I will prove that if $F$ is the the point in ray $CA$ such that $AF = AC$ then $ADEF$ is cyclic. Let $C_1$ be the circle through $C$ that touches $AB$ at $A$ and $C_2$ the circle through $B$ that touches $AC$ at $A$. Invert about $A$. This sends $B$ and $C$ to $B'$ and $C'$. Also, this inversion sends $C_1$ to the lines that passes through $C$ and is parallel to $AB$ and it sends $C_2$ to the lines through $B$ parallel to $AC$. $D'$ is the intersection of this lines so $AB'C'D'$ is a parallelogram. We see that $AE \cdot AE' = r^2$ so $AE' = \frac{r^2}{AE}$. Also, $B'E' = \frac{BE \cdot r^2}{AB \cdot AE} = \frac{r^2}{AE} = AE'$ so $E'$ is midpoint of $AB'$ Since $AF \cdot AF' =r^2 = AC \cdot AC'$ we have $AF' = AC'$ so $A$ is the midpoint of $C'F'$ Clearly, since $AC' \parallel B'D'$ we have $\angle F'AE' = \angle E'B'D'$. Also $AF' = AC' = B'D'$ and $AE' = B'E'$ so triangles $F'AE'$ and $D'B'E'$ are congruent so $\angle F'E'A = \angle D'E'B'$ and therefore $D',E',F'$ are collinear. Hence, $ADEF$ is cyclic.

Here is a simpler solution:- Note that triangles $ADB$ and $ADC$ are similar because $\angle DAB=\angle ACD$ and $\angle ABD=\angle DAC$.Now $CA$ meets $\odot ADE$ at $F$.So $\angle DFA=\angle DEA\Longrightarrow \triangle DFC\sim \triangle ADE$. Hence $A\cap\triangle DFC\sim B\cap\triangle ADE$.So we conclude $AF=AC$.

By tangencies, we have $$\angle BAD=\angle DCA \quad \text{and}\quad \angle ABD=\angle DAC$$which implies $\triangle ABD\sim \triangle CAD$. Thus, $$\dfrac{AB}{AC}=\dfrac{BD}{AD}\Longleftrightarrow AC=\dfrac{AB\cdot CD}{AD}$$On the other hand, observe that $$\angle CDF=\angle DCA-\angle DFC=\angle BAD-\angle AED=\angle ADE$$Therefore $\triangle AED\sim \triangle CFD$. Thus, $$\dfrac{CF}{AE}=\dfrac{CD}{AD}\Longleftrightarrow CF=\dfrac{AE\cdot CD}{AD}$$To sum up $$AC=\dfrac{AB\cdot CD}{AD}=\dfrac{AE\cdot CD}{AD}=CF$$as desired.

$\angle BAD = \angle ACD, \angle ABD = \angle DAC \implies \triangle DAB \sim \triangle DCA$ $\angle BED = \angle DFA, \angle FDE = \angle FAE = \angle ADC \implies DCAF \sim DABE$. Since $B$ is midpoint of $AE$, hence $A$ is midpoint of $FC$ and we are done. $\blacksquare$