In triangle $ABC$, $\angle A = 20º$ and $AM = CN = CB$. Find angle $MBN$. Recently, I stumbled across this question which had some interesting geometry questions from a primary school competition. Here is one of the questions:

In the diagram below, $ABC$ is an isosceles triangle, where $\angle A = 20^{\circ}$. Let $M$ and $N$ be two points on $AC$ such that $AM = CN = CB$. What is the measure, in degrees, of $\angle MBN$?


After drawing this diagram on paper, I realised that simple angle chasing would not do. $CN = CB$ means that $\Delta CBN$ is isosceles of course, but I was having trouble connecting this to side $AM$. I also did not want to use trigonometry and yet I could not see a way to apply theorems such as similar triangles, circle theorems or power of a point. I then used GeoGebra and gradually worked my way up to this diagram (link):

I noticed a few curious things when I constructed this diagram. It turned out that $BC = BD$ as well, and in turn $BD = MD$ where $D$ is constructed such that $AM = MD$ and $D$ lies on $AB$. Then I realised in a flash of lighting that I could make use of $AM = CN = CB$ and reflect the triangle where the base $AB$ is constant. Yet more odd things popped up: the circle passing through $B', C'$ and centred at $A$ also passed through $M$; $ND$ was parallel to $CB$.
Using all of these insights, I am very close to a solution but not there yet.
 A: 
We draw an equilateral triangle with side $AB$ as shown. Then $\triangle PAC$ is isosceles with $\angle PAC = 40^\circ$. That leads to $\angle BPC = 10^\circ$. Also, $\angle PBC = 20^\circ$
Now $\triangle PBC \cong \triangle BAM$ (by S-A-S)
So it follows that $\angle ABM = 10^\circ$.
$\therefore \angle MBN = 80^\circ - 50^\circ - 10^\circ = 20^\circ$
A: Using angle chasing, $\angle ACB = \angle ABC = \frac{180º-20º}{2} = 80º$. Since $\Delta NCB$ is isosceles, $\angle BNC = \angle NBC = 50º$.
Now we have to make use of the information $AM = CN = CB$, focusing in on $AM$ in particular. Thus, let us reflect triangle $ABC$ horizontally around the middle, so that base $AB$ stays in the same position. Thus $CN = CB = AC' = AM$. Since $\angle B' = 80º$ by symmetry, $\angle CAM = 80º - 20º = 60º$. Since $\Delta CAM$ is also iscoceles, this triangle must be equilateral! Therefore $\angle AC'M = \angle AMC' = 60º$ as well.
Again by symmetry, we have that $AE = EB$. Thus $\angle EBA = 20º$ also and $\angle AEB = 140º$. Now if we construct $D$ such that $AM = MD$ and $D$ lies on $AB$, then $\angle MDA = 20º$ as well. Thus by AA similarity, $\Delta AMD \sim \Delta AEB$.
