Find the $MD$ value in the pentagon below

Question

In the graph, $AB=a$, $AE=b$ and $MC=c$. Compute $MD$ in terms of $a$, $b$ and $c$
_{(Answer:$\frac{ac}{b}$)}

I find:

$\triangle CAD(\text{Isosceles}) \therefore AC = AD\\ \angle CMB = \alpha + \theta\implies \triangle CBM(\text{Isosceles})\therefore BC = BM\\ \angle DNA = \alpha +2\theta \implies \triangle DNA \sim \triangle CED\\ \therefore \frac{AN}{ED}=\frac{AD}{CE}=\frac{DN}{CD}$

Answer 1

Since $\angle BCE=\angle BDE$, $BCDE$ is cyclic.
Therefore $\angle CEB=\angle CDB=\theta$.
Since $\angle ACE=\angle ADE$, $ACDE$ is cyclic.
Therefore $\angle CEA=\angle CDA=\theta+\alpha$.
Thus $\angle BEA=\alpha$.
Since $\angle AEB=\angle ACB=\alpha$, $ABCE$ is cyclic.
Therefore $\angle ABE=\angle ACE=\theta$.
Hence $\triangle MCD\sim\triangle AEB$.
$\implies MD=\dfrac{ac}b.$

Answer 2

The quadrilateral $ABCD$ is circunscribable because $BA$ is "seen" by the same angles $ADB$ and $ACB$. Hence, there's a circle which passes through $A$, $B$, $C$ and $D$. By using the same argument in quadrilateral $ACDE$ using angles $ECA$ and $EDA$, there's a unique circle which passes through $A$, $C$, $D$ and $E$. From here, its clear that there's an unique circle which goes through $A$, $B$, $C$, $D$ and $E$. Equal angles on a circle "see" segments with the same length: $BA=DE=a$ (from angles $ACB$ and $DCE$) and $EA=CB=b$ (from angles $CDB$ and $ACE$).

Now, $\Delta CMB$ is similar to $\Delta MDE$ (notice that the external angle of $\Delta CMD$ through $CM$ (and $DM$) is $\alpha + \theta$ so they have two equal angles and are isosceles) which means that $\dfrac{MD}{MC} = \dfrac{DE}{MB} = \dfrac{DE}{CB}$ (since $\Delta CMB$ is isosceles) which means $\dfrac{MD}{c} = \dfrac{a}{b}$ so $MD = \dfrac{ac}{b}$, as desired.

There might be a much cleaner solution though.