelscalefactor); dot((-2.9993751879215687,3.586625329960073),linewidth(4pt) + dotstyle); label("$Q$", (-2.850744780091752,3.8857183758827807), NE * labelscalefactor); dot((-11.49291255479232,-1.5644761264366276),linewidth(4pt) + dotstyle); label("$K$", (-11.419465182466347,-2.6759143646743313), NE * labelscalefactor); dot((-15.75661080980647,-1.7659106896656425),linewidth(4pt) + dotstyle); label("$L$", (-16.63017353526171,-2.6373165250239956), NE * labelscalefactor); dot((-11.767160966015583,3.7044814464317337),linewidth(4pt) + dotstyle); label("$X$", (-12.615998211626763,3.847120536232445), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy]