Angle of triangle Let $ABC$ be a triangle such that $\angle ABC=20^\circ$, $AB=BC$. Let $D \in AB$, $E \in BC$, $AD=BE=AC$. Find angle $\angle BDE$. Does this problem have a nice geometric solution?
I don't like my trigonometric solution. Let $AC=b$, $AB=a$, $\angle BDE=x$. From triangle $BDE$ we have: $$\frac{b}{\sin x}=\frac{a-b}{\sin\left(20^\circ+x\right)}$$
Then $\sin20^\circ \cot x+\cos20^\circ=\frac{a}{b}-1$.
But $$\frac{a}{b}-1=\frac{\sin80^\circ }{\sin20^\circ}-1=2\cos20^\circ $$
Then $\cot x =\cot 20^\circ $. Then $\angle BDE=20^\circ$.

 A: The given figure can be treated as a part of a "bigger figure" i.e., as part of a star-shaped 9-gon:

Fig. 1.
making it accessible to a rather natural proof also based on angle chasing but seemingly in a more logical way.
Two steps:
Step 1: Why is the red triangle in fig. 1 identical to the initial figure ?

*

*$a:=\angle{B}=20°$ because the associated center angle is $2a = 40°=\frac19 360°$,


*$BE=AD$ (all the "spikes" of the star are identical),


*$AC=AD$ because triangle $ACF$ is equilateral. Indeed $\angle{CAF}=\angle{CAF}=60°$ because they both sustain 3 times arc $a$.
Step 2: Now, let us prove that $\angle{BDE} = 20°$. It  follows from the fact that lines $DE$ and $AF$ are parallel, as a consequence of the symmetry of the figure with respect to the axis of symmetry $OG$ where $O$ is the center of the circle.
Remark: The nine-pointed star is an "enneagram (9/4)" if you want to explain at a dinner what you are doing.
A: An alternative path with Euclidean geometry. Let $F$ be the point in $\triangle ABC$ such that $\triangle ACF$ is equilateral.

*

*Show that $\triangle BDE \cong \triangle CEF$ (SAS criterion).

*Noting that $ABEF$ is an isosceles trapezoid, and therefore $EF\parallel AB$, conclude that $\measuredangle FEC = 20^\circ$.

*Obtain the thesis from the fact that $\triangle CEF$ is isosceles and point 1., which yields $ED \cong EB$.


A: 
Reflect $\triangle ABC$ over line $CA$. Let the reflections of points $D$ and $E$ be $D'$ and $E'$ respectively. Draw $DD'$ and $BD'$.
Observe that $\triangle DCD'$ is equilateral and thereafter quadrilateral $D'DBC$ is a kite.
Hence, $\angle ABD'=\frac {1}{2} \angle DBC=40^{\circ}$. Thus $\triangle ABD'$ is isosceles and thereafter $BD'=AD'$. Also, $AD'=AD=AB-BD=AC-AE=CE$ and hence $BD'=CE$.
Now, observe that, in $\triangle ED'C$ and $\triangle BDD'$, $D'C=DD'$, $CE=BD'$ an $\angle ECD=30^{\circ}=\angle BD'D$ and therefore $\triangle ED'C\cong \triangle BDD'$ by $S-A-S$ criterion of congruence and hence $BD=ED'$.
Also, since $ED'=DE$ and $BD=AE$, $AE=DE$ and therefore $\angle BDE=\angle DAE=20^{\circ}$
