Find angle $x$ in the isosceles $\triangle ABC$. Find angle $x$ in the isosceles $\triangle ABC$.

Can anyone help me with this problem?
I was stuck. I made my findings in this diagram but I didn't know how can I spot the angle that's needed.

Can anyone solve it?
 A: We identify the key angle $\angle AFB = 100^{\circ}$.
Construct an equilateral $\triangle HBF$ as shown. Join $HA$.

$\angle HFB = 60 \Rightarrow \angle HFA = 40 = \angle HBA \Rightarrow HBFA$ is cyclic.
Then $\angle HAB = \angle HFB = 60 \Rightarrow H$ lies on line $AC$.
Now $\angle HGB = \angle AGB = 50 = \angle HBG \Rightarrow HG=HB=HF$
Thus $H$ is center of circle through $B,F,G$.
Hence $\angle BGF = \frac{1}{2}\angle BHF=30^{\circ}$.
A: Here is a solution using trigonometry.

Using law of sines,
In $\triangle CDF, \displaystyle \frac{CF}{\sin (130^0-x)} = \frac{DF}{\sin20^0} \,$ (as $\angle BDC = 130^0$)
In $\triangle BDF, \displaystyle \frac{BF}{\sin x} = \frac{DF}{\sin10^0}$
As $BF = CF$, we have
$\displaystyle \frac{\sin (130^0-x)}{\sin20^0} = \frac{\sin x}{\sin10^0}$
$\displaystyle \frac{\sin (50^0+x)}{\sin20^0} = \frac{\sin x}{\sin10^0}$
$\displaystyle \frac{1}{2}\sin (50^0+x)= \sin x \cos10^0$
$\displaystyle \sin 30^0 \sin (50^0+x)= \sin x \sin80^0$
Now using the identity $2 \sin A \sin B = \cos (A-B) - \cos (A+B)$, we have
$\cos (20^0 + x) - \cos (80^0 + x) = \cos (80^0 -x) - \cos (80^0 + x)$
$\implies 2x = 60^0, x = 30^0$
