Let $G=AE\cap CF, K=CF\cap \odot (ABE), L$ is such point on ray $GA$ that $|GL|^2=|GF|\cdot |GC|,$ it's suffice to prove that $CDFL$ is cyclic. Observe that $|GK|\cdot |GL|=\sqrt{|GA|\cdot |GE|\cdot |GF|\cdot |GC|}=|GA|\cdot |GC|=|GF|\cdot |GE|,$ therefore $FL\parallel KE, LC\parallel AK$ and $\angle CLF=\angle AKE=\pi-\angle ABC=\pi-\angle CDF$ as desired.

Let $CF$ touch circle of $ABE$ at $S$ and $AE,CF$ meet at $K$. Assume $AFS$ meet $AE$ at $T$ ( Case that $AFS$ is tangent at $A$ to $AE$ will give contradiction easily with angle chasing). we will prove $T$ is the tangency point so we need to prove $CDFT$ is cyclic and tangent to $AE$. Claim $: TSCE$ is cyclic. Proof $:$ Note that $AF || EC$ and $AFST$ is cyclic. Claim $: FT || SE$ and $AS || TC$. Proof $:$ Note that $\angle AES = \angle ASF = \angle ATF \implies FT || SE$ and $\angle TCS = \angle TES = \angle ASF \implies AS || TC$. Now Note that $\angle FTS = \angle 180 - \angle TFC - \angle TCF = \angle 180 - \angle AES - \angle EAS = \angle ASE = \angle 180 - \angle ABE = \angle 180 - \angle CDF$ so $CDFT$ is cyclic and Note that $\angle TCF = \angle TES = \angle ATF$ so $CDFT$ is tangent to $AE$ is well. we're Done.

Let $CF$ touch $(ABE)$ at $T$, the second intersection between $(AFT)$ and $AE$ be $T_1$, and $F = AE \cap CF$. By Reim's, we know $CET_1T$ is also cyclic, so $$RCT_1 \overset{-}{\sim} RET \overset{-}{\sim} RTA \overset{-}{\sim} RT_1F$$which means $\overline{AT_1E}$ is tangent to $(CFT_1)$. Now, angle chasing yields $$\angle CT_1F = \angle CT_1T + \angle TT_1F = \angle CET + \angle TAF$$$$= \angle BAT + \angle TAF = \angle BAF = 180^{\circ} - \angle CDF$$so $CDFT_1$ is cyclic, as required. $\blacksquare$ Remark: This problem is very minimalistic and beautiful!

Let $X$ denote the tangency point of $CF$ with $\odot (ABE)$. Let the line through $F$ parallel to $XE$, hit $AE$ at $Y$. We show that $Y$ is the requested tangency point. Angle chase as follows : $$\angle YFX = \angle EXC = \angle EAX = \angle YAX \implies Y \in \odot (AFX)$$$$\angle YXC = \angle YAF = \angle AEB = \angle YEB \implies Y \in \odot (CEX)$$$$\angle AYX = \angle YEX = \angle YCX$$This implies $\odot (CDF)$ is tangent to $AE$ at $Y$, as desired. $\square$

Nice problem! Let $Y= (ABE) \cap CF$ and $X$ be a point on $AE$ such a $(CFX), AE$ are tangent. So, we need to prove that $\angle FXC + \angle FDC =^? 180^{\circ} = \angle ABE + \angle AYE = \angle ADC + \angle AYE$ or $\angle AYE = \angle CXF$. Let $K = AX \cap CY$. Since tangents and $AF \parallel CE$ we have $KY^2=KA \cdot KE, KX^2= KF \cdot KC, KE \cdot KF =KA \cdot KC$, and so $(KY\cdot KX)^2 = KA \cdot KC \cdot KF \cdot KE = (KE \cdot KF)^2$, so $\frac{KE}{KX}=\frac{KY}{KF}$ and $XF \parallel EY$. Now note that $\angle AYF =\angle AEY =\angle AXF$ and by analogy $\angle EXC = \angle EYC$. So, $\angle CXF = \angle AYE$ $\blacksquare$.

Let $X$ be the point of tangency of $(ABX)$ and $CF$ and let $Y$ be on $(CDF)$ such that $FY \parallel XE$. Then: $$\angle AEX = \angle AXF = \angle AYF$$which implies $AYXF$ is cyclic. Similarly, $CEFX$ is cyclic. Notice that: $$\angle AYX = 180^\circ \angle AFX = \angle XCE = 180^\circ - \angle XYE.$$Therefore $A, Y, E$ are collinear. Notice that: $$\angle YFX = \angle EXC = \angle CYE$$which implies $Y$ is the point of tangency of $(CDF)$ with $AE$.

Let $AE\cap CF=P$. Let $(ABE)$ be tangent to $CF$ at $Y$, and construct the circle through $C,E$ tangent to $AE$ at $X$. Consider force-overlay inversion at $P$ swapping pairs $(A,C)$ and $(E,F)$. This must also swap $(X,Y)$. Now \[\measuredangle ABC=\measuredangle AYE=\measuredangle AYP+\measuredangle PYE=\measuredangle CXP+\measuredangle PXF=\measuredangle CXF=\measuredangle CDA,\]implying $Y\in(CDF)$ as desired.