Very nice problem! Let $M,N$ be midpoints of minor and major arcs $AB$. Let $CM \cap AB = X$. Finally, let $Z = NX \cap LM$. Note that since $X$ is the orthocenter of $\triangle NLM$, we have $Z \in (ABC)$. Also, see that $CIFL$ is cyclic. Observe that that $(LBZ)$ is tangent to $MB$ since $\angle MLB = \angle MNZ = \angle MBZ$. So by PoP, $MZ.ML = MB^2 = MI^2$ so $(LIZ)$ is tangent to $CI$. Redefine $K$ as $(LIZ) \cap (ABC)$ now. Let $CK \cap (LIZ) = P$. To show that $K$ is the one in the problem statement, we need to show that $\angle IKC = \angle IKL$, or that $IL = IP$ or $LP || CI$. This is equivalent to $\angle PLI = \angle LIC$, or $(CKI)$ tangent to $LI$. But see that $\angle KIL = \angle KZL = \angle KAM = \angle KAI$ so this is true. Now, to finish the problem, see that $\angle CKI = 180 - \angle CIL = 180 - \angle CFL = \angle CFB$. Let $CF \cap (ABC) = Q$. Then $\angle CFA = \angle CKD = \angle CQD$, so $QD || AB$. Therefore, we have $ABQD$ is a isosceles trapezium so $\angle ACF = \angle ACQ = \angle ABQ = \angle BAD = \angle DCB$, as desired. $\blacksquare$

Redefine $D$ as intersection of $C$-isogonal of $\overline{CF}$ and $(ABC)$ and redefine $K$ as intersection of $\overline{DI}$ and $(ABC)$. I'll claim that $\measuredangle LKI=\measuredangle IKC$. Let $M$ be the midpoint of arc $AB$, not containing $C$. Let $R=\overline{MK}\cap\overline{AB}$. By the shooting lemma, $MR\cdot MK=MI^2$, which yields that $\overline{MI}$ is tangent to $(KIR)$. Point $L$ lies on $(KIR)$, since $\measuredangle RLI=\measuredangle FLI=\measuredangle FCI=\measuredangle MCD=\measuredangle MKD=\measuredangle RKI$, where we used the fact that $LCIF$ is cyclic as $\measuredangle LCI=90^\circ=\measuredangle LFI$. Therefore, $$\measuredangle LKI=\measuredangle LIM=\measuredangle LIC=\measuredangle LFC=\measuredangle DAC=\measuredangle IKC,$$as desired.

rafaello wrote: Redefine $D$ as intersection of $C$-isogonal of $\overline{CF}$ and $(ABC)$ and redefine $K$ as intersection of $\overline{DI}$ and $(ABC)$. I'll claim that $\measuredangle LKI=\measuredangle IKC$. Let $M$ be the midpoint of arc $AB$, not containing $C$. Let $R=\overline{MK}\cap\overline{AB}$. By the shooting lemma, $MR\cdot MK=MI^2$, which yields that $\overline{MI}$ is tangent to $(KIR)$. Point $L$ lies on $(KIR)$, since $\measuredangle RLI=\measuredangle FLI=\measuredangle FCI=\measuredangle MCD=\measuredangle MKD=\measuredangle RKI$, where we used the fact that $LCIF$ is cyclic as $\measuredangle LCI=90^\circ=\measuredangle LFI$. Therefore, $$\measuredangle LKI=\measuredangle LIM=\measuredangle LIC=\measuredangle LFC=\measuredangle DAC=\measuredangle IKC,$$as desired. Can you give me some source where I can find the "Shooting Lemma"?

Consider an "$\sqrt{ab}$" inversion $\Psi$ (centered at $C$ with power $CA\cdot CB$ followed by a reflection around the angle bisector of $\angle ACB$). Redefine $D=\Psi(F)$, $K=ID\cap (ABC)\neq D$. We want to show that these definitions coincide with the original ones; to do so I prove that we still have $\angle CKI=\angle IKL$. Let $J$ be the $C$-excenter of $\triangle ACB$. Under these definitions, we see that $C$ is the center of the spiral similarity taking $IF$ to $DJ$; hence, it is well known that the point $X=ID\cap JF$ lies on both $(CIF)$ and $(CDJ)$. We now prove the following two lemmas, which together resolve the problem. Lemma 1: $K$ is the midpoint of $IX$ Pf: By Power of a Point we get $$ID\cdot IK=IC\cdot IM$$$$ID\cdot IX=IC\cdot IJ$$dividing yields $$\frac{IK}{IX}=\frac{IM}{IJ}=\frac{1}{2}$$as desired. $\blacksquare$. Lemma 2 The quadrilateral $CLIX$ is harmonic. Pf: Clearly $L\in (CIF)$. Then $$(C, L; I, X)\overset{F}{=}(C, AB\cap CI; I, J)=B(C, AB\cap CI; I, J)=-1$$as desired. $\blacksquare$ This means that $K$ is the $X$-dumpty point in $\triangle CXL$. Hence $\triangle KCX\sim \triangle KXL$ and, in particular $$\angle CKI=\angle IKL$$.