Find missing angle in triangle 
In this triangle, we are given that $AB=AD=CE$, $BE=CD$ and angle  $ECD=2*AED=2a$.
We are asked to find $AED=a$.
I have seen several problems similar to this but can't follow any of the solution methods.
The only thing which is obvious to me is that since $AD+CD = BE+EC$, the triangle is isosceles.
Therefore its base angles (the ones I have drawn in green) are equal.
Also triangle ABD is isosceles, therefore its base angles are equal.
I don't see any other relation to use.
(FYI this is not homework or anything - just practice and self-motivation).
Thank you very much.
EDIT: If this is of any help, through Geogebra I get that $a=10^{\circ}$, therefore $ECD=20^{\circ}$ and we have a 80-80-20 triangle.
 A: Choose $X$ so that $ADEX$ is a parallelogram. Then $XE=AD=CE$ and since $XE \parallel AC$, we get $\angle XEC = 180^\circ - 2a$. It follows that $a=\angle XCE$.
On the other hand, $\angle XAE = \angle DEA = a$. Therefore $\angle XCE = \angle XAE$ which shows that $AXEC$ is cyclic. Then we see that $\angle ACX = a = \angle XCE$, so the corresponding chords are equal: $AX=XE$. Also note that $X$ lies on the bisector of angle $ACB$ and since $ACB$ is isosceles, it follows that $AX=XB$.
Then we find $AX=XB=AB$ so $AXB$ is equilateral. Clearly $\angle CBX = XAC = 2a$. Writing down the sum of angles of triangle $ABC$ leads to $2a+2\cdot(2a+60^\circ)=180^\circ$ which yields $a=10^\circ$.


Below is another solution.
First we're going to show that $DE=AD$. To do so we exclude the possibilities that $DE>AD$ and $DE<AD$.
If $DE>AD$ then $\angle EAD > \angle DEA = a$. This means that $\angle EDC = a+\angle EAD > 2a = \angle DCE$ which implies that $CE>DE$ and we have a contradiction: $CE>DE>AD=CE$. If $DE<AD$ then all inequalities get reversed and we have a contradiction again. Hence $DE=AD$.
Then let $F$ be the circumcenter of $ADE$. Then $\angle DFA = 2\angle DEA = \angle ACB$, so the isosceles triangles $ADF$ and $BAC$ are similar. They are actually congruent because $AD=AB$. Since $AD=DE$, it also follows that $DEF$ is congruent to both $ADF$ and $BAC$. We can also see that $F$ lies on $AB$ because $\angle FAD = \angle BAC$.
Let $G$ be the point on $DF$ such that $DG=BE$. Then triangles $BAE$, $ADG$ and $EDG$ are congruent. In particular, $AE=EG=GA$, so triangle $AEG$ is equilateral.

What is left to do is some angle chasing. We have $\angle EGA = 60^\circ$, so $\angle DGA = 30^\circ$ as $DF$ is the axis of symmetry of $AEG$. Then $30^\circ = \angle DGA = \angle GFA + \angle FAG = \angle ACE + \angle EAC = 2a+\angle DEA = 2a+a=3a$, so $a=10^\circ$.
A: Using Sine law,
Say, $\angle DAE = u, AB = x, DE = y, AE = z$
In $\displaystyle \triangle ADE, \frac{\sin u}{y} = \frac{\sin a}{x} \implies y = \frac{x \sin u}{\sin a}$
In $\displaystyle \triangle CDE, \frac{\sin 2a}{y} = \frac{\sin (a+u)}{x} \implies y = \frac{x \sin 2a}{\sin (a+u)}$
This leads to, $\, \sin u \sin(a+u) = \sin a \sin 2a$
Using $\cos(A - B) - \cos(A+B) = 2 \sin A \sin B,$ we get $u = a$.
Now in $\triangle ADE,$
$\displaystyle \frac{\sin 2a}{z} = \frac{\sin a}{x} \implies z = 2x \cos a$
In $\triangle ABE,$
$\angle ABE = 90^0 - a, \angle BAE = (90^0 - a) - a = 90^0 - 2a, \angle BEA = 3a$
$\displaystyle \frac{\sin 3a}{x} = \frac{\sin (90^0 - a)}{z}$
$\displaystyle \sin 3a = \frac{1}{2} \implies a = 10^0$
A: First, from Euclid: an inscribed angle is half of the central angle that subtends the same arc on the circle.
Second, if you already have the angle 2a finding the angle a is trivial with a compass you can easily find the bisector, if you want to do it "pretty", if you are allowed to replicate C (with AD as a base instead of AB) and call that point F, and draw a circunference of center F and radius AF, the intersection with BC should give you point E.
A: Using brute force.
Calling $u$ the segment with two marks and $v$ the segment with three marks we have
$$
\cases{
v^2=2(u+v)^2-2(u+v)^2\cos(2a)\Rightarrow a=\frac 12\arccos\left(\frac{2u^2+4u v+v^2}{2(u+v)^2}\right)\\
v^2=ED^2+AE^2-2ED\cdot AE\cos(a)\\
ED^2=u^2+v^2-2u v\cos(2a)\\
AE^2=u^2+v^2-2u v\cos(\phi)\\
\phi = \frac{\pi}{2}-a
}
$$
making substitutions we get at
$$
\sqrt{\frac{(2 u+v) (2 u+3 v) \left(u^3+u^2 v+v^3\right) \left(u^4-2 u^2 v^2+u v^3+v^4\right)}{(u+v)^5}}-\frac{v^3 (2 u+v)}{(u+v)^2}+2 u v-2 u^2=0
$$
making now $v = \lambda u$ we obtain
$$
\left(\sqrt{\frac{(\lambda +2) (3 \lambda +2) \left(\lambda ^3+\lambda +1\right) \left(\left(\lambda ^2+\lambda -2\right) \lambda ^2+1\right)}{(\lambda
   +1)^5}}-\frac{\lambda  \left(\lambda ^3-2 \lambda +2\right)+2}{(\lambda +1)^2}\right) u^2=0
$$
and the $\lambda$ factor reduces to
$$
\frac{\lambda ^2 \left(2 (\lambda +2) \lambda ^3+\lambda ^2+\lambda +1\right) \left(\lambda ^2 (\lambda +3)-1\right)}{(\lambda +1)^5}=0
$$
with the only feasible root
$$
\lambda = 0.532088886237956
$$
which corresponds to $a=10^{\circ}$
