euclidean-geometrygeometryplane-geometry
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
Let $\measuredangle FEH=\measuredangle EAF=\alpha.$
Thus, by your work and by law of sines we obtain: $$\frac{x}{\sin{\alpha}}=\frac{4}{\sin108^{\circ}}$$ and $$\frac{x}{\sin108^{\circ}}=\frac{1}{\sin\alpha},$$ which gives $$x^2=4$$ and $$x=2.$$
With this symmetrical figure, focus on $\triangle AOK$ , $\triangle ABK$,$\triangle AOB$.
For the simpliticy of calulation, first let us assume BK=1 $\to$AB=2,AK=OK=$\sqrt 3$:
\begin{align} \frac{S_{ABCDEFGH}}{S_{ACEG}} =\frac{S_{\triangle AOB}}{S_{\triangle AOK}}=\frac{S_{\triangle AOK}-S_{\triangle ABK}}{S_{\triangle AOK}}=1-\frac{S_{\triangle ABK}}{S_{\triangle AOK}}=1-\frac{BK}{OK}=1-\frac{1}{\sqrt{3}}\end{align}
Now scale back with $S_{ACEG}=4$, $S_{ABCDEFGH}$= 4$(1-\frac{1}{\sqrt{3}})$
Best Answer
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 $M$.