Find the angle $x$ in the quadrilateral $ABCD$ below

Question

Calculate the angle x in the figure below (Answer:$80^o$)

Is it possible to resolve it just with the data provided? I think I would need auxiliary lines but I couldn't identify it

$\angle A = 180-80 - 100^o $

$\angle C = 180-130 = 50^o $

Answer 1

$\angle EBC = 60^{\circ}$

$\angle FDC = 70^{\circ}$

then

$\angle EBC = 60^{\circ}=\angle GBC = 60^{\circ}$ ( bisector of exterior angle)

$\angle FDC = 70^{\circ}=\angle GDC = 70^{\circ}$ ( bisector of exterior angle)

Since In a $\triangle$ two exterior angle bisectors and one interior angle bisector are meet the same point that is point $C$, $AG$ is the bisector of $\angle A$

then $\angle A = 180^{\circ} - (60^{\circ} + 40^{\circ})=80^{\circ}$

and $x=80^{\circ}$