How to find the angle formed by an isosceles triangle next to another one? The problem is as follows:

Using the figure from below: Find the unknown angle indicated as $x$.
Assume $AD=BC$ and $BD=DC$


The alternatives given in my book are as follows:
$\begin{array}{ll}
1.&12^{\circ}\\
2.&10^{\circ}\\
3.&15^{\circ}\\
4.&16^{\circ}\\
4.&14^{\circ}\\
\end{array}$
What I attempted to do here was to add the angles in the isosceles which adds up to $4x$, this can be added to the $3x$ triangle but that's where I'm stuck.
In other words the only thing which I could spot was:
$\angle BDA = 2x+2x$
How exactly can it be used congruency to solve this problem?. Can someone help me here?. The intended approach is relying in euclidean geometry postulates, but I don't know exactly which sort of congruency of triangles identity should be used.
Please include a drawing in your answer because this part is difficult for me to spot with accuracy. Can you please use an explanation step-by-step.
 A: Draw a line $DF$ such that $\angle ADF = 3x$. Draw lines $AE$ and $DE$ such that $\angle DAE = \angle ADE = 2x$.

So $\triangle AED \cong \triangle BDC$ (by A-A-S, as $BC = AD$)
So, $AE = DE = BD = DC$
Now $AF = DF$ so $\triangle AFE \cong \triangle DFE$ (by S-A-S).
Similarly, $\triangle DFB \cong \triangle DFE$
So, we have $\angle AFE = \angle EFD = \angle DFB = 60^0$
So $\angle FAD = \angle FDA = 3x = 30^0$
So, $x = 10^0$.
A: 
In the figure we have $$\dfrac{a}{\sin(7x)}=\dfrac{b}{\sin(3x)}\\\dfrac{a}{\sin(4x)}=\dfrac{b}{\sin(2x)}$$ It follows $$\sin(7x)=2\sin(3x)\cos(2x)$$
You have two options: solve this last equation or check which of the five answers given is correct. By both means (the second is preferable) you find $x = 10^{\circ}$(in boths sides you have $0.93969262$).
►If you have difficulty to solve the equation note that $$2\sin(3x)\cos(2x)=\sin(5x)+\sin(x)$$ and$$\sin(7x)-\sin(5x)=2\cos(6x)\sin(x)$$
Thus $$2\cos(6x)\sin(x)=\sin(x)\Rightarrow \cos(6x)=\dfrac12\Rightarrow  x=10^{\circ}$$
