Difficult problem in elementary euclidean geometry

Question

A point D is chosen inside an equilateral triangle $ABC$ such that $AD$ = $BD$. A point $E$ outside the triangle is chosen such that $\angle DBE$ = $\angle DBC$ and $BE$ = $AB$. Find the degree measure of angle $\angle DEB$.

My attempt:

I first tried letting $\angle EBD$= x and trying to find the other angles in terms of x, in the hope of getting a congruent triangle. But that didn't go anywhere, or lead to anything useful. It seems that this problem requires some construction to find something equal to $E$ but I can't figure that out

Answer 1

Construct $DF\perp BC, E\in BC$, and $DG\perp EB, E\in EB$. Then triangles $BFD$ and $BGD$ are congruent, and so are $DGE$ and $DFC$. So $\angle DEB=\angle DCB = 30^\circ$. The construction idea comes from the fact that $D$ is by construction the incenter of the triangle formed by $BE$, $BC$, and $AC$.

Answer 2

Note that $\angle DCB = \frac12\angle C = 30^\circ$ because of $AD = BD$. Since $\angle EBD = \angle CBD $, $EB = AB = BC$ and $DB=DB$, the triangles $CBD$ and $EBD$ are congruent, which yields $\angle DEB = \angle DCB = 30^\circ$.

Answer 3

Maybe you can see this as a proof with almost no words. Just think of reflection in the line through $B$ and $D$: