TITLE: Find missing angle in triangle
QUESTION [8 upvotes]: 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.

REPLY [2 votes]: 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 $\triangle ABE,$
$\angle ABE = 90^0 - a, \angle BAE = (90^0 - a) - a = 90^0 - 2a, \angle BEA = 3a$
$\displaystyle \frac{\sin 3a}{x} = \frac{\sin (90^0 - a)}{z}$
$\displaystyle \sin 3a = \frac{1}{2} \implies a = 10^0$