Let the tangent to $\omega_1$ at $K$, meet $AC$ at $E$. It's clear that $KE=EA$ and since $ABDC$ is cyclic , we get that $\widehat{LCA}=\widehat{ADC}=\widehat{ABC}=\widehat{AKE}$, thus if $P$ is the intersection of $AK$ and $CL$, $EPKC$ is cyclic, and let the intersection of $BC$ and $AD$ be $Q$, by angle chasing, it's clear that $QLPK$ is also cyclic, so if we look at the triangle $ACQ$ we get that $AEPL$ is also cyclic, so $CE=LE$, thus the tangents at $K$ and $L$ meet on $AC$. Thus we only need to prove that $X$, $E$ and $Z$ are collinear. Let the intersection of $XZ$ and $AC$ be $E'$. Since $AXCZ$ is cyclic, we get that $AE'.CE'=XE'.ZE'$, thus the triangles $E'YC$ and $E'AX$ are similar. Thus $\frac{AE'}{CE'}=\frac{sin\widehat{AYE'}}{sin\widehat{CYE'}}.\frac{AY}{CY}=\frac{AE'}{LE'}$ since the triangles $AE'L$ and $CE'K$ are similar, thus $LE'=CE'$, hence $E\equiv E'$ and we are done.

ayeen_izady wrote: Let the tangent to $\omega_1$ at $K$, meet $AC$ at $E$. It's clear that $KE=EA$ and since $ABDC$ is cyclic , we get that $\widehat{LCA}=\widehat{ADC}=\widehat{ABC}=\widehat{AKE}$, thus if $P$ is the intersection of $AK$ and $CL$, $EPKC$ is cyclic, and let the intersection of $BC$ and $AD$ be $Q$, by angle chasing, it's clear that $QLPK$ is also cyclic, so if we look at the triangle $ACQ$ we get that $AEPL$ is also cyclic, so $CE=LE$, thus the tangents at $K$ and $L$ meet on $AC$. Thus we only need to prove that $X$, $E$ and $Z$ are collinear. Let the intersection of $XZ$ and $AC$ be $E'$. Since $AXCZ$ is cyclic, we get that $AE'.CE'=XE'.ZE'$, thus the triangles $E'YC$ and $E'AX$ are similar. Thus $\frac{AE'}{CE'}=\frac{sin\widehat{AYE'}}{sin\widehat{CYE'}}.\frac{AY}{CY}=\frac{AE'}{LE'}$ since the triangles $AE'L$ and $CE'K$ are similar, thus $LE'=CE'$, hence $E\equiv E'$ and we are done. Why $\triangle AE'L\sim\triangle CE'K$? I don't think it's trivial.

Claim: The tangents at $K$ and $L$ intersect on $AC$. Proof: Let the tangent at $L$ to $\omega_2$ intersect $AC$ at $S$. We'll prove that $SK$ is tangent to $\omega_1$. Let $AL\cap BC=V$ and $AK\cap LC=U$. $\angle LCA=\angle LDC=\angle KBA=\angle CAU\implies \angle LUA=2\angle ABC=\angle AVK$ Hence $L,U,K,V$ are cyclic. Also $\angle SAU=\angle ABC=\angle LDC=\angle SLU$ so $A,S,U,L$ are cyclic. These two give that $C,S,U,K$ are also cyclic. $\angle UKS=\angle UCS=\angle CAK$ so $SA=SK$ which gives that $SK$ is tangent to $\omega_1$. We get that $S\in AC$. Let $T$ be the intersection of $AC$ and $BD$. $AD\cap BC=O, AL\cap \omega_1=Q$ By Pascal at $QKKBAA$, we get $QK\cap BA, S, O$ are collienar which gives that $SO\parallel AB\parallel DC$. It means $TO\perp OS$. $\angle AOS=\angle SOC$ and $\angle TOS=90\implies (T,S;A,C)=-1$. Let $XY$ intersect $AC$ at $M$ where is the midpoint of $AC$. We have $TA.TC=TS.TM$ Take the inversion centered at $T$ with radius $\sqrt{TA.TC}$. $A \leftrightarrow C$ and $B\leftrightarrow D$ so $X,Y$ stay where they were. This gives that $TX^2=TA.TC=TS.TM$ The radius of this inversion is $TX=TY$. $M,X,Y$ are collinear and $TM.TS=TX^2$ so $T,S,X,Y$ are cyclic. \[\angle XZY=\frac{\angle XTY}{2}=\angle XTO=\angle STO-\angle STX=\angle STO-\angle SYX=\angle ZYX-\angle SYX=\angle ZYS=\angle SZY\]$\angle SZY=\angle XZY\implies X,S,Z$ are collinear as desired.$\blacksquare$

AlperenINAN wrote: Let $\omega_1$ and $\omega_2$ be two different circles that intersect at two different points, $X$ and $Y$. Let lines $l_1$ and $l_2$ be common tangent lines of these circles such that $l_1$ is tangent $\omega_1$ at $A$ and $\omega_2$ at $C$ and $l_2$ is tangent $\omega_1$ at $B$ and $\omega_2$ at $D$. Let $Z$ be the reflection of $Y$ respect to $l_1$ and let $BC$ and $\omega_1$ meet at $K$ for the second time. Let $AD$ and $\omega_2$ meet at $L$ for the second time. Prove that the line tangent to $\omega_1$ and passes through $K$ and the line tangent to $\omega_2$ and passes through $L$ meet on the line $XZ$. Let $S$, $S'$ are intersections of the tangent at $K$ to $\omega_1$, the tangent at $L$ to $\omega_2$ with line $AC$, respectively First, we will prove that $\frac{SC}{SA} = \frac{CD}{AB}$ : $\, \,$Cuz $ACDB$ is an isosceles trapezoid, $AC$ is the tangent at $C$ to $\omega_2$ so : $\hspace{0.5cm}\angle LAC = \angle LCD = \angle BCD \Leftrightarrow \angle ACB = \angle LCD$ $\, \,$Similar, $\angle ABC = \angle LDC$ $\, \,$Therefore $\triangle BAC \sim \triangle DLC$. Lead to : $\hspace{0.5cm}\frac{AC}{AB}. \frac{LC}{LD} \Leftrightarrow \frac{CD}{AB} = \frac{CD}{AC}. \frac{LC}{LD} = \frac{\sin \angle CAD}{\sin \angle ADC}. \frac{\sin \angle LDC}{\sin \angle LCD} = \frac{\sin \angle CAD}{\sin \angle LCD} = \frac{\sin \angle SAL}{\sin \angle ALS} = \frac{SL}{SA} = \frac{SC}{SA}$ (1) $\hspace{0.5cm}$(Use Sin theorem for triangles $ACD$, $LCD$, $LAS$) Similar, we have $\frac{S'A}{S'C} = \frac{AB}{CD}$ (2) Next, let $M$ is the midpoint of segment $AC$, $S''$ is the intersection of lines $XZ$ and $AC$, we will show that $\frac{S''A}{S''C} = \frac{AB}{CD}$ : $\, \,$Power of $M$ respect to circles $\omega_1$, $\omega_2$ tells us that $M$, $X$, $Y$ are colinear and $MA^2 = MC^2 = MX. MY$ $\, \,$Therefore, $\triangle MXA \sim \triangle MAY$, $\triangle MXC \sim \triangle MCY$. Lead to $\frac{XC}{YC} = \frac{MC}{MY} = \frac{MA}{MY} = \frac{XA}{YA}$ $\, \,$In the order hand, cuz $\angle AZC = \angle AYC = \angle XYA + \angle XYC = \angle XAC + \angle XCA = 180^\circ - \angle AXC$, $\, \,$so $A$, $C$, $Z$, $X$ are concyclic $\, \,$Thus, $\frac{S''A}{S''C} = \frac{S''A}{S''Z}. \frac{S''Z}{S''C} = \frac{XA}{ZC}. \frac{ZA}{XC} = \frac{XA^2}{XC^2} = \frac{XA}{\sin \frac{1}{2} \angle AO_1X}. \frac{\sin \frac{1}{2} \angle XO_2C}{XC} = \frac{O_1A}{O_2C} = \frac{AB}{CD}$ (3) From (1), (2), (3) we see that $S \equiv S' \equiv S''$, the tangent at $K$ to $\omega_1$, the tangent at $L$ to $\omega_2$ and line $XZ$ concur at $S$

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Turkey-egmo-2024.pdf (139kb)

Same main steps with #4 but little bit shorter. Let the tangents at $K$ and at $L$ meet at $M$. And WLOG, let $X$ be further than $Y$ from $l_1$. Claim 1: $M \in [AC]$. Proof: Let $M'$ be the intersection of $AC$ and tangent line to $\omega_1$ at $K$. First, we know that $AL.AD = AC^2 = CK.CB$ and $AD = CB$ hence $CK = AL$ We also know $M'A = M'K$. Now, we are going to show $\angle M'KC = \angle M'AL$. Let $AD$ meets $\omega_1$ at $T$ different from $A$. $\angle M'KC = \angle BAK = \angle BAT + \angle TAK = \angle KAM' + \angle TAK = \angle M'AL$ Hence, we conclude $\triangle M'KC \cong \triangle M'AL$. That means $M'C = M'L$ so $M'L$ is tangent. So actually $M' \equiv M$. $\square$ Claim 2: $\angle XMA = \angle YMC$. Proof: Take an inversion around $M$ which has radius $\sqrt{MA.MC}$. We are going to show the image of any thing $Q$ as $Q'$. See second figure below. One can conclude that $MC' = MA$ and $MA' = MC$. Note that $M$ sees the circles in equal angles in both figures because of inversion and congruency of the triangles $M'KC$ and $M'AL$. Hence, $\omega_1'$ and $\omega_2$ have same radius. Similarly, $\omega_2'$ and $\omega_1$ have so. Also $\omega_1'$ and $\omega_2'$ intersect at $X'$ and $Y'$. ($Y'$ is the further one from $M$.) Now, consider the first figure and the figure after inversion. Actually, these figures are symmetric. Because the circles and the distance of these circles are same. By symmetry, $\angle Y'MC' = \angle XMA$ and by inversion, $\angle Y'MC' = \angle YMC$. Hence, $\angle XMA = \angle YMC$. $\square$ By last claim, we can conclude $M \in YZ$. Hence, we are done. $\blacksquare$

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