How to find an opening angle in a concave quadrilateral polygon? The problem is as follows:

In the figure find $\angle{ABC}$. It is known $AB=BC=AD$.


The choices given in my book are:
$\begin{array}{cc}
1.110^{\circ}-3\alpha\\
2.115^{\circ}-2\alpha\\
3.100^{\circ}-\alpha\\
4.150^{\circ}-3\alpha\\
5.120^{\circ}-2\alpha\\
\end{array}$
According to the official answers sheet the answer for this problem is choice 5. But how do you get there?
What kind of construction is needed here?. Since this problem belongs to a chapter that has not yet introduced circumpherence I think it can be solved without that. So how to solve this without using trigonometry.
The only thing I can remember is that in these cases you can say:
$\angle{ADC}=\angle{BAD}+\angle{ABC}+\angle{BCD}$
But other than that I am out of ideas. I don't know what to do here. Can someone help me please?.
 A: A synthetic approach that does not require trigonometry.
First, consider a more basic scenario. You have a circle centered at $O$, with a chord given by the points $A, B$ describing a minor arc. Let the arc measure be $2X$. Let's say you have a point $C$ somewhere on the minor arc described by these points. Then, the claim is that $\angle AOC = 2\angle CBA$. There are a lot of isoceles triangles here. $\triangle AOB$, $\triangle AOC$ and $\triangle COB$. Then it follows that:
$$ \angle CBA = \angle CBO - \angle ABO = (90 - X + \frac{1}{2} \angle AOC)- (90 - X) = \frac{1}{2}\angle AOC$$ from which the result follows.

With this in mind, let $\omega$ be a circle centered at  $E$ passing through $B, C$ and $D$. Then, the angle $\angle DEB = 2\alpha$. Note that this angle is the angle $\angle BAD$. What's more, $AB = AD$ and $EB = ED$ together with the opposite angles equal (why?) implies that the shape $ABED$ is a rhombus! Thus, $AB = AD = EB = ED = BC = EC$. So $\triangle EBC$ is equilateral. That means that $\angle BEC = 60$. This therefore implies that $\angle DBC = 30 - \alpha$. (due to the result about the point on a minor arc.) Now, you already know that $AB = AD$ gives $\triangle ABD$ isoceles and therefore $\angle ABD = 90 - \alpha$.

$\angle ABC = \angle ABD + \angle DBC = (90 - \alpha) + (30 - \alpha) = 120 - 2 \alpha$, and you're done.
A: I found this to be a very challenging problem, which is why I am providing an answer, even though the OP (i.e. original poster) hasn't really shown any work.
Further, the OP specifically indicated that Trigonometry is not to be used.  The only way that I could solve the problem was to use Trigonometry.
It could well be that there is some elegant purely geometric construction that proves the problem.  Anyway...
Tools

*

*Law of Sines


*Sum and Difference Formulas

In the diagram below, which is not drawn to scale, the challenge is to show that $(y + z) = 120^\circ - 2a.$
$(y + z) = 180^\circ - 4a - 2b = (120^\circ - 2a) + (60^\circ - 2a - 2b).$
So, the problem has been reduced to showing that 
$$(60^\circ - 2a - 2b) = 0 \iff a + b = 30^\circ \iff \sin(a + b) = \dfrac{1}{2}. \tag0 $$

Since $\triangle ABC$ is isosceles,
$$S = 2R\cos(2a+b). \tag1 $$
Let $T$ denote the length of line segment $\overline{CD}.$ 
Then,
$$S = R\cos(b) + T\cos(a + b). \tag2 $$
Using the Law of Sines, you have that
$$\frac{T}{\sin(b)} = \frac{R}{\sin(a + b)} \implies 
T = \frac{R\sin(b)}{\sin(a + b)}. \tag3 $$
Combining (2) and (3) gives
$$S = R\cos(b) + \frac{R\cos(a+b)\sin(b)}{\sin(a + b)} \implies $$
$$ S = \frac{R\cos(b)\sin(a+b) + R\cos(a+b)\sin(b)}{\sin(a + b)}. \tag4$$
Using sum and difference formulas, (4) above simplifies to
$$ S = \frac{R\sin(a+2b)}{\sin(a + b)}. \tag5$$
Combining (1) and (5) gives:
$$\cos(2a + b) \times [2 \times \sin(a + b)] = \sin(a + 2b). \tag6 $$
So, the entire problem has been reduced to showing that (6) above implies the RHS of (0) above.

Let $\theta = (a + b) \implies $

*

*$(a + 2b) = (2\theta - a)$

*$(2a + b) = (\theta + a)$.

Also, using sum and difference formulas,

*

*$\sin(2\theta) = 2\sin(\theta)\cos(\theta).$


*$\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) = 1 - 2\sin^2(\theta).$
So, (6) above may be re-expressed as
$$\cos(\theta + a) \times 2 \times \sin(\theta) = \sin(2\theta - a). \tag7 $$
Using sum and difference formulas, the LHS of (7) above becomes
$$[\cos(\theta)\cos(a) - \sin(\theta)\sin(a)] \times 2\sin(\theta) $$
$$ = 2\sin(\theta)\cos(\theta)\cos(a) - 2\sin^2(\theta)\sin(a). \tag8 $$
Similarly, the RHS of (7) above becomes
$$\sin(2\theta)\cos(a) - \cos(2\theta)\sin(a) $$
$$ = 2\sin(\theta)\cos(\theta)\cos(a) - [1 - 2\sin^2(\theta)]\sin(a). \tag9 $$
Comparing (8) and (9) above, notice that the first term in each expression is identical.  Therefore, as a direct implication of (7) above, you have that
$$ - 2\sin^2(\theta)\sin(a) = - [1 - 2\sin^2(\theta)]\sin(a) \implies $$
$$ 2\sin^2(\theta) = [1 - 2\sin^2(\theta)] \implies $$
$$ 4\sin^2(\theta) = 1 \implies  \sin^2(\theta) = \dfrac{1}{4} \implies $$
$$\sin(\theta) = \dfrac{1}{2} \implies \sin(a + b) = \dfrac{1}{2}.$$
Therefore, the equation on the RHS of (0) above has been established.

