Find the angle $DBM$, in the pentagon below In the figure below, calculate the angle $DBM$, knowing that $ABCDE$ is a pentagon where $B=D=90°$, $AB=BC$, $CD=DE$ and that $M$ is the midpoint of the side $AE$

My progress

$\triangle DEC(isosceles)\implies \angle DEC  = \angle ECD  =45^0\\
\triangle ABC(isosceles)\implies \angle BCA= \angle CAB  =45^0$
My idea would be to demonstrate that the triangle DMB is isosceles and the angle DMB is rightbut I couldn't find the way
 A: 
Reflect point $A$ about $BC$. Then by midpoint theorem,
$BM \parallel A'E$
Now note that $\triangle A'CE \sim \triangle BCD $ and so we have $\angle CA'E = \angle CBD$
Also, $\angle ABM = \angle AA'E = 45^\circ - \angle CA'E = 45^\circ - \angle CBD$
That shows $\angle DBM = 45^\circ$

As a side note, if you also reflect point $E$ about $CD$, you can easily show $AE' = A'E$ and that means $BM = DM$. So, $\triangle DMB$ is indeed an isosceles triangle, right at $M$.
A: Let $X$ be the point lying on the other side of the line $AE$ than $C$ such that $\angle AEX = \angle XAE = 45^\circ$.
Note that $\angle CEX = \angle DEM$ and $\dfrac{CE}{DE} = \sqrt 2 = \dfrac{EX}{EM}$. Hence by SAS triangles $CEX$ and $DEM$ are similar. Hence $\dfrac{CX}{DM} = \sqrt 2$ and it is easy to see that $\angle(CX, DM)=45^\circ$.
Analogusly $\dfrac{CX}{BM}=\sqrt 2$ and $\angle(BM, CX) = 45^\circ$.
Therefore $DM = \frac{1}{\sqrt 2} CX = BM$ and $\angle DMB = 45^\circ+45^\circ=90^\circ$. As a consequence, triangle $DBM$ is the $45-45-90$ triangle.

A: You can use a following theorem:
If $R_B$ and $R_D$ are two rotations around $B$ and $D$ for angles $\alpha $ and $\beta$ then their compostion $R_D\circ R_B$ is another rotation around new point $Q$ for an angle $\alpha +\beta$ where $\angle BDQ = \beta /2$ and $\angle DBQ =\alpha/2$.
A: 
As shown in the image above let $CD=y$ and $AB=x$. And construct $\overline{EC}$ and $\overline{AC}$ and let $K$ and $L$ be midpoints of $EC$ and $AC$ respectively. Finally construct $\overline{MD}$ That's all for construction!
Now you get $EC=y\sqrt{2}$ and $EK=KC=\frac{y}{\sqrt{2}}$ and similarly, $AC = x\sqrt{2}$ and $AL=LC=\frac{x}{\sqrt{2}}$
By midpoint theorem, you can find out that $MKCL$ is a parallelogram. Then $MK=\frac{x}{\sqrt{2}}$ and $ML=\frac{y}{\sqrt{2}}$ and $\angle DKM=90+\angle AEC=\angle MLB$
With the above information $\triangle DKM \cong \triangle BLM$. With the congruence, $DM=MB$
Now, look at $\triangle {DKM}$ and $\triangle {DCB}$ As they have a common angle and the other sides that are used to form the common angle have a constant ratio They are similar. With triangle similarity, you can get the following relationship,
$\frac{y}{\frac{y}{\sqrt{2}}}=\frac{BD}{MD}$ Finally, $MD=\frac{BDD}{\sqrt{2}}$
Now you can write the lengths of sides $MD,MB,DB$ of $\triangle MBD$ as $\frac{BD}{\sqrt{2}},\frac{BD}{\sqrt{2}},DB$ respectively.
With simple trigonometry you can find $\angle MBD=45$
Hope my answer was helpful.
