Simple angle chasing gives that $\overleftrightarrow{APQ}$ is the tangent line of $A$ WRT $\odot(ABC)$. Now let $K$ be the center of sprial similarity sending $\triangle BCD$ to $\triangle EAP$. Then $\measuredangle KBC=\measuredangle KEA$, hence $\odot(KBE)$ is tangent to $BC$; $\measuredangle KCD=\measuredangle KAP$, hence $K$ lies on $\odot(ABC)$; $\measuredangle KDB=\measuredangle KPE$, and since $\angle BDC=\angle APE \implies X=PE \cap BD \in \odot(APD)$, $K$ lies on $\odot(APD)$. Thus $\angle AKE=\angle CKB=\angle BAC$. Since there is a unique point $R$ lying on $\odot(APD)$ such that $\angle ARE=\angle BAC$ and $R, C$ are on different sides of $AB$, we have $K \equiv R$. Similarly if $L$ is the center of sprial similarity sending $\triangle CBD$ to $\triangle AEP$, then $L \equiv S$. Construct points $X,Y \in \odot(ABC)$ such that $BX \parallel AC$ and $CY \parallel AB$. Let $J=KY \cap LX$. From $\angle BKY=\angle ABC$, $(K, E, Y)$ are collinear; so $J=RE \cap DS$. Now note that $AX=BC=AY \implies AO \perp XY$, so $X, Y$ are symmetric WRT $\overleftrightarrow{AO}$. Also $K$ and $L$ are symmetric WRT $AO$ since $KL \parallel PQ \perp AO$. Thus, by symmetry, $J \equiv KY \cap LX$ lies on $AO$. We're done.

IMO 2000, BB'//AC, CC'//AB, B',C' are on circiumcircle of ABC, B'C'//A tangent line. If, R,S are on circiumcircle, REB', SDC' colinear and RS//PQ//B'C'-> RE,DS,AO meet one point! And I prove (R is on circiumcircle) use center of spiral similarity! I think this problem is so beautiful!

Let $l$ be tangent line of $A$ WRT $\odot(ABC)$. Let $Z\in l$ and $T\in l$ such that $\triangle ZAD \sim \triangle ABC$ and $\triangle TEA \sim \triangle ABC$(In that order). According to the problem $P, Q\in l$. Claim: $\odot(ZAD),\odot(ABC),\odot(AQE)$ are concurrent in $S$. Proof: Note that $\odot(ZAD) \cap \odot(AQE) \equiv S$; and $S$ is unique. Let $\odot(ABC) \cap \odot(AZD)\equiv S^*$. $\angle ZAS^*=\angle ABS^*$, so $S^*$ is a corresponding point on $\odot(ZAD)$ and $\odot(ABC)$, where $\triangle ZAD \sim \triangle ABC$. $(\star)$ $\triangle ABC \sim \triangle ZAD , \triangle BEC \sim \triangle AQD , (\star) \implies \triangle QZS^* \sim \triangle EAS^* \implies \angle ZQS^*=\angle AES^* \implies AQS^*E$ is cyclic. So $S^* \equiv S$. And we’re done with the claim. And similarly $\odot(TEA),\odot(ABC),\odot(APD)$ are concurrent in $R$. $RS \parallel PQ \implies RS \parallel l \implies AS=AR \implies AO$ is perpendicular bisector of $RS$. $\implies \angle ARS =\angle ASR$. So; $\angle ERS=\angle ERA-\angle SRA= \angle DSA-\angle RSA=\angle DSR$. So $(SD \cap RE) \in AO$ (cause $AO$ is perpendicular bisector of $RS$). And we’re done. $\square$

Can someone draw a diagram, please?