Since $\angle CDF = 90^\circ - 45^\circ = 45^\circ$, the line $DF$ bisects $\angle CDA$, and so $F$ lies on the perpendicular bisector of segment $AE$, which meets $AB$ at $G$. Let $\angle ABC = 2\beta$. Since $ADEF$ is cyclic, $\angle AFE = 90^\circ$, and hence $\angle FAE = 45^\circ$. Further, as $BF$ bisects $\angle ABC$, we have $\angle FAB = 90^\circ - \beta$, and thus $\angle EAB = \angle AEG = 45^\circ - \beta$ and $\angle AED = 45^\circ + \beta$, so $\angle GED = 2\beta$. This implies that right-angled triangles $\triangle EDG$ and $\triangle BDC$ are similar, and so we have $\frac{|GD|}{|CD|} = \frac{|DE|}{|DB|}$. Thus the right-angled triangle $\triangle DEB$ and $\triangle DGC$ are similar, whence $\angle GCD = \angle DBE = \beta$. But $\angle DFE = \angle DAE = 45^\circ - \beta$, then $\angle GFD = 45^\circ - \angle DFE = \beta$. Hence $GDCF$ is cyclic, so $\angle GFC = 90^\circ$, whence $CF$ is perpendicular to the radius $FG$ of $\omega$. It follows that $CF$ is a tangent to $\omega$, as required.

As $\angle ADF = 45^\circ$ line $DF$ is an exterior bisector of $\angle CDB$. Since $BF$ bisects $\angle DBC$ line $CF$ is an exterior bisector of $\angle BCD$. Let $\angle ABC = 2\beta$, so $\angle ECF = (\angle DBC + \angle CDB)/2 = 45^\circ + \beta$. Hence $\angle CFE = 180^\circ - \angle ECF - \angle BCE - \angle EBC = 180^\circ - (45^\circ + \beta + 90^\circ - 2\beta + \beta) = 45^\circ$. It follows that $\angle FDC = \angle CFE$, then $CF$ is tangent to $\omega$.

Note that $AE$ is a diameter of circumcircle of $\triangle ABC$ since $\angle CDF = 90^\circ$. From $\angle AEF = \angle ADF = 45^\circ$ it follows that triangle $\triangle AFE$ is right-angled and isosceles. Without loss of generality, let points $A, E$ and $F$ have coordinates $(-1,0), (1,0)$ and $(0,1)$ respectively. Points $F, E, B$ are collinear, hence $B$ have coordinates $(b,1-b)$ for some $b \neq -1$. Let point $C'$ be intersection of line tangent to circumcircle of $\triangle AFE$ at $F$ with line $ED$. Thus $C'$ have coordinates $(c,1)$ and from $C' \cdot E \perp AB$ we get $c = \frac{2b}{b+1}$. Now vector $\vec{BC'} = (\frac{2b}{b+1} - b, b) = \frac{b}{b+1}(1-b, b+1)$, vector $\vec{BF} = (-b, b) = b(-1, 1)$ and vector $\vec{BA} = -(b+1), -(1-b))$. Its clear that $(1-b, b+1)$ and $(-b+1), -(1-b))$ are symmetric with respect to $\vec{FE} = (-1, 1)$, hence $BF$ bisects $\angle C'BA$ and $C' = C$ which completes the proof.

Again $F$ lies on the perpendicular bisector of segment $AE$, so $\triangle AFE$ is right-angled and isosceles. Let $M$ be an intersection of $BC$ and $AF$. Note that $\triangle AMB$ is isosceles since $BF$ is a bisector and altitude in this triangle. Thus $BF$ is a symmetry line of $\triangle AMB$. Then $\angle FDE = \angle FEA = \angle MEF = 45^\circ$, $AF = FE = FM$ and $\angle DAE = \angle EMC$. Let us show that $EC = CM$. Indeed,
$$\angle CEM = 180^\circ - (\angle AED + \angle FEA + \angle MEF) = 90^\circ - \angle AED = DAE = \angle EMC.$$
It follows that $FMCE$ is a kite, since $EF = FM$ and $MC = CE$. Hence $\angle EFC = \angle CFM = \angle EDF = 45^\circ$, so $FC$ is tangent to $\omega$.

Let the tangent to $\omega$ at $F$ intersect $CD$ at $C'$. Let $\angle ABF = \angle FBC = \beta$. It follows that $\angle C'FE = 45^\circ$ since $C'F$ is tangent. We have
$$\frac{\sin \angle BDC}{\sin \angle CDF} \cdot \frac{\sin \angle DFC'}{\sin \angle C'FB} \cdot \frac{\sin \angle FBC}{\sin \angle CBD} = \frac{\sin 90^\circ}{\sin 45^\circ} \cdot \frac{\sin(90^\circ - \beta)}{\sin 45^\circ} \cdot \frac{\sin \beta}{\sin 2\beta} = \frac{2\sin \beta \cos \beta}{\sin 2\beta} = 1.$$
So by trig Cheva on triangle $\triangle BDF$, lines $FC', DC$ and $BC$ are concurrent (at $C$), so $C = C'$. Hence $CF$ is tangent to $\omega$.