and to Strudan_Borisov for figuring out the details. Here is a quick write up from me.

@above I think that works. Unfortunately for me, I didn't see that. Let $L$ be the midpoint of $\widehat{AB}$ that doesn't contain $D$. Then we can eliminate $C$ and $E$ by realizing that the problem is equivalent to showing that $L$ is on the radical axis of the $A,B$ mixtilinear incircles of $\triangle ADF,\triangle BDF$ (the power to both of these is $LC^2$ iff the problem statement is true). It now suffices to prove this. Let $M$ be the midpoint of the arc $\widehat{FD}$ not containing $A$. Let $I_A,O_A$ be the incenter of $\triangle ADF$ and the center of the $A$-mixtilinear incircle of $\triangle ADF$ respectively. Define $I_B$ and $O_B$ similarly. Let $P_A$ be the foot from $O_A$ to $\overline{AD}$ and let $P_B$ be the foot from $O_B$ to $\overline{BD}$. Then $A-I_A-O_A-M$ and $B-I_B-O_B-M$, $\angle O_AAP_A=\angle MAD=\angle MBD=\angle O_BBP_B$, and $\angle BP_BO_B=90^{\circ}=\angle AP_AO_A$. $\angle O_AI_AP_A=90^{\circ}=\angle O_BI_BP_B$ due to mixtilinear incircle properties, so the figures made by $A,I_A,O_A,P_A$ and $B,I_B,O_B,P_B$ are similar. Let $L_O$ and $L_I$ be such that $\triangle ALB\stackrel{+}{\sim}\triangle I_AL_II_B\stackrel{+}{\sim}\triangle O_AL_OO_B$. By gliding principle, the figure made by $L,L_I,L_O$ is similar to the figure made by $A,I_A,O_A$. Rename the mixtilinear incircles to $\omega_A$ and $\omega_B$. Then $L_O$ is on the perpendicular bisector of $O_AO_B$, so $\text{Pow}_{\omega_A}(L_O)-\text{Pow}_{\omega_B}(L_O)=O_BP_B^2-O_AP_A^2$. By angle chasing $L_I=\overline{I_AP_A}\cap\overline{I_BP_B}$. $\text{Pow}_{\omega_A}(L_I)=O_AL_I^2-O_AP_A^2=I_AL_I^2+I_AO_A^2-I_AP_A^2-I_AO_A^2=I_AL_I^2-I_AP_A^2$. Since $MI_A=MI_B$, and $\overline{I_AL_I}$ and $\overline{I_BL_I}$ are tangent to the circle centered at $M$ passing through $I_A$ and $I_B$, $I_AL_I=I_BL_I$. Thus $\text{Pow}_{\omega_A}(L_I)-\text{Pow}_{\omega_B}(L_I)=I_BP_B^2-I_AP_A^2=\sin^2(\angle MAD)(\text{Pow}_{\omega_A}(L_O)-\text{Pow}_{\omega_B}(L_O))$. Since $AI_A=AO_A\cdot \sin^2(\angle MAD)$, by Linearity of PoP, $\text{Pow}_{\omega_A}(L)-\text{Pow}_{\omega_B}(L)=0$, as desired.

This problem is equivalent to Problem: Cyclic quadrilateral $ABCD$ is given. Prove that the one of the external common tangents of $A$-mixtilinear and $B$-mixtilinear circle in $DAC, DBC$ is parallel to $CD$. Solution: $K$ is the midpoint of arc $AB$ not containing $C$. $N$ is the midpoint of arc $DABC$, M is the midpoint of arc $DC$ not containing $A$. $I_A, I_B$ are the incenters of $DAC$ and $DBC$. $P$ and $Q$ are the tangency points of the mentioned mixtilinear circles with $(ABCD)$. Lemma: In a triangle $ABC$ with incenter $I$ radius $R$, $N$ is the midpoint of arc $BAC$. $O_A$ is the center of $A$-mixtilinear circle, $R_A$ is the radius of it. Then $$\frac{NI^2}{1-\frac{R_A}{R}}=2+4R^2-\frac{4a^2}{\cos^2 \frac{A}{2}}$$is true. Proof: $O$ is the center of $(ABC)$, $M$ is the antipode of $N, X$ is the $A$-mixtilinear circle (with center $O_A$) tangency point. From Menelaus Theorem, $$\frac{MO}{MN} \frac{NI}{IX} \frac{XO_A}{O_AO} \implies \frac{R}{R-R_A}=2\frac{IX}{NI}+1$$Thus $\frac{NI^2}{1-\frac{R_A}{R}}=2IX \cdot IN+IN^2=2AI \cdot IM+(2R)^2+AI^2-AM^2$. By ptolemy $AM=\frac{b+c}{b+c-a} AI$ and $\frac{AI}{u-a}=\frac{1}{\cos \frac{A}{2}}$. The rest is obvious. $\square$ It is enough to prove that $K$ is on the radical axis of mixtilinear circles. $\triangle KPQ \sim \triangle NI_BI_A$ because $\angle NI_BA= 180- \angle I_AI_BM-\angle MI_BQ=90+\frac{1}{2} \angle AMB-\angle NMB-\angle MNQ=\frac{1}{2} \angle AMB+\angle MNB-\angle MNQ$ $=\angle KPB+\angle BPQ= \angle KPQ$ $KU \cdot KP=KP^2-PU \cdot PK=KP^2-\frac{R_A}{R} KP^2=KQ^2-\frac{R_B}{R} KQ^2=KV \cdot KQ \iff \frac{NI_A^2}{1-\frac{R_A}{R}}=\frac{NI_B^2}{1-\frac{R_B}{R}}$ which is true by the lemma.

The problem is closely related to these lemmas.

The problem can be rephrased in the following way: Given cyclic quadrilateral $ABDF$, the $A-$ and $B-$ mixtillinear incircles of $\triangle{AFD}, \triangle{BFD}$ respectively have a common tangent parallel to $AB$. Solution: Let $M$ be the midpoint of $\widehat{DF}$, and $I_A, I_B$ be the incenters of $\triangle{AFD}, \triangle{BFD}$; then if $\angle{FAM} = \theta$, the distances from $A$ and $B$ to their corresponding mixillinear incircle centers are $AI_A \cdot \left(\frac{1}{\cos \theta}\right)^2$ and $BI_B \cdot \left(\frac{1}{\cos \theta}\right)^2$, respectively, and the radii of these circles are $AI_A \cdot \frac{\tan \theta}{\cos \theta}$, $BI_B \cdot \frac{\tan \theta}{\cos \theta}$ respectively, so it suffices to show that $$\frac{AI_A}{\cos^2 \theta} (\sin \angle{MAB} - \sin \theta) = \frac{BI_B}{\cos^2 \theta} (\sin \angle{MBA} - \sin \theta).$$If $d$ is the diameter then $AI_A, BI_B = MA - d \sin \theta, MB - d \sin \theta$ respectively, so plugging this in, we get $$MA \sin \angle{MAB} - \sin \theta (d \sin \angle{MAB} - MA) = MB \sin \angle{MBA} - \sin \theta (d \sin \angle{MBA} - MB)$$which is clearly true since the first terms cancel by Law of Sines and $d \sin \angle{MAB} - MB, d \sin \angle{MBA} - MA$. $\square$

My write down of the inversion: Defining the inversion Define an inversion centered at $A$ with $r=\sqrt{AE\cdot AC}$ followed by a reflection across the angle bisector of $\measuredangle EAC$, and then a reflection across the perpendicular bisector of $CE$. $\newline$ Properties of this inversion Since $ABCE$ is an isosceles trapezoid since they share the same circumscribed circle, $AB$ and $CE$ share the same perpendicular bisector. Through the inversion $C\longmapsto C$ and $E\longmapsto E$. Thus, $\tau \longmapsto CE$ and conversely also $CE\longmapsto \tau$. Let $D\longmapsto D'$ and $F\longmapsto F'$ $\newline$ Define $K$ as the reflection of $D'$ over the perpendicular bisector of $CE$. Now through an angle chase we get: $\measuredangle D'BE=\measuredangle KAC=\measuredangle DAE=\measuredangle DBE$, and thus $D-D'-B$ are collinear. We can now define as $D'=BD\cap BF$, and symmetrically $F'=BF\cap EC$. $\newline$ Inverting and finishing Here, we notice that $AD\longmapsto BD$ and $AF\longmapsto BF$, since $AB$ and $CE$ share the same perpendicular bisector. Since $k$ is tangent to $AD$,$AF$, $CE$, and $\tau$, if $k\longmapsto k'$ we notice $k'$ must be tangent to $BD$, $BF$, $CE$, and $\tau$. Which finishes the statement. $\blacksquare$

Let $l$ be the circle tangent to $CE,BD,\tau$. We will show that $l$ is tangent to $BF$. Set $k\cap \tau=u,l\cap \tau=V,CE\cap k=P,l\cap CE=Q,AD\cap K=R,BD\cap l=S$. Let the tangent lines to $\tau$ at $U,V$ intersect $EC$ at $X,Y$ respectively. Let $M$ be the midpoint of arc $AB$. Let $I$ be the incenter of $\triangle DCE$. Shooting Lemma: $\omega$ is a circle tangent to $\Gamma$ at $K$ and lies inside of it. Let $AB$ be a chord of $\Gamma$ tangent to $\omega$ at $L$. If $M$ is the midpoint of arc $AB$ which lies on different sides with $\omega$ according to $AB$, then $K,L,M$ are collinear. Proof: Inversion at $M$ proves this. Note that by shooting lemma, $U,P,M$ and $V,Q,M$ are collinear. By Sawayama-Thebault Theorem, $P,R,I$ and $Q,S,I$ are collinear. Claim: $A,P,V$ and $B,Q,U$ are collinear. Proof: Let $N_a$ be the midpoint of arc $AD$. By shooting lemma, $U,R,N_a$ are collinear. Since $MI^2=MQ.MV$, we have $ \measuredangle QVI= \measuredangle QIM$. \[ \measuredangle DSV= \measuredangle SQV= \measuredangle DIV\]Hence $D,S,I,V$ are concyclic. By utilising this, \[ \measuredangle IVD=\measuredangle IVS+ \measuredangle SVD= \measuredangle IDS+ \measuredangle SID= \measuredangle MDB+ \measuredangle QIM= \measuredangle AVM+ \measuredangle QVI= \measuredangle AVI\]Thus, $V,I,N_a$ are collinear. Apply Pascal on $DAVN_aUM$ to get that $R,AV\cap UM,I$ are collinear which implies $A,P,V$ are collinear. Apply Pascal on $AVMMUB$ to see that $P,VM\cap UB,CE_{\infty}$ hence $B,Q,U$ are collinear.$\square$ Claim: $XA$ and $YB$ intersect on $\tau$. Proof: Let $XA$ intersect $\tau$ at $Z$. Pascal on $ZAVUUM$ gives $X,P,UV\cap MZ$ are collienar. Pascal on $ZBUVVM$ yields $ZB\cap VV,Q,UV\cap MZ$ are collinear. Since $UV\cap MZ,Q\in CE$, we observe that $ZB\cap VV$ is on $CE$ which must be $Y$.$\square$ Claim: $BF$ is tangent to $l$. Proof: Since $AF,AD$ are tangent to $k$, by DDIT $(\overline{AF},\overline{AD}),(\overline{AU},\overline{AP}),(\overline{AX},\overline{AX})$ is an involution. Projecting this onto $\tau$ yields the concurrency of $DF,UV,ZZ$. Hence $(\overline{BD},\overline{BF}),(\overline{BU},\overline{BV}),(\overline{BZ},\overline{BZ})$ is an involution which is equavilent to $(\overline{BS},\overline{BF}),(\overline{BQ},\overline{BV}),(\overline{BY},\overline{BY})$. Thus, $BF$ is tangent to $l$ as desired.$\blacksquare$