Let $PQ$ cuts $BC,\odot(ABC)$ at $R,X,X'$ ; $AP,AQ$ cut $\odot(ABC)$ at $P'Q'$ , we have $BC\parallel P'Q'$; Let $P'Q'$ cut $PQ$ at $R'$. First recall two properties of isogonal conjugates: $1)$ Let $U,V$ be two isogonal conjugates wrt $ABC$, if $V'$ is the symmetric of $V$ in $BC$ then $A,V'$ are isogonal conjugates wrt $UBC$. $2)$ if $P,Q$ and $ U,V$ are two isogonal conjugates pairs then $\odot (APU),\odot (AQV)$ and $ \odot (ABC)$ are concurrent. ________________________________________________________________________________________________ Let $RE$ cuts $AB,AC,AX'$ at $ C', B',F$ ; more applying the property 1 with $B,D$ isogonal conjugates wrt $APQ$ we get $A,D'$ isogonal wrt $BPQ$ (where $D'$ is the reflection of $D$ in $PQ$) thus $\angle QBD'=\angle ABP=\angle QBC$ which means $D'\in BC$ hence $D=B'$ or $B'$ is the isogonal of $B$ wrt $APQ $; idem $C'$ is the isogonal of $C$ whence $\odot (ABC),\odot (AB'C')$ and $ \odot (APQ)$ share a common point say $M$; further $\angle MXE =\angle MBA=\angle MBC' =\angle MRC'=\angle MRE$ i.e. $MRXE$ is cyclic similarly we get $MRFX'$ cyclic whence $M$ is the Miquel point of $XX'FE$ we deduce $ MEFA$ cyclic; we claim that $APQEFM$ is cyclic indeed: since the circle $\odot(ABC)$ is the image of $\odot(AEF)$ by the similarity of the center $M$ that sends $E\to X$ and $\odot(ABC)$ is also the image of $\odot(APQ)$ by the similarity of the center $M$ that sends $P\to P'$ it suffices to prove that $MEX\sim MPP'$: we have $\angle XME=\angle XRE=\angle BRX=\angle P'R'P=\angle P'MP$ besides $\angle MXE=\angle MXA=\angle MP'A=\angle MP'P$ which completes the proof. Best regards RH HAS

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