Inverting trough $A$, we can reformulate as follows: Let $C^*B^*D^*$ be a triangle, $A\in CD, E^*$ such that $AE^*||B^*D^*$ and $F^*\in B^*D^*$ such that $C^*F^*$ is the internal bisector of $\angle B^*C^*D^*$. Then $(AC^*E^*),(AB^*D^*)$ and the circle trough $A$ and $F^*$ tangent to $B^*D^*$ are concurrent at a point $X$ (in other words $A\neq X=(C^*E^*A)\cap (AB^*D^*)$) Consider the circle $(C^*B^*D^*)$, which is tangent to $(C^*E^*A)$, by the parallelism. By radical axes theorem, we have that the tangent to  $(C^*E^*A)$ at $C^*$, $AX$ and $B^*C^*$ concur at a point $Z$. Since $\measuredangle ZC^*F^*=\measuredangle ZC^*B^*+\measuredangle B^*C^*F^*=\measuredangle C^*D^*B^*+\measuredangle F^*C^*D^*=\measuredangle C^*F^*Z$, it follows that $ZC^*=ZF^*$. Therefore $ZA\cdot ZX=(ZC^*)^2=(ZF^*)^2$ therefore, $(AXC^*)$ is tangent to $B^*D^*$, which implies our concurrence. @2below: corrected lol

#2 this isn't problem 3

Here's a solution without inversion: Let $O$ be the circumcenter of $\triangle ACF$ and $N = \overline{AF} \cap \Gamma_1 \ne A$. Then $O \in \Gamma_1$ as $\angle AOC = 2 \angle AFC = \angle ABC$. Also, $$\angle FCN = \angle ANC - \angle NFC = \angle AOC - \angle AFC = \angle AFC = \angle NFC$$so $NF = NC$. In particular, we obtain that line $ON$ is the perpendicular bisector of segment $CF$ (let this be $(1)$). [asy][asy] size(200); pair A=dir(90),B=dir(-90),O=B+0.7*dir(-160); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$O$",O,dir(-60)); path c1=circumcircle(A,B,O); draw(c1,red); pair C=2*foot(A,O,circumcenter(A,B,O)) - A; dot("$C$",C,dir(C)); pair N = waypoint(c1,0.05); dot("$N$",N,dir(N)); pair F=2*foot(C,O,N)-C; path c2=circumcircle(A,B,F); dot("$F$",F,dir(-90)); draw(c2,red); pair D=IP(1.1*A-0.1*C--3*A-2*C,c2); dot("$D$",D,dir(D)); path c=circumcircle(O,B,F); pair T= intersectionpoint(1.2*B-0.2*D--10*B-9*D,circumcircle(O,B,F)); dot("$T$",T,dir(-130)); pair E=IP(0.9*C+0.1*T--T,c1); dot("$E$",E,dir(E)); draw(CP(circumcenter(T,B,F),B,25,190),red); draw(T--C^^T--D^^T^^T--1.2*F-0.2*T,dashed+brown); draw(N--C--D^^E--1.2*A-0.2*E^^A--F,royalblue); draw(T--N^^C--F,purple); draw(C--B--A^^C--O--A,green); label("$\Gamma_1$",E+A-N,red); label("$\Gamma_2$",5.2*B-4.2*O,red); [/asy][/asy] Now define $T$ as the second intersection point of line $ON$ and $\odot(FBO)$. We will prove that the three lines concur at $T$. \begin{align*} \angle TFB &= 180^\circ - \angle BOT = \angle NOB = \angle NAB = \angle FAB \\ &~ \implies ~ \boxed{\text{line } TF \text{ is tangent to }\Gamma_2 \text{ at } F} \\ \angle FBD &= \angle FAD = 180^\circ - \angle CAN = \angle NOC \stackrel{(1)}{=} \angle FON = 180^\circ - \angle FOT = 180^\circ- \angle FBT \\ &~ \implies ~ \boxed{T \text{ lies on line } BD} \\ \angle ECN &= \angle ECB + \angle BCN = \angle EAB + \angle BON = \angle AFB + \angle BFT = \angle AFT = \angle NFT \stackrel{(1)}{=} \angle TCN \\ &~ \implies \boxed{T \text{ lies on line } CE} \end{align*}This completes the proof of the problem. $\blacksquare$

Proposed by Tak Wing Ching - Hong Kong

Let $T=CE\cap DB$ and let $F\in\Gamma_2$ such that $TF$ is tangent to $\Gamma_2$ and $F$ is in the opostite side of $A$ w.r.t $DB$, $M$ is the midpoint of the arc $ABC$ in $\Gamma_1$. Finally let $K=AF\cap\Gamma_1$. 1-$\angle ECB=\angle EAB=\angle ADB$ then $TC^2=TB\cdot TD= TF^2$, meaning that $TC=TF.$ 2- $\angle KFC=\angle AFT-\angle CFT=\angle ADF-\angle CFT=\angle EAK-\angle FCE=\angle KCF$, meaning that $KC=KF$. 3- $CKFT$ is a kite, then $KT$ bisects $\angle CKF$, meaning that $M\in KT$. 4- Conclude noticing that $M$ is in the perpendicular bisector of $CF$ and $CA$ then $M$ is the circumcenter of $\triangle ACK$ and then $2\angle AFC=\angle AMC=\angle ABC$.

$Q=CE \cap BD$. Let $QF$ be tangent to $\Gamma_2$ and $P=QF \cap AE, CF,FA \cap \Gamma_1=R,S$. $\triangle QCF$ is an isosceles triangle because $\angle QCB=\angle EAB=\angle CDQ \implies QC^2=QB \cdot QD=QF^2$. $\triangle PAF$ is an isosceles triangle. Thus $\angle CFA=\angle PFA-\angle QFC=\angle EAF-\angle ECF=\angle RAS \implies \angle ABC=\angle ASC=2 \angle CFA$ as desired.

Let $M$ be the midpoint of the minor arc $AC$.Let $BM$ intersect $\Gamma_2$ at $X$.The given conditions tell us that $C,X,F$ are collinear. Now , let $BM \cap AF=P$ CLAIM I: $P,C,E$ are collinear Proof:Note that $$\angle CFP=\angle XBA =\angle CBX \implies (CBFP)$$Then $$\angle CPB+\angle CBP= \angle BFC+ \angle CBP= \angle BFC+ \angle XFA =\angle BFA =\angle BDA =\angle BAE=\angle BCE$$implying the desired conclusion . Finish : In degenerate hexagon $FFXBDA$ , pascal gives $$FF \cap BD , FX \cap DA=C , FA \cap BX $$are collinear .Also , we just showed that $C,E, FA \cap BX$ are collinear.Thus , we are done

Let $CF$ meet $\Gamma_2$ at $K$ and $EC$ and tangent at $F$ meet at $S$. Claim $: SC$ is tangent to $CBD$. Proof $:$ Note that $\angle CDB = \angle ADB = \angle EAB = \angle ECB$. Claim $: SC = SF$. Proof $:$ Note that $\angle CFS = \angle KFS = \angle KAF = \angle EAF - \angle EAK = \angle EAD + \angle DAF - \angle AFK = \angle ECD + \angle DKF - \angle ABK = \angle ECD + \angle DKF - \angle KBC = \angle ECD + \angle BCK = \angle SCF$. Now that $SC^2 = SF^2$ so $S$ has same power wrt $\Gamma_2$ and $DBC$ so $S$ lies on Radical Axis which is $BD$.

Inverting around $A$ handles the angle condition (angle bisector after inversion), so why not inversion. Invert everything around $A$ with arbitrary radius. Let $X'$ denotes the inverted point of $X$. Therefore we get $AE'\parallel B'D'$, $C'F'$ is the bisector of $\angle B'C'D'$, a circle $\omega$ is tangent to $B'D'$ at $F'$. Then it is equivalent to prove $\omega, (AB'D')$ and $(AC'E')$ are coaxial. Now remove the $(')$ sign from the inverted problem (its really annoying). Let $\omega_1$ be $(ACE)$. Our problem is equivalent to show \begin{align*} \frac{P(B, \omega_1)}{P(B, \omega)}=\frac{P(D, \omega_1)}{P(D, \omega)}=\frac{BE\cdot BC}{BF^2}=\frac{DA\cdot DC}{DF^2}\iff \left(\frac{BF}{DF}\right)^2=\frac{BE}{DA}\cdot \frac{BC}{DC} \end{align*}which is true because of the angle bisector and parallelism.

Assume that $BD$ and $CE$ meet at $K$. Then draw a tangent from $K$ to $\Gamma_2$. Let this tangent meets $\Gamma_2$ at point F', we can show by angle chasing that $2\angle AF'C=\angle ABC$, which means that F'=F, here is the trick that redefinition of point $F$.