First, recall that we have $DI=DB=DC=DI_A$. Then, we have $\angle DCF = \frac{\overarc{DC}-\overarc{FC}}{2}=\frac{\overarc{BD}-\overarc{FC}}{2}=\angle CEF$ and hence $ID^2=DC^2=DF \cdot DE$. Let $E'$ be the reflection of $E$ wrt $D$. Notice that $DE' \cdot DF = ID \cdot DI_A \implies E'IFI_A$ cyclic $\implies \angle IEI_A=\angle IE'I_A = 180 - \angle IFI_A = 180 - \angle PFQ$ $\implies PEQF$ cyclic. Now, let $P'$ be the reflection of $P$ wrt $D$. Since $PEQF$ is cyclic we have $\angle IP'I_A + \angle IQI_A = \angle IPI_A + \angle IQI_A = 360 - \angle EPF - \angle EQF = 180 \implies IP'I_AQ$ cyclic. This implies that $(PII_A)$ is the reflection of $(QII_A)$ wrt $D$. Note that $D$ lies on the radical axis of $(PII_A)$ and $(QII_A)$. Let $X, Y$ be the intersection points of $DS, (PII_A)$ and $DR, (QII_A)$ respectively. Observe that the reflections of $X, Y$ wrt $D$, $X', Y'$ lie on $QII_A$ and $PII_A$ respectively. It follows that $DX \cdot DS = DX' \cdot DS = DI \cdot DI_A = DE \cdot DF = DY' \cdot DR = DY \cdot DR$. Therefore, the quadrilaterals $XYRS$, $EFYR$ and $EFXS$ are all cyclic. Finally, observe that $IY=I_AY'$ which combined with the fact that the circles $(PII_A)$ and $(QII_A)$ have the same radius, implies that $\angle DRI_A=\angle Y'RI_A = \angle II_AY = \angle ISY \implies SYRE$ cyclic. Therefore, the six points $E, F, R, S, X, Y$ all lie on the same circle.

Here is my solution. $Claim 1$ : $QEPF$ is cyclic! $Proof$ : It's well known that $D$ is midpoint of $II_a$. Also it's easy to see that $DI^2=DE.DF$. It shows that $E$ is $F$-humpty point in $\triangle FII_a$ and $\angle QEP+\angle QFP= 180^{\circ}$ $\implies$ so $QEPF$ is cyclic! $Claim 2$ : $\triangle QIR\sim\triangle PI_aS$! $Proof$ : $\angle PSI_a=\angle IQR=\angle IPF=\angle IRQ$ $\implies$ so $\triangle QIR$ and $\triangle PI_aS$ are isosceles and $\triangle QIR\sim\triangle PI_aS$! $Claim 3(Last Claim)$ : $\triangle FQR\sim\triangle FPS$ and $FRES$ is cyclic! $Proof$ : $\angle FQR=\angle FPS$ so it suffices to show $\frac{FQ}{FP}=\frac{QR}{PS}$. By second claim $\frac{QR}{PS}=\frac{IQ}{I_aP}$. Also by $Ceva$ $theorem$ it's easy to see $\frac{FQ}{FP}=\frac{IQ}{I_aP}$(because $D$ is midpoint of $II_a$). $\implies$ so $\triangle FQR\sim\triangle FPS$ and $FRES$ is cyclic!

Initially, E lies on radical axis of $(ABC)$ and $(BICI_a)$: $DE \cdot EF = DI^2-DE^2$ $\implies DE \cdot (EF+DE) = DI^2$ $\implies DE \cdot DF = DI^2$ $\implies E$ is the $F$-Humpty in $\triangle FII_a$!!! Now, let's prove that $\triangle FQR\sim\triangle FPS$. It's easy to obtain $\angle FQR = \angle FPS$, so we just need $\frac{FQ}{QR} = \frac{FP}{PS}$. $\leftrightarrow \frac{FQ}{FP} = \frac{QR}{PS}$ $\leftrightarrow \frac{FI}{FI_a} = \frac{RE}{PE}$ (Because $PR \parallel SQ$) By POP, we have $RE = \frac{IE \cdot PE}{EI_a}$, substituting: $\leftrightarrow \frac{FI}{FI_a} = \frac{IE}{EI_a}$, which is true since the reflection of the Humpty is the intersection of symmedian with circumcircle!! Notice that by spiral similarity, $\triangle FQR\sim\triangle FPS$ implies $\triangle FQP\sim\triangle FRS$, so $\angle IFI_a = \angle RPS \implies FERS$ cyclic, as desired (Again by reflection fact).

We have $\angle{DEB} = \angle{DBC} - \angle{BDF} = \angle{DCB} - \angle{BCF} = \angle{DCF}$. Then $DI^2 = DB^2 = DF \cdot DE$. So $F$ is $E$ - Humpty of $\triangle EII_a$. Hence $\angle{PFQ} = \angle{IFI_a} = 180^{\circ} - \angle{IEI_a}$ or $E, P, F, Q$ lie on a circle. Let $Y$ be second intersection of $DP$ and $(EPFQ)$. Note that $PQ \parallel II_a$. We have $DI^2 = DF \cdot DE = DY \cdot DP,$ then $\angle{DYI} = \angle{DIP} = \angle{QPE} = \angle{QFE} = \angle{DFI},$ hence $Y \in (DFI)$. Therefore, $Y$ is Miquel point of completed quadrilateral $DFQI_a.EI$. This leads to $Y \in (QII_a)$. Suppose that $Z$ is second intersection of $DS$ with $(DFI)$. Consider inversion $\mathcal{I}^{k = DI^2}_D: I \longleftrightarrow I, I_a \longleftrightarrow I_a, F \longleftrightarrow E$. From this, we have $\mathcal{I}^{k = DI^2}_D: (DFI) \longleftrightarrow EI$. Then $\mathcal{I}^{k = DI^2}_D: Z \longleftrightarrow S$. So $Z \in (DFI)$. But $\mathcal{I}^{k = DI^2}_D: (PII_a) \longleftrightarrow (YII_a)$ and $S \in (YII_a)$ hence $Z \in (PII_a)$. Note that $DZ \cdot DS = DI^2 = DF \cdot DE,$ we have $E, F, Z, S$ lie on a circle. Since $\angle{SZR} = \angle{SZI} + \angle{IZR} = 180^{\circ} - \angle{IZD} + \angle{II_aR} = \angle{EII_a} + \angle{II_aR} = 180^{\circ} - \angle{REI},$ we have $E, S, Z, R$ lie on a circle. Hence $E, R, F, Z, S$ lie on a circle

Claim: $E$, $F$, $P$, and $Q$ are concyclic Notice that by the Shooting lemma $DF\cdot DE=DB^2=DI^2$, thus $(IEF)$ is tangent to $AD$. Since $D$ is the midpoint of $II_a$, Menelaus on $II_aF$ yields $PQ\parallel II_a$. Finally $$\angle QFE=\angle IFE=\angle I_aIE=\angle EQP$$Claim: $QE\cdot QF=PE\cdot PF$ This is a well known result given that $PQEF$ is cyclic and $EF$ bisects $PQ$. Claim: $F$ is the spiral center of $QR\mapsto PS$ First notice that $\angle FQR=\angle FQE=\angle FPS$. Notice by the converse of Reim's Theorem $RP\parallel QS$. Also $$\frac{QR}{PS}=\frac{QE}{ES}=\frac{IE}{EI_a}=\frac{EP}{QE}=\frac{QF}{FP}$$ To finish, $\angle RFS=\angle QFP=180^{\circ}-\angle RES$.

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