First note that $KAYC$ and $KBYD$ are cyclic by Miquel's theorem. Next, $PDYC$ is cyclic because $\angle PDY=180- \angle DXY=180- \angle PCY$. Likewise $QDYC$ is cyclic, so $(PQY)=(DCY)$, and likewise, $(YWZ)=(ABY)$. Let $\ell_a, \ell_b,\ell_c, \ell_d$ be the tangent line of $A, B, C, D$ wrt to their respective circles ($C_1$ or $C_2$), and let $M=\ell_a\cap \ell_d, N=\ell_b \cap \ell_c$. Obviously the spiral similarity centered $Y$ sends $BNC$ to $DMA$, so $\angle (MS,SN)=\angle (AY,YC)=\angle (MY,YN)$, so $MSNY$ is cyclic. Likewise, $MRNY$ is cyclic, so $MRSNY$ is cyclic, so $(RSY)=(MNY)$. Final step is to prove that $(DCY), (ABY), (MNY)$ are coaxial, which is sufficient to prove that $$\frac{pow(M, (ABY))}{pow(M, (DCY))}=\frac{pow(N, (ABY))}{pow(N, (DCY))}$$$$\iff \frac{DM\cdot MP}{MA\cdot MZ}=\frac{CN\cdot NQ}{NB\cdot NW}$$$$\iff \frac{PM}{MZ}=\frac{QN}{NW}$$But this is because $\angle (PM,MZ)=\angle (WN,NQ)$ since $RMNS$ is cyclic, and $\angle (MP,PZ)=\angle (WQ,QN)$ because $DPQC$ is cyclic. So $\triangle MPZ\sim \triangle WQN$, hence the ratio equality and we are done.

navi_09220114 wrote: First note that $KAYC$ and $KBYD$ are cyclic by Miquel's theorem. Next, $PDYC$ is cyclic because $\angle PDY=180- \angle DXY=180- \angle PCY$. Likewise $QDYC$ is cyclic, so $(PQY)=(DCY)$, and likewise, $(YWZ)=(ABY)$. Let $\ell_a, \ell_b,\ell_c, \ell_d$ be the tangent line of $A, B, C, D$ wrt to their respective circles ($C_1$ or $C_2$), and let $M=\ell_a\cap \ell_d, N=\ell_b \cap \ell_c$. Obviously the spiral similarity centered $Y$ sends $BNC$ to $DMA$, so $\angle (MS,SN)=\angle (AY,YC)=\angle (MY,YN)$, so $MSNY$ is cyclic. Likewise, $MRNY$ is cyclic, so $MRSNY$ is cyclic, so $(RSY)=(MNY)$. Final step is to prove that $(DCY), (ABY), (MNY)$ are coaxial, which is sufficient to prove that $$\frac{pow(M, (ABY))}{pow(M, (DCY))}=\frac{pow(N, (ABY))}{pow(N, (DCY))}$$$$\iff \frac{DM\cdot MP}{MA\cdot MZ}=\frac{CN\cdot NQ}{NB\cdot NW}$$$$\iff \frac{PM}{MZ}=\frac{QN}{NW}$$But this is because $\angle (PM,MZ)=\angle (WN,NQ)$ since $RMNS$ is cyclic, and $\angle (MP,PZ)=\angle (WQ,QN)$ because $DPQC$ is cyclic. So $\triangle MPZ\sim \triangle WQN$, hence the ratio equality and we are done. Can you please explain slightly more with a diagram

Note that $Y$ is the center of spiral similarity which takes $AC$ to $DB$, and so we have $\triangle YAD \stackrel{+}{\sim} YCB$. This gives us $$\measuredangle WAY=\measuredangle DAY=\measuredangle BCY=\measuredangle WBY \Rightarrow W \in \odot (YAB)$$Similarly, $Z \in \odot (YAB)$, and so we have $\odot (YWZ) \equiv \odot (YAB)$. Also, $$\measuredangle QCY=\measuredangle CBY=\measuredangle ADY=\measuredangle QDY \Rightarrow Q \in \odot (CDY)$$Similarly, $P \in \odot (CDY)$, and so we have $\odot (PQY) \equiv \odot (CDY)$. Now, let $T=AD \cap BC$ and $U$ (resp. $V$) be the point where the tangents to $C_1$ (resp. $C_2$) at $A$ and $D$ (resp. $B$ and $C$) meet. Then $$\measuredangle SAT=\measuredangle ZAW=\measuredangle CBW=\measuredangle SCT \Rightarrow S \in \odot (ACT)$$But, as $Y$ is also the center of spiral similarity taking $AD$ to $CB$, so $AYTC$ is cyclic. This means that $SAYTC$ and $BRYDT$ are cyclic pentagons (the second one following in a similar one as the first one). Thus, we have $$\measuredangle SYR=\measuredangle SYT+\measuredangle TYR=\measuredangle SCT+\measuredangle TBR=\measuredangle VCB+\measuredangle CBV=\measuredangle CVB=\measuredangle SVR$$which means that $V \in \odot (RSY)$. Similarly, $U \in \odot (RSY)$, and so we must have $\odot (RSY) \equiv \odot (UVY)$. Thus, it suffices to show that $$\frac{UP \cdot UD}{UZ \cdot UA}=\frac{VC \cdot VQ}{VW \cdot VB} \Leftrightarrow \frac{UP}{UZ}=\frac{VQ}{VW} \Leftrightarrow \frac{\sin \angle UZP}{\sin \angle UPZ}=\frac{\sin \angle VWQ}{\sin \angle VQW}$$But, we have $$\measuredangle UZP=\measuredangle AZB=\measuredangle AWB=\measuredangle QWV \Rightarrow \sin \angle UZP=\sin \angle VWQ$$Similarly, we have $\sin \angle UPZ=\sin \angle VQW$. Combining the above two equalities, we get the desired result. $\blacksquare$

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Nice and Easy!!! steppewolf wrote: Let $C_{1}$, $C_{2}$ be circles intersecting in $X$, $Y$ . Let $A$, $D$ be points on $C_{1}$ and $B$, $C$ on $C_2$ such that $A$, $X$, $C$ are collinear and $D$, $X$, $B$ are collinear. The tangent to circle $C_{1}$ at $D$ intersects $BC$ and the tangent to $C_{2}$ at $B$ in $P$, $R$ respectively. The tangent to $C_2$ at $C$ intersects $AD$ and tangent to $C_1$ at $A$, in $Q$, $S$ respectively. Let $W$ be the intersection of $AD$ with the tangent to $C_{2}$ at $B$ and $Z$ the intersection of $BC$ with the tangent to $C_1$ at $A$. Prove that the circumcircles of triangles $YWZ$, $RSY$ and $PQY$ have two points in common, or are tangent in the same point. Proposed by Misiakos Panagiotis Let $CQ\cap\{AZ,BW\}=\{S,T\}$ respectively. Claim 1:- $\{C,D\}\in\odot(YPQ)$ Consider the Spiral Similarity $\sigma$ at $Y$ mapping $\triangle ADY\mapsto\triangle YCB$. So, $\angle QCY=\angle CBY=\angle ADY$ and $\angle PDY=\angle DAY=\angle YCB$. So, $\{C,D\}\in\odot(YPQ)$ Claim 2:- $\{S,T\}\in\odot(RSY)$ Now notice that $\sigma:AA\cap DD\mapsto CC\cap BB=T$. Hence, $\triangle YKA\stackrel{+}{\sim}\triangle YTC$. Hence, $\angle AKY=\angle YTS$. Similarly $RKYT$ is a cyclic quad $\implies \{S,T\}\in\odot(RSY)$ Now by Coaxility Lemma it suffices to prove that \begin{align*} \frac{Pow_{\odot(YQP)}W}{Pow_{\odot(YRS)}W}=\frac{Pow_{\odot(YQP)}Z}{Pow_{\odot(YRS)}Z} &\iff \frac{WQ\cdot WD}{WR\cdot WT}=\frac{ZP\cdot ZC}{ZS\cdot ZK} \\ &\iff\frac{WD}{WR}\cdot\frac{WQ}{WT}=\frac{ZC}{ZS}\cdot\frac{ZP}{ZK} \\ &\iff\frac{\sin\angle WTQ}{\sin\angle WQT}\cdot\frac{\sin\angle WRD}{\sin\angle WDR}=\frac{\angle ZKP}{\sin\angle ZPK}\cdot\frac{\sin\angle ZSC}{\sin\angle ZCS}\cdots\cdots\cdots(\star)\end{align*} $(\star)$ is true due to Claim 1 and Claim 2. Hence $\{\odot(YWZ),\odot(RSY),\odot(YPQ)\}$ are coaxial. $\blacksquare$