Let $X \in \odot (ABC)$ with $BX \parallel AC$. By reflection about the perpendicular bisector of $BC$ it suffices to show that $XDB_1E$ is cyclic. Invert about $B$ with radius $\sqrt{BA \cdot BC}$ and reflect the inverted figure in the angle bisector of $\angle ABC$ to get the following problem- Inverted problem wrote: Let $M$ be the midpoint of arc $\widehat{AC}$ not containing $B$, and suppose the tangent at $B$ to $\odot (ABC)$ meets $AC$ at $X$. Suppose $\odot (M,MA)$ meets the $B$-Apollonius circle $\omega$ at $D,E$. Then $MDXE$ is cyclic. Since $X$ is the center of $\omega$, so the result directly follows from the fact that $\omega$ and $\odot (M,MA)$ are orthogonal. EDIT: To show that $\omega$ and $\odot (M,MA)$ are orthogonal, just invert about $\odot (M,MA)$ and invoke Shooting Lemma to see that $\omega$ remains fixed.

Let $N$ be the midpoint of smaller arc $AC$. Lemma: $NB_1 \cdot BN = ND^2=NE^2=NI^2$ because $\triangle AB_1N \sim \triangle BAN$ From this it follows $\angle B_1EN =\angle NBE$ and $\angle B_1DN= \angle NBD$ so basically $B_1$ is the Humpty point of triangle $EDB$. (key insight) Since $F$ is the reflection of the Humpty point it follows $F$ is the intersection of the symmedian with the circumcircle of triangle $EDB$ so $E,D,B,F$ are concyclic points and are harmonic at that!

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Let $X$ be the midpoint of arc $AC$ of $(ABC)$ not containing $B$.Note that $D,X,F$ are collinear.Easy to see that $XB.XB_1=XD^2=XE^2$.[inversion in $(AIC)$ swaps $B,B_1$ because it swaps $AC$ and $(ABC)$]. So $\angle XEF=\angle XEB_1=\angle XBE$ [from $XB_1.XB=XE^2$] and $\angle XDF=\angle XDB_1=\angle XBD$[from $XD^2=XB.XB_1$] SO $\angle EBD= \angle XBE+\angle XBD= \angle XEF+\angle XDF=180^\circ-\angle DFE$.So $E,B,D,F$ are concyclic.

Cute XbenX wrote: $BB_1$ is the angle bisector of $\triangle ABC$, and $I$ is its incenter. The perpendicular bisector of segment $AC$ intersects the circumcircle of $\triangle AIC$ at $D$ and $E$. Point $F$ is on the segment $B_1C$ such that $AB_1=CF$.Prove that the four points $B, D, E$ and $F$ are concyclic. Let $\overline{BB_1}$ meet $\odot(ABC)$ again at $M \ne B$. Note that $\overline{MB}, \overline{MF}$ are symmetric in $\overline{DE}$ hence the reflection $K$ of $F$ in the perpendicular bisector of $\overline{DE}$ (whose midpoint is $M$) lies on $\overline{BB_1}$. Since $MB \cdot MK=MB \cdot MF=MI^2=MD \cdot ME$, the conclusion follows.

Let $O$ be the circumcenter of $\triangle AIC$. It is well known that $O$ lies on $BB_1$. Let $J$ be the reflection of $I$ over $O$, and let $AB$ hit the circumcircle of $\triangle AIC$ a second time at $K$. Using directed angles, it is easy to see that $\measuredangle IAC = \measuredangle KAI$. From here, we use complex numbers with $(AIC)$ as the unit circle. Let $I = x^2$, $A = y^2$ and $C = z^2$. Because $DE$ is the perpendicular bisector of $AC$, we have that $\{ D, E \} = \{ yz, -yz \}$. WLOG, let $D = yz$ and $E = -yz$. Since $B_1 = IJ \cap AC$, we obtain that \[ B_1 = \frac{ AC(I+J) - IJ(A+C)}{AC - IJ} = \frac{ x^4(y^2+z^2)}{x^4 + y^2z^2}. \]Because $F$ is the reflection of $B_1$ over the midpoint of $AC$, we have \[ F = A + C - B_1 = \frac{ y^2z^2(y^2+z^2)}{x^4 + y^2z^2}. \] Because $\measuredangle IKC = \measuredangle KCI$, we get $K = \frac{x^4}{z^2}$. Then $B = IJ \cap AK$ implies \[ B = \frac{ AK(I+J) - IJ(A+K)}{AC - IJ} = \frac{ x^4(y^2 + \frac{x^4}{z^2})}{y^2 \frac{x^4}{z^2} + x^4}= \frac{x^4 + y^2z^2}{y^2 + z^2} \]To check that $B,D,E,F$ are concyclic, we just have to check that the quantity \[ \frac{B - D}{F-D} \div \frac{B-E}{F-E} \in \mathbb{R} \]We just simplify: \begin{align*} \frac{B - D}{F-D} \div \frac{B-E}{F-E} &= \frac{ \frac{x^4 + y^2z^2}{y^2+z^2} - yz}{\frac{y^2z^2(y^2+z^2)}{x^4+y^2z^2} - yz} \cdot \frac{ \frac{y^2z^2(y^2+z^2)}{x^4 + y^2z^2} + yz}{\frac{x^4+y^2z^2}{y^2+z^2}+ yz} \\ &= \frac{ (x^4+ y^2z^2 - yz(y^2 + z^2))(yz(y^2+z^2) + (x^4 + y^2z^2))}{(yz(y^2+z^2) - (x^4+y^2z^2) ) (x^4 + y^2z^2 + yz(y^2+z^2))} \\ &= 1, \end{align*}which is obviously a real number.

See my solution on my Youtube channel in the link below. It is similar to anantmudgal09's solution, but somewhat different in the end. https://www.youtube.com/watch?v=Bv_HFRmVBio

Let $BI$ meet $\odot (ABC)$ at $M$. Inversion $(M; MI)$ followed by a reflection about $MD$ swaps $(ABC)\mapsto AC$ , $B\mapsto F$, $D\mapsto D$. Hence $\odot (BDF)$ is fixed. So, second point of intersecion $\odot (BDF)$ with $\odot (M)$ lays on $MD$, which is $E$.

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Let $N = BI \cap (ABC)$, which is the circumcenter of $IAC$. Let $B_2$ be the reflexion of $B_1$ w.r.t $N$. On one hand $ NB_2 \cdot NB= NB_1 \cdot NB = NI^2 = ND \cdot NE$, implying that $B,D,B_2,E$ are concyclic. On the other hand, $DEB_2F$ is an isosceles trapezoid, which are concyclic, which completes the proof.

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I'll explain math_pi_rate's solution, because in its essence it's more simple than it looks and quite inversion-instructive. We don't even need to construct inversive diagram to understand it. Let $\odot (AIC)$ be $\omega$ and $M$ is where $BI$ hit $(ABC)$. Let $X\in (ABC), BX\parallel AC$. Perform inversion $B_1\mapsto M$ with center $B$ (bisector-flip actually don't even needed here). $\omega$ is fixed. Since $DE$ orthogonal to $\omega$ it goes to circle (say, $\Omega$) orthogonal to $\omega$. Since $X$ is a reflection of $B$ over $DE$ it goes to circumcenter of $\Omega$ (say, $X'$). $D,E$ obviously go to $\omega \cap \Omega=[D',E']$. With $M,X'$ being centers of orthogonal circles we immediately see that $X'D'ME'$ is cyclic. Inverting back it makes $XDB_1E$ cyclic. And finally reflect it about $DE$ to get what we need.

Let $M$ and $N$ be the midpoint of $BC$ and arc $BC$. $AM$ meets $(ABC)$ at $X$. Then as $XM$ and $NB_{1}$ are concurrent on $(ABC)$, by butterfly theorem, $XF$ and $NM=DE$ are also concurrent on $(ABC)$. Let this concurrency point be $Y$. Then by power of a point, \[MX\cdot MB=MA\cdot MC=MD\cdot ME\]and so $B,D,X,E$ are concyclic. Consider an inversion at $Y$ with radius $YA=YC$. This swaps $AC$ with $(ABC)$, and preserves the circle $(AIC)$. Therefore, $D\rightarrow E$ and $F\rightarrow X$, and consequently, \[YD\cdot YE=YA^2=YF\cdot YX\]which shows that $F$ also lies on $(DEX)$.

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Very nice problem: Let us consider $N$ the intersection of the median of $AC$ with the circle. Clearly, $N$ (radical axis argument) lies in the circle $(BDE)$. Now we prove the following Lemma: Claim: $N,F,L$ are collinear where $L$ is the midpoint of arc $ABC$. Proof: Apply Angle Bissector Theorem in $ANC$ and you'll be good to go. In conclusion, notice that $LF\times LN = LA^2=LB^2 $ because $\angle ANF=\angle FAL$. However, $LA^2= LE.LD$ which implies that $LE.LD = LF.LN$ $\implies$ $(EDFN)$ is cyclic desired.

let $N,R,M$ be the midpoint of arc $BC$ , the intersection of the external bisector of $\angle ABC$ with $BC$ and the midpoint of $BC$ let $B' $ be the reflection of $B_1$ beyond $N$ $\angle RBB'=90=\angle B_1FB'$ so $RBFB'$ is cyclic since $(C,A;B_1,R)=-1$ we have $MD.ME=MA.MC=MB_1.MR=MF.MR$ so $DEFR$ is cyclic $NB'.NB=NB_1.NB=NA^2=ND.NE$ so $DB'EB$ IS cyclic so $BDFB'ER$ is cyclic

Let $G$ be the midpoint of $AC$ and $H$ be the intersection of $BG$ and $(ABC)$. $$DG \cdot GE = AG \cdot GC = BG \cdot GH$$So $BDHE$ is cyclic. Let $J$ be the intersection of perpendicular bisector and $HF$. Since $G$ is midpoint of $B_1F$ by butterfly it follows that $J$ lies on $(ABC)$. $$\angle JFD = \angle GDF - \angle DJF = \angle B_1BD - \angle B_1BG = \angle GBD = \angle HBD = \angle HED$$Which is the exterior angle of the quadrilateral $DFHE$ so $DFHE$ is cyclic. Combining both results we get that $BDFE$ is cyclic.

Let $BB_1$ meets the circumcircle of $ABC$ again at $X$. We know that $X$ is the center of $(AIC)$ that is the midpoint of $DE$. Assume that $XF$ meets the circumcircle of $ABC$ again at $P$. Then, the points $B$ and $P$, $B_1$ and $F$ are symmetrical with respect to the line $DE$. Considering an inversion centred at $X$ with radius $XI$, we obtain that $B\leftrightarrow B_1$, $P\leftrightarrow F$ and the points $B,C,D,E$ are fixed. So the circle $(BDF)$ goes to the circle $(PDB_1)$. Let $E'$ be the second intersection point of the circles $(BDF)$ and $(PDB_1)$. As they are mapped each to other by the inversion, we get that $E'$ should be fixed under the inversion, i.e. $E'$ lies on $(AIC)$. On the other hand, the triangles $BDF$ and $PDB_1$ are symmetrical with respect to the line $DE$, so their circumcenters are too. That means $DE$ be the radical axis of their circumcircles. Hence $E\equiv E'$ and $B$, $D$, $F$, $E$ are concyclic.