Regular Pentagon, Geometry

Question

Can someone help me to find the value of $x$? I wish I could share my attempted solution but I really couldn't develop something interesting to share. I know that the value of the internal angle are $108^°$, because the sum of the internal angles of a convex polygon is $S=180^°(n-2)$, and they are all the same, because the it's a regular polygon. But that's all I figured out.

Answer 1

Consider the figure below, where $ABCDE$ is a regular pentagon, $F \in BC$ and $G \in AB$ with $EG \cong FG$, and $\measuredangle BFG = 30^\circ$. Also, $G'$ is taken in such a way that $\triangle FGG'$ is equilateral.

1. Show that $\triangle BGG'$ is isosceles, with $\measuredangle BGG' = \measuredangle BG'G = 18^\circ$.
2. Use Exterior Angle Theorem on $\triangle BGG'$, plus the fact that $\triangle ABE$ is isosceles, to show that $E$, $B$, and $G'$ are aligned.
3. Observing that $\triangle EGG'$ is isosceles, prove that $$\boxed{\measuredangle DEG = 90^\circ}.$$